Ac Circuits 3bs
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H12 AC Circuits (31.1-6) Due: 11:59pm on Monday, December 2, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy
Problem 31.1 Description: The voltage across the terminals of an ac power supply varies with time according to V=V_0cos(omega t). The voltage amplitude is V_0. (a) What is the root-mean-square potential difference V_rms? (b) What is the average potential difference V_(a(v))... The voltage across the terminals of an ac power supply varies with time according to V
= V0 cos(ωt) . The voltage amplitude is V0 = 44.0V .
Part A What is the root-mean-square potential difference Vrms ? ANSWER: = 31.1 V
Part B What is the average potential difference Vav between the two terminals of the power supply? ANSWER: 0 V
Phasors Explained Description: Introduces phasors with questions both general and in the context of AC circuits. Recommended after resistance, inductance, and capacitance in AC circuits have been discussed. Includes a part on series LCR circuits and another part on parallel LCR circuits. Learning Goal: To understand the concept of phasor diagrams and be able to use them to analyze AC circuits (those with sinusoidally varying current and voltage). Phasor diagrams provide a convenient graphical way of representing the quantities that change with time along with cos(ωt + ϕ) , which makes such diagrams useful for analyzing AC circuits with their inherent phase shifts between voltage and current. You have studied the behavior of an isolated resistor, inductor, and capacitor connected to an AC source. However, when a circuit contains more than one element (for instance, a resistor and a capacitor or a resistor and an inductor or all three elements), phasors become a useful tool that allows us to calculate currents and voltages rather easily and also to visualize some important processes taking place in the AC circuit, such as resonance. Let us assume that a certain quantity I(t) changes over time as I(t) = I 0 cos(ωt) . A phasor is a vector whose length represents the amplitude I 0 (see the diagram ).This vector is assumed to rotate counterclockwise with angular frequency ω; that way, the horizontal component of the vector represents the actual value I(t) at any given moment. In this problem, you will answer some basic questions about phasors and prepare to use them in the analysis of various AC circuits.
In parts A - C consider the four phasors shown in the diagram . Assume that all four phasors have the same angular frequency ω.
Part A At the moment t depicted in the diagram, which of the following statements is true? ANSWER:
I 2 leads I 1 by π . I 1 leads I 2 by π . I 2 leads I 1 by π . 2
I 1 leads I 2 by π2 .
Part B At the moment shown in the diagram, which of the following statements is true? ANSWER:
I 2 lags I 3 by π . I 3 lags I 2 by 2π . I 2 lags I 3 by π2 . I 3 lags I 2 by π2 .
Part C At the moment shown in the diagram, which of the following statements is true? ANSWER:
I 2 leads I 4 by π . I 2 lags I 4 by 2π . I 2 leads I 4 by π2 . I 2 lags I 4 by π2 .
Let us now consider some basic applications of phasors to AC circuits. For a resistor, the current and the voltage are always in phase. For an inductor, the current lags the voltage by π . 2 For a capacitor, the current leads the voltage by π . 2
Part D Consider this diagram. Let us assume that it describes a series circuit containing a resistor, a capacitor, and an inductor. The current in the circuit has amplitude I , as indicated in the figure. Which of the following choices gives the correct respective labels of the voltages across the resistor, the capacitor, and the inductor?
ANSWER:
V1 ; V2 ; V3 V1 ; V2 ; V4 V1 ; V4 ; V2 V3 ; V2 ; V4 V3 ; V4 ; V2
Part E Now consider a diagram describing a parallel AC circuit containing a resistor, a capacitor, and an inductor. This time, the voltage across each of these elements of the circuit is the same; on the diagram, it is represented by the vector labeled V0 . The currents in the resistor, the capacitor, and the inductor are represented respectively by which vectors?
ANSWER:
I1 ; I2 ; I3 I1 ; I3 ; I4 I1 ; I4 ; I2 I1 ; I2 ; I4 I1 ; I3 ; I2
Problem 31.5 Description: (a) What is the reactance of an inductor with an inductance of L at a frequency of f? (b) What is the inductance of an inductor whose reactance is X_L at a frequency of f? (c) What is the reactance of a capacitor with a capacitance of C at a...
Part A What is the reactance of an inductor with an inductance of 3.30H at a frequency of 76.0Hz ? ANSWER: = 1580
Ω
Part B What is the inductance of an inductor whose reactance is 11.7Ω at a frequency of 76.0Hz ? ANSWER: = 2.45×10−2 H
Part C What is the reactance of a capacitor with a capacitance of 4.38×10−6F at a frequency of 76.0Hz ? ANSWER: = 478
Ω
Part D What is the capacitance of a capacitor whose reactance is 119Ω at a frequency of 76.0Hz ?
ANSWER: = 1.76×10−5 F
Problem 31.8 Description: A Radio Inductor. You want the current amplitude through a inductor with an inductance of L (part of the circuitry for a radio receiver) to be I when a sinusoidal voltage with an amplitude of V is applied across the inductor. (a) What frequency is... A Radio Inductor. You want the current amplitude through a inductor with an inductance of 4.50 mH (part of the circuitry for a radio receiver) to be 2.00 mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor.
Part A What frequency is required? ANSWER: = 2.12×105 Hz
Problem 31.13 Description: You have a resistor of resistance 200 Omega and a 6.00 mu F capacitor. Suppose you take the resistor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 (rad)/s. ... You have a resistor of resistance 200 Ω and a 6.00 µF capacitor. Suppose you take the resistor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s .
Part A What is the impedance of the circuit? ANSWER: 696
Ω
Part B What is the current amplitude? ANSWER: 4.31×10−2 A
Part C What is the voltage amplitude across the resistor? ANSWER: 8.62 V
Part D What is the voltage amplitudes across the capacitor? ANSWER: 28.7 V
Part E What is the phase angle ϕ of the source voltage with respect to the current? ANSWER: -73.3 ∘
Phasors: Analyzing a Parallel AC Circuit Description: This problem contains several questions, both qualitative and quantitative, directed at analyzing a parallel AC circuit using phasors. The circuit contains only a resistor and capacitor in parallel. It is recommended that the students solve the problem Phasors Explained before trying this one. Learning Goal: To understand the use of phasors in analyzing a parallel AC circuit. Phasor diagrams, or simply phasors, provide a convenient graphical way of representing the quantities that change with time along with cos(ωt + ϕ) . This makes them useful for analyzing AC circuits with their inherent phase shifts between voltage and current. If a quantity I(t) changes with time as I(t) = I 0 cos(ωt) , a phasor is a vector whose length represents the amplitude I 0 (see the diagram). This vector is assumed to rotate counterclockwise with angular speed ω; that way, the horizontal component of the vector represents the actual value I(t) at any given moment. In this problem, you will use the phasor approach to analyze an AC circuit. In answering the questions of this problem, keep the following in mind: For a resistor, the current and the voltage are always in phase. For an inductor, the current lags the voltage by π . 2 For a capacitor, the current leads the voltage by π . 2
Part A Phasors are helpful in determining the values of current and voltage in complex AC circuits. Consider this phasor diagram: The diagram describes a circuit that contains two elements connected in parallel to an AC source. The vector labeled V0 corresponds to the voltage across both elements of the circuit. Based on the diagram, what elements can the circuit contain?
ANSWER:
two resistors two inductors two capacitors a capacitor and a resistor an inductor and a resistor a capacitor and an inductor
Part B The phasor diagram from Part A describes a circuit that looks like the one in the figure: What are the respective amplitudes of the currents in the capacitor and the resistor in the diagram for Part A?
ANSWER:
I 1 ;I 2 I 2 ;I 1
Part C Find the amplitude of the current I 0 through the voltage source. Express your answer in of the magnitudes of the individual currents I 1 and I 2 .
Hint 1. The parallel connection In a parallel circuit, the current through the voltage source is the sum of the individual currents. That rule, learned first for DC circuits, holds true for AC circuits, too: The law of conservation of charge is still valid! Therefore, at any instant, the current through the voltage source is the sum of the currents through the resistor and the capacitor. However, you have to be careful in adding these currents.
Hint 2. The current through the voltage source as a vector sum The current through the voltage source may also be represented by the component of a rotating vector whose length represents its amplitude I 0 : You should find the value of I 0 as the magnitude of the vector sum of the individual currents. Note that the phase angle between I 1 and I 2 is π . 2
ANSWER:
I0 =
Part D What is the tangent of the phase angle ϕ between the voltage and the current through the voltage source? Express tan(ϕ) in of I 1 and I 2 .
Hint 1. The current through the voltage source as a vector sum The current through the voltage source may also be represented by the component of a rotating vector whose length represents its amplitude I 0 : You should find the value of I 0 as the magnitude of the vector sum of the individual currents. Note that the phase angle between I 1 and I 2 is π . 2
ANSWER:
tan(ϕ) =
As you can see, phasors can be helpful in visualizing the processes in AC circuits.
Problem 31.14 Description: You have a 0.400 H inductor and a 6.00 mu F capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 (rad)/s. (a) What is... You have a 0.400 H inductor and a 6.00 µF capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s .
Part A What is the impedance of the curcuit? ANSWER: 567
Ω
Part B What is the current amplitude? ANSWER:
5.29×10−2 A
Part C What is the voltage amplitude across the capacitor. ANSWER: 35.3 V
Part D What is the voltage amplitude across the inductor. ANSWER: 5.29 V
Part E What is the phase angle ϕ of the source voltage with respect to the current? ANSWER: -90.0 ∘
Exercise 31.19 Description: The power of a certain CD player operating at ## V rms is ## W. Assume that the CD player behaves like a pure resistance. (a) Find the maximum instantaneous power. (b) Find the rms current. (c) Find the resistance of this player. The power of a certain CD player operating at 120V rms is 20.0 W . Assume that the CD player behaves like a pure resistance.
Part A Find the maximum instantaneous power. ANSWER:
p max = 40.0 {\rm W}
Part B Find the rms current. ANSWER: I_{\rm rms} = 0.167 {\rm A}
Part C Find the resistance of this player. ANSWER: R = 720 {\rm {}}\Omega
Problem 31.25 Description: An L-R-C series circuit with an inductance of L, a resistance of R, and a capacitance of C carries an rms current of I with a frequency of f.
(a) What is the phase angle? (b) What is the power factor for this circuit? (c) What is the impedance of... An L-R-C series circuit with an inductance of 0.118{\rm H} , a resistance of 241{\rm \Omega} , and a capacitance of 7.26{\rm \mu F} carries an rms current of 0.455{\rm A} with a frequency of 392{\rm Hz} .
Part A What is the phase angle? ANSWER:
= 0.772 radians
Part B What is the power factor for this circuit? ANSWER:
= 0.716
Part C What is the impedance of the circuit? ANSWER:
= 336 \Omega
Part D What is the rms voltage of the source? ANSWER:
= 153 V
Part E What average power is delivered by the source? ANSWER:
= 49.9 W
Part F What is the average rate at which electrical energy is converted to thermal energy in the resistor?
ANSWER:
= 49.9 W
Part G What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? ANSWER: 0 W
Part H What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor? ANSWER: 0 W
Problem 31.33 Description: A Step-Down Transformer. A transformer connected to an AC line with an rms voltage of V_source is to supply an rms voltage of V_rms to a portable electronic device. The load resistance in the secondary is R. (a) What should the ratio of primary to ... A Step-Down Transformer. A transformer connected to an AC line with an rms voltage of 120{\rm V} is to supply an rms voltage of 12.0{\rm V} to a portable electronic device. The load resistance in the secondary is \texttip{R}{R} = 4.50{\rm \Omega} .
Part A What should the ratio of primary to secondary turns of the transformer be? ANSWER: = 10.0
Part B What rms current must the secondary supply? ANSWER: = 2.67 A
Part C What average power is delivered to the load? ANSWER: = 32.0 W
Part D What resistance connected directly across the source line (which has a voltage of 120{\rm V} ) would draw the same power as the transformer? ANSWER: = 450 {\rm \Omega}
A Low- Filter Description: In this problem, the student will derive an expression for the ratio of the voltage out to the voltage in (from source) as a function of source frequency for an L-R-C circuit where the output is taken over the capacitor. The figure shows a low- filter; the output voltage is taken across the capacitor in an L-R-C series circuit.
Part A Derive an expression for V_{\rm out}/V_{\rm s}, the ratio of the output and source voltage amplitudes, as a function of the angular frequency \texttip{\omega }{omega} of the source. Express your answer in of the capacitance \texttip{C}{C}, resistance \texttip{R}{R}, inductance \texttip{L}{L}, and angular frequency \texttip{\omega }{omega} of the circuit.
Hint 1. Voltage over a capacitor The voltage over a capacitor in an L-R-C circuit is given by the equation V_C = I X_C, where \texttip{V_{\mit C}}{V_C} is the voltage across the capacitor (also equal to \texttip{V_{\rm out}}{V_out} in this case), \texttip{I}{I} is the current flowing through the capacitor, and \large{X_C = \frac{1}{\omega C}} is the capacitive reactance.
Hint 2. Current through the circuit Recall that the current through an L-R-C circuit is given by the equation \large{I = \frac{V}{Z}}, where \texttip{I}{I} is the current, \texttip{V}{V} is the source voltage, and \texttip{Z}{Z} is the circuit impedance.
Hint 3. What is an expression for impedance? Choose the correct expression for the impedance ( \texttip{Z}{Z}) for a series L-R-C circuit. ANSWER:
\large{Z = \sqrt{R^2+\left(L\omega-\frac{1}{C\omega}\right)^2}} \large{Z = \sqrt{R^2+\left(L\omega+\frac{1}{C\omega}\right)^2}} \large{Z = \sqrt{R^2+\left(\frac{1}{L\omega}+C\omega\right)^2}} \large{Z = \sqrt{R^2+\left(\frac{1}{L\omega}-C\omega\right)^2}}
ANSWER:
\large{\frac{V_{\rm out}}{V_{\rm s}}} =
You can see from the shape of the graph (see the figure) of V_{\rm out}/V_{\rm s} why the circuit in this problem is called a low- filter: Low frequencies through (i.e., have output voltage) at a much higher multple of their source voltage than high frequencies. No units are shown, since this graph is simply meant to give a general idea of how the graph is shaped. The exact details depend on the inductance, resistance, and
capacitance of the components of the circuit.
Problem 31.35 Description: The internal resistance of an ac source is R_1. (a) What should the ratio of primary to secondary turns of a transformer be to match the source to a load with a resistance of R_2? ("Matching" means that the effective load resistance equals the... The internal resistance of an ac source is 12.3{\rm k \Omega} .
Part A What should the ratio of primary to secondary turns of a transformer be to match the source to a load with a resistance of 8.05{\rm \Omega} ? ("Matching" means that the effective load resistance equals the internal resistance of the source.) ANSWER: = 39.1
Part B If the voltage amplitude of the source is 60.0{\rm V} , what is the voltage amplitude in the secondary circuit under open circuit conditions? ANSWER:
= 1.53 V
Voltage and Current in AC Circuits
Description: Find the current when an AC voltage is applied to a resistor, a capacitor, and an inductor in turn. Conceptual questions about reactance. Learning Goal: To understand the relationship between AC voltage and current in resistors, inductors, and capacitors, especially the phase shift between the voltage and the current. In this problem, we consider the behavior of resistors, inductors, and capacitors driven individually by a sinusoidally alternating voltage source, for which the voltage is given as a function of time by V(t)=V_0 \cos (\omega t). The main challenge is to apply your knowledge of the basic properties of resistors, inductors, and capacitors to these "single-element" AC circuits to find the current \texttip{I\left(t\right)}{I(t)} through each. The key is to understand the phase difference, also known as the phase angle, between the voltage and the current. It is important to take into the sign of the current, which will be called positive when it flows clockwise from the b terminal (which has positive voltage relative to the a terminal) to the a terminal (see figure). The sign is critical in the analysis of circuits containing combinations of resistors, capacitors, and inductors.
Part A First, let us consider a resistor with resistance \texttip{R}{R} connected to an AC source (diagram 1). If the AC source provides a voltage V_R(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit R}\left(t\right)}{I_R(t)} through the resistor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{R}{R}, \texttip{\omega }{omega}, and \texttip{t}{t}.
Hint 1. Ohm's law Ohm's law \large{I=\frac{V}{R}} is still true at any moment in time.
ANSWER: \texttip{I_{\mit R}\left(t\right)}{I_R(t)} =
Note that the voltage and the current are in phase; that is, in the expressions for \texttip{V_{\mit R}\left(t\right)}{V_R(t)} and \texttip{I_{\mit R}\left(t\right)}{I_R(t)}, the arguments of the cosine functions are the same at any moment of time. This will not be the case for the capacitor and inductor.
Part B Now consider an inductor with inductance \texttip{L}{L} in an AC circuit (diagram 2). Assuming that the current in the inductor varies as I_L(t) = I_0 \cos(\omega t), find the voltage \texttip{V_{\mit L}\left(t\right)}{V_L(t)} that must be driving the inductor. Express your answer in of \texttip{I_{\rm 0}}{I_0}, \texttip{L}{L}, \texttip{\omega }{omega}, and \texttip{t}{t}. Use the cosine function, not the sine function, in your answer.
Hint 1. Kirchhoff's loop rule Apply Kirchhoff's loop rule to this circuit, going clockwise around the circuit (the same direction as the current arrow). Write down the sum of the voltage drops across each of the circuit elements.
Use \texttip{V_{\mit L}\left(t\right)}{V_L(t)} for the voltage from the source, and describe the voltage drop across the inductor in of \texttip{L}{L} and \texttip{dI_{\rm L(t)/dt}}{dI_L(t)/dt} ANSWER: \Sigma V_i =
Hint 2. The derivative of \cos(\omega t) What is the derivative of \texttip{\cos(\omega t)}{cos(omega*t)}, with respect to time? Express your answer in of any or all of \texttip{\sin(\omega t)}{sin(omega*t)}, \texttip{\omega }{omega}, and \texttip{t}{t}. ANSWER: \large{\frac{d\cos(\omega t)}{dt}} =
Hint 3. The phase relationship between sine and cosine Note that \large{\sin(x) = - \cos\left(x + \frac{\pi}{2}\right)}
ANSWER: \texttip{V_{\mit L}\left(t\right)}{V_L(t)} =
Graphs of \texttip{V_{\mit L}\left(t\right)}{V_L(t)} and \texttip{I\left(t\right)}{I(t)} are shown below. As you can see, for an inductor, the voltage leads (i.e., reaches its maximum before) the current by \large{\frac{\pi}{2}}; in other words, the current lags the voltage by \large{\frac{\pi}{2}}. This can be conceptually understood by thinking of inductance as giving the current inertia: The voltage "tries" to push current through the inductor, but some sort of inertia resists the change in current. This is another manifestation of Lenz's law. The difference \large{\frac{\pi}{2}} is called the phase angle.
Part C Again consider an inductor with inductance \texttip{L}{L} connected to an AC source. If the AC source provides a voltage V_L(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit L}\left(t\right)}{I_L(t)} through the inductor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{L}{L}, \texttip{\omega }{omega}, and \texttip{t}{t}. Use the cosine function, not the sine function, in your answer.
Hint 1. Using Part B You can obtain the answer almost immediately if you consider the results of Part B: The amplitude of the voltage ( \texttip{V_{\rm 0}}{V_0}) is I_0 \omega L; the frequency is the same as in Part B, and the phase difference is \large{\frac{\pi}{2}}. Do you what leads and what
lags? ANSWER: \texttip{I_{\mit L}\left(t\right)}{I_L(t)} =
For the amplitudes (magnitudes) of voltage and current, one can write V_0=I_0 R (for the resistor) and V_0=I_0 \omega L (for the inductor). If one compares these expressions, it should not come as a surprise that the quantity \omega L, measured in ohms, is called inductive reactance; it is denoted by \texttip{X_{\mit L}}{X_L} (sometimes \texttip{\chi _{\mit L}}{chi_L}). It is called reactance rather than resistance to emphasize that there is no dissipation of energy. Using this notation, we can write V_0=I_0 R (for a resistor) and V_0=I_0 X_L (for an inductor). Also, notice that the current is in phase with voltage when a resistor is connected to an AC source; in the case of an inductor, the current lags the voltage by \large{\frac{\pi}{2}}. What will happen if we replace the inductor with a capacitor? We will soon see.
Part D Consider the potentials of points a and b on the inductor in diagram 2. If the voltage at point b is greater than that at point a, which of the following statements is true?
Hint 1. How to approach the problem Try drawing graphs of the current through the inductor and voltage across the inductor as functions of time. ANSWER: The current \texttip{I\left(t\right)}{I(t)} must be positive (clockwise). The current \texttip{I\left(t\right)}{I(t)} must be directed counterclockwise. The derivative of the current \large{\frac{dI(t)}{dt}} must be negative. The derivative of the current \large{\frac{dI(t)}{dt}} must be positive.
It may help to think of the current as having inertia and the voltage as exerting a force that overcomes this inertia. This viewpoint also explains the lag of the current relative to the voltage.
Part E Assume that at time \texttip{t_{\rm m}}{t_m} , the current in the inductor is at a maximum; at that time, the current flows from point b to point a. At time \texttip{t_{\rm m}}{t_m}, which of the following statements is true?
Hint 1. How to approach the problem Try drawing graphs of the current through the inductor and voltage across the inductor as functions of time. ANSWER: The voltage across the inductor must be zero and increasing. The voltage across the inductor must be zero and decreasing. The voltage across the inductor must be positive and momentarily constant. The voltage across the inductor must be negative and momentarily constant.
Part F Now consider a capacitor with capacitance \texttip{C}{C} connected to an AC source (diagram 3). If the AC source provides a voltage V_C(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit C}\left(t\right)}{I_C(t)} through the capacitor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{C}{C}, \texttip{\omega }{omega} and \texttip{t}{t}. Use the cosine function
with a phase shift, not the sine function, in your answer.
Hint 1. The relationship between charge and voltage for a capacitor Recall that by definition of capacitance, the relationship q(t)=CV(t) among charge, capacitance, and voltage is true at all times. The convention here is that the voltage is that on the plate with charge +q relative to the other plate (which has charge -q).
Hint 2. The relationship between charge and current No charged particles "flow" through the capacitor. Instead, the current deposits or drains charge from the plates. If the voltage \texttip{V_{\mit C}\left(t\right)}{V_C(t)} is the voltage on terminal b (appropriate for Kirchhoff's law in this circuit), then if the current flow is positive (clockwise here as usual), the charge will increase. Hence \large{I(t)=\frac{dq(t)}{dt}}, where \texttip{q\left(t\right)}{q(t)} is the charge on the capacitor plate b.
Hint 3. Mathematical details To obtain the current, obtain \texttip{q\left(t\right)}{q(t)} from the voltage and then obtain the current from the time derivative of \texttip{q\left(t\right)}{q(t)}. Finally, rewrite the expression you obtain in of the cosine (instead of sine) function.
ANSWER: \texttip{I_{\mit C}\left(t\right)}{I_C(t)} =
For the amplitude values of voltage and current, one can write \large{V_0=\frac{I_0} {\omega C}}. If one compares this expression with a similar one for the resistor, it should come as no surprise that the quantity \large{\frac{1}{\omega C}}, measured in ohms, is called capacitive reactance; it is denoted by \texttip{X_{\mit C}}{X_C} (sometimes \texttip{\chi _{\mit C}}{chi_C}). It is called reactance rather than resistance to emphasize that there is no dissipation of energy. Using this notation, we can write V_0=I_0 X_C, and voltage lags current by \pi/2 radians (or 90 degrees). The notation is analogous to V_0=I_0 R for a resistor, where voltage and current are in phase, and V_0=I_0 X_L for an inductor, where voltage leads current by \pi/2 radians (or 90 degrees). We see, then, that in a capacitor, the voltage lags the current by \large{\frac{\pi}{2}}, while in the case of an inductor, the current lags the voltage by the same quantity \large{\frac{\pi}{2}}. In a capacitor, where voltage lags the current, you may think of the current as driving the change in the voltage.
Part G Consider the capacitor in diagram 3. Which of the following statements is true at the moment the alternating voltage across the capacitor is zero?
Hint 1. How to approach the problem Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)}
ANSWER:
The current must be directed clockwise. The current must be directed counterclockwise. The magnitude of the current must be a maximum. The current must be zero.
Part H Consider the capacitor in diagram 3. Which of the following statements is true at the moment the charge of the capacitor is at a maximum?
Hint 1. How to approach this problem Since the voltage is directly proportional to the charge, when the charge is maximum, so is the voltage. Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time. Find the current when the voltage drop is maximum.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)}
ANSWER: The current must be directed clockwise. The current must be directed counterclockwise. The magnitude of the current must be a maximum. The current must be zero.
Part I Consider the capacitor in diagram 3. Which of the following statements is true if the voltage at point b is greater than that at point a?
Hint 1. How to approach the problem Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)} Look at the region where \texttip{V_{\mit C}\left(t\right)}{V_C(t)} is positive. Is the current positive or negative or both?
ANSWER: The current must be directed clockwise. The current must be directed counterclockwise. The current may be directed either clockwise or counterclockwise.
Part J Consider a circuit in which a capacitor and an inductor are connected in parallel to an AC source. Which of the following statements about the magnitude of the current through the voltage source is true?
Hint 1. Driven AC parallel circuits The voltage across each element is the same at every moment in time. However, the magnitudes of the currents in an AC circuit cannot be added without consideration of the phase angle between the currents. ANSWER: It is always larger than the sum of the magnitudes of the currents in the capacitor and inductor. It is always less than the sum of the magnitudes of the currents in the capacitor and inductor. At very high frequencies it will become small. At very low frequencies it will become small.
This surprising result occurs because the currents in inductor and capacitor are exactly out of phase with each other (i.e., one lags and the other leads the voltage), and hence they cancel to some extent. At a particular frequency, called the resonant frequency, the currents have exactly the same amplitude, and they cancel exactly; that is, no current flows from the voltage source to the circuit. (Lots of current flows around the loop made by the inductor and capacitor, however.) If an L-C parallel circuit like this one connects the wire between amplifier stages in a radio, it will allow frequencies near the resonance frequency to easily, but will tend to short those at other freqeuncies to ground. This is the basic mechanism for selecting a radio station.
Alternating Voltages and Currents Description: Find the rms voltage from the graph of the instantaneous voltage; then find the rms current and average power from a given value of R. The voltage supplied by a wall socket varies with time, reversing its polarity with a constant frequency, as shown in the graph.
Part A What is the rms value \texttip{V_{\rm rms}}{V_rms} of the voltage plotted in the graph? Express your answer in volts.
Hint 1. RMS value of a quantity with sinusoidal time dependence A quantity that varies with time as x=x_{\max}\sin \omega t (or x=x_{\max}\cos \omega t) has a maximum value equal to x_{\max} and an rms value given by \large{x_{\rm rms}=\frac{x_{\max}}{\sqrt{2}}}.
Hint 2. Find the maximum value of the voltage What is the maximum value \texttip{V_{\rm max}}{V_max} of the voltage plotted in the graph? Express your answer in volts. ANSWER: \texttip{V_{\rm max}}{V_max} = 170 \rm V
ANSWER: \texttip{V_{\rm rms}}{V_rms} = 120 \rm V
This is the standard rms voltage supplied to a typical household in North America.
Part B When a lamp is connected to a wall plug, the resulting circuit can be represented by a simplified AC circuit, as shown in the figure. Here the lamp has been replaced by a resistor with an equivalent resistance \texttip{R}{R} = 120{\rm \Omega} . What is the rms value \texttip{I_{\rm rms}}{I_rms} of the current flowing through the circuit? Express your answer in amperes.
Hint 1. Ohm's law in AC circuits Ohm's law can still be applied to an AC circuit, provided the values used to describe all physical quantities are consistent. For example, Ohm's law can be written using maximum values of voltage and current, or alternatively using rms quantities. ANSWER:
\texttip{I_{\rm rms}}{I_rms} =
= 1.00 \rm A
Part C What is the average power \texttip{P_{\rm avg}}{P_avg} dissipated in the resistor? Express your answer in watts.
Hint 1. Average power in an AC circuit The average power \texttip{P_{\rm avg}}{P_avg} dissipated in a resistor with resistance \texttip{R}{R} is given by P_{\rm avg}=I_{\rm rms}^2R, where \texttip{I_{\rm rms}}{I_rms} is the rms value of the current flowing through the resistor. Note that the average power can also be expressed as \large{P_{\rm avg}=\frac{V_{\rm rms}^2}{R}}, where \texttip{V_{\rm rms}}{V_rms} is the rms voltage supplied to the resistor .
ANSWER:
\texttip{P_{\rm avg}}{P_avg} =
= 120 \rm W
The instantaneous power dissipated in the resistor can be substantially higher than the average power. However, since the voltage supplied to the resistor varies in time, so does the instantaneous power. Therefore, a better estimate of the energy dissipated in an AC circuit is given by the average power. For example, the power rating on light bulbs is in fact the average power dissipated in the bulb.
The Resonance Peak Description: Short quantitative problem on resonance in a series RLC circuit. Based on Young/Geller Quantitative Analysis 22.5. In this problem we consider the resonance curve for a circuit with electrical components \texttip{L}{L}, \texttip{R}{R}, and \texttip{C}{C} and resonant frequency \texttip{\omega _{\rm 0}}{omega_0}.
Part A For given values of \texttip{R}{R} and \texttip{C}{C}, if you double the value of \texttip{L}{L}, how does the new resonance curve differ from the original one?
Hint 1. How to approach the problem In a series R-L-C circuit, the resonance frequency is determined by \texttip{L}{L} and \texttip{C}{C}, whereas the peak value of the current in the circuit depends only on \texttip{R}{R}. Since the value of \texttip{R}{R} is being kept constant, the peak of the resonance curve is unchanged. To solve this problem, use proportional reasoning to find a relation between the resonance frequency \texttip{f_{\rm 0}}{f_0} and inductance \texttip{L}{L}. Find the simplest equation that contains these variables and other known quantities from the problem. Write this equation twice, once to describe the original circuit and again for the circuit with greater inductance. You then need to write each equation with all the constants on one side and the variables on the other. Since the variable is \texttip{f_{\rm 0}}{f_0} in this problem, you will write the equations in the form f_0=\cdots. To finish the problem you need to compare the two cases presented. For this question you should find the ratio of the resonance frequency of the original circuit to that of the circuit with greater inductance.
Hint 2. Resonance frequency In a series R-L-C circuit, the resonance frequency \texttip{f_{\rm 0}}{f_0} is given by \large{f_0=\frac{1}{2\pi\sqrt{LC}}}.
Hint 3. Find an expression for the new resonance frequency Write an expression for the resonance frequency \texttip{f_{\rm 1}}{f_1} of the circuit when the inductance \texttip{L}{L} is doubled? Express your answer in of some or all of the variables \texttip{R}{R}, \texttip{L}{L}, and \texttip{C}{C}. ANSWER: \texttip{f_{\rm 1}}{f_1} =
ANSWER: Both the peak height and peak frequency double. The peak height will be half as great, and the peak frequency will double. The peak height won't change, and the peak frequency will be 1/\sqrt 2 times as great. The peak height won't change, and the peak frequency will be half as great.
In an R-L-C circuit, the resonance peak depends only on \texttip{R}{R}, while the resonant frequency is determined by both \texttip{L}{L} and \texttip{C}{C}. In particular, for a given value of \texttip{C}{C}, the resonance frequency is inversely proportional to the square root of \texttip{L}{L}. Similarly, for a given value of \texttip{L}{L}, the resonance frequency is inversely proportional to the square root of \texttip{C}{C}.
Part B For given values of \texttip{R}{R} and \texttip{C}{C}, if you double the value of \texttip{L}{L}, how does the new rms current at resonance \texttip{I_{\rm rms}}{I_rms} differ from its original value? Assume that the voltage amplitude of the ac source is the same.
Hint 1. How to approach the problem Recall that in an ac circuit, for a given voltage, the rms current is always inversely proportional to the impedance of the circuit. This holds also at resonance. To solve this problem, use proportional reasoning and find a relation between the rms current \texttip{I_{\rm rms}}{I_rms} and the impedance \texttip{Z}{Z}.
Find the simplest equation that contains these variables and other known quantities from the problem. Write this equation twice, once to describe the original circuit and again for the circuit with greater inductance. You need to write each equation so that all the constants are on one side and the variables are on the other. Since the variable is \texttip{I_{\rm rms}}{I_rms} in this problem, you will write the equations in the form I_{\rm rms}=\cdots. To finish the problem you need compare the two cases presented. For this question you should find the ratio of the rms current in the original circuit to that in the circuit with greater inductance.
Hint 2. \texttip{V_{\rm rms}}{V_rms} across an ac circuit In a series R-L-C circuit, the rms voltage \texttip{V_{\rm rms}}{V_rms} across the circuit is always proportional to the rms current \texttip{I_{\rm rms}}{I_rms} in it. If \texttip{Z}{Z} is the impedance of the circuit, then V_{\rm rms}=I_{\rm rms}Z.
Hint 3. Find the impedance of the circuit What is the impedance \texttip{Z}{Z} of a series R-L-C circuit at resonance? Express your answer in of some or all of the variables \texttip{R}{R}, \texttip{L}{L}, \texttip{C}{C}, and \texttip{\omega _{\rm 0}}{omega_0}.
Hint 1. Impedance of a series R-L-C circuit The impedance \texttip{Z}{Z} of a series R-L-C circuit is a function of \texttip{R}{R}, \texttip{L}{L}, \texttip{C}{C}, and the angular frequency \texttip{\omega }{omega}. It can be expressed as Z=\sqrt{R^2 + (X_L-X_C)^2}, where \texttip{X_{\mit L}}{X_L} and \texttip{X_{\mit C}}{X_C} are the respective inductive and capacitive reactances of the circuit.
Hint 2. Condition for resonance At resonance, the inductive and capacitive reactances must be equal. That is, X_L=X_C .
ANSWER: \texttip{Z}{Z} =
At resonance, the impedance of the circuit does not depend on \texttip{L}{L}. Thus, for a given voltage, will \texttip{I_{\rm rms}}{I_rms} change when \texttip{L}{L} is increased?
ANSWER: \texttip{I_{\rm rms}}{I_rms} is twice as great. \texttip{I_{\rm rms}}{I_rms} is half as great. \texttip{I_{\rm rms}}{I_rms} is 1/\sqrt 2 times as great. \texttip{I_{\rm rms}}{I_rms} is unchanged.
In a series R-L-C circuit, for a given voltage, the rms current is always inversely proportional to the circuit impedance. Since at resonance the impedance depends only on \texttip{R}{R}, the rms current in the circuit remains constant when either \texttip{L}{L} or \texttip{C}{C} is changed. Note that this is true only at resonance.
Secondary Voltage and Current in a Transformer Ranking Task Description: Rank the current and voltage in the secondary coil of different transformers. (ranking task)
Six transformers have the rms primary voltages (\texttip{V_{\rm p}}{V_p}), number of primary turns (\texttip{N_{\rm p}}{N_p}), and number of secondary turns (\texttip{N_{\rm s}}{N_s}) listed below.
Part A Which of the transformers are step-up transformers? Which of the transformers are step-down transformers? Place the appropriate transformers into the two categories listed below.
Hint 1. Step-up and step-down transformers A transformer is referred to as step-up if the secondary voltage is larger than the primary voltage. Similarly, a transformer is referred to as stepdown if the secondary voltage is smaller than the primary voltage.
Hint 2. Turns ratio The turns ratio, N_{\rm s}/N_{\rm p}, is the factor that determines the characteristics of a transformer. If the ratio is greater than one, the transformer "steps up" the voltage by this factor. If it is less than one, it "steps down" the voltage. ANSWER:
Part B Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Voltage and number of turns The proportionality constant between voltage and number of turns is the same on both sides of the transformer.
Hint 2. Secondary voltage Since the proportionality constant between voltage and number of turns is the same on both sides of the transformer, we can set the ratios V/N on both sides equal to each other: \large{\frac {V_{\rm p}}{N_{\rm p}} = \frac {V_{\rm s}}{N_{\rm s}}}, or
\large{V_{\rm s} = \left(\frac {N_{\rm s}}{N_{\rm p}}\right)V_{\rm p}}. Therefore, the secondary voltage is just the product of the primary voltage and the turns ratio. ANSWER:
Part C 100 \rm A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Energy conservation Since energy flow (power) in a circuit is given by P = IV, energy conservation requires that any increase in voltage be accompanied by a corresponding decrease in current, and vice versa. Thus, a step-up transformer will step down current.
Hint 2. Stepping down current in a step-up transformer Since voltage is stepped up by the turns ratio, \large{V_{\rm s} = \left(\frac {N_{\rm s}}{N_{\rm p}}\right)V_{\rm p}}, conservation of energy ensures that current is stepped down by the same ratio. This can be written as \large{I_{\rm s} = \left(\frac {N_{\rm p}}{N_{\rm s}}\right)I_{\rm p}}.
ANSWER:
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H12 AC Circuits (31.1-6) Due: 11:59pm on Monday, December 2, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy
Problem 31.1 Description: The voltage across the terminals of an ac power supply varies with time according to V=V_0cos(omega t). The voltage amplitude is V_0. (a) What is the root-mean-square potential difference V_rms? (b) What is the average potential difference V_(a(v))... The voltage across the terminals of an ac power supply varies with time according to V
= V0 cos(ωt) . The voltage amplitude is V0 = 44.0V .
Part A What is the root-mean-square potential difference Vrms ? ANSWER: = 31.1 V
Part B What is the average potential difference Vav between the two terminals of the power supply? ANSWER: 0 V
Phasors Explained Description: Introduces phasors with questions both general and in the context of AC circuits. Recommended after resistance, inductance, and capacitance in AC circuits have been discussed. Includes a part on series LCR circuits and another part on parallel LCR circuits. Learning Goal: To understand the concept of phasor diagrams and be able to use them to analyze AC circuits (those with sinusoidally varying current and voltage). Phasor diagrams provide a convenient graphical way of representing the quantities that change with time along with cos(ωt + ϕ) , which makes such diagrams useful for analyzing AC circuits with their inherent phase shifts between voltage and current. You have studied the behavior of an isolated resistor, inductor, and capacitor connected to an AC source. However, when a circuit contains more than one element (for instance, a resistor and a capacitor or a resistor and an inductor or all three elements), phasors become a useful tool that allows us to calculate currents and voltages rather easily and also to visualize some important processes taking place in the AC circuit, such as resonance. Let us assume that a certain quantity I(t) changes over time as I(t) = I 0 cos(ωt) . A phasor is a vector whose length represents the amplitude I 0 (see the diagram ).This vector is assumed to rotate counterclockwise with angular frequency ω; that way, the horizontal component of the vector represents the actual value I(t) at any given moment. In this problem, you will answer some basic questions about phasors and prepare to use them in the analysis of various AC circuits.
In parts A - C consider the four phasors shown in the diagram . Assume that all four phasors have the same angular frequency ω.
Part A At the moment t depicted in the diagram, which of the following statements is true? ANSWER:
I 2 leads I 1 by π . I 1 leads I 2 by π . I 2 leads I 1 by π . 2
I 1 leads I 2 by π2 .
Part B At the moment shown in the diagram, which of the following statements is true? ANSWER:
I 2 lags I 3 by π . I 3 lags I 2 by 2π . I 2 lags I 3 by π2 . I 3 lags I 2 by π2 .
Part C At the moment shown in the diagram, which of the following statements is true? ANSWER:
I 2 leads I 4 by π . I 2 lags I 4 by 2π . I 2 leads I 4 by π2 . I 2 lags I 4 by π2 .
Let us now consider some basic applications of phasors to AC circuits. For a resistor, the current and the voltage are always in phase. For an inductor, the current lags the voltage by π . 2 For a capacitor, the current leads the voltage by π . 2
Part D Consider this diagram. Let us assume that it describes a series circuit containing a resistor, a capacitor, and an inductor. The current in the circuit has amplitude I , as indicated in the figure. Which of the following choices gives the correct respective labels of the voltages across the resistor, the capacitor, and the inductor?
ANSWER:
V1 ; V2 ; V3 V1 ; V2 ; V4 V1 ; V4 ; V2 V3 ; V2 ; V4 V3 ; V4 ; V2
Part E Now consider a diagram describing a parallel AC circuit containing a resistor, a capacitor, and an inductor. This time, the voltage across each of these elements of the circuit is the same; on the diagram, it is represented by the vector labeled V0 . The currents in the resistor, the capacitor, and the inductor are represented respectively by which vectors?
ANSWER:
I1 ; I2 ; I3 I1 ; I3 ; I4 I1 ; I4 ; I2 I1 ; I2 ; I4 I1 ; I3 ; I2
Problem 31.5 Description: (a) What is the reactance of an inductor with an inductance of L at a frequency of f? (b) What is the inductance of an inductor whose reactance is X_L at a frequency of f? (c) What is the reactance of a capacitor with a capacitance of C at a...
Part A What is the reactance of an inductor with an inductance of 3.30H at a frequency of 76.0Hz ? ANSWER: = 1580
Ω
Part B What is the inductance of an inductor whose reactance is 11.7Ω at a frequency of 76.0Hz ? ANSWER: = 2.45×10−2 H
Part C What is the reactance of a capacitor with a capacitance of 4.38×10−6F at a frequency of 76.0Hz ? ANSWER: = 478
Ω
Part D What is the capacitance of a capacitor whose reactance is 119Ω at a frequency of 76.0Hz ?
ANSWER: = 1.76×10−5 F
Problem 31.8 Description: A Radio Inductor. You want the current amplitude through a inductor with an inductance of L (part of the circuitry for a radio receiver) to be I when a sinusoidal voltage with an amplitude of V is applied across the inductor. (a) What frequency is... A Radio Inductor. You want the current amplitude through a inductor with an inductance of 4.50 mH (part of the circuitry for a radio receiver) to be 2.00 mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor.
Part A What frequency is required? ANSWER: = 2.12×105 Hz
Problem 31.13 Description: You have a resistor of resistance 200 Omega and a 6.00 mu F capacitor. Suppose you take the resistor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 (rad)/s. ... You have a resistor of resistance 200 Ω and a 6.00 µF capacitor. Suppose you take the resistor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s .
Part A What is the impedance of the circuit? ANSWER: 696
Ω
Part B What is the current amplitude? ANSWER: 4.31×10−2 A
Part C What is the voltage amplitude across the resistor? ANSWER: 8.62 V
Part D What is the voltage amplitudes across the capacitor? ANSWER: 28.7 V
Part E What is the phase angle ϕ of the source voltage with respect to the current? ANSWER: -73.3 ∘
Phasors: Analyzing a Parallel AC Circuit Description: This problem contains several questions, both qualitative and quantitative, directed at analyzing a parallel AC circuit using phasors. The circuit contains only a resistor and capacitor in parallel. It is recommended that the students solve the problem Phasors Explained before trying this one. Learning Goal: To understand the use of phasors in analyzing a parallel AC circuit. Phasor diagrams, or simply phasors, provide a convenient graphical way of representing the quantities that change with time along with cos(ωt + ϕ) . This makes them useful for analyzing AC circuits with their inherent phase shifts between voltage and current. If a quantity I(t) changes with time as I(t) = I 0 cos(ωt) , a phasor is a vector whose length represents the amplitude I 0 (see the diagram). This vector is assumed to rotate counterclockwise with angular speed ω; that way, the horizontal component of the vector represents the actual value I(t) at any given moment. In this problem, you will use the phasor approach to analyze an AC circuit. In answering the questions of this problem, keep the following in mind: For a resistor, the current and the voltage are always in phase. For an inductor, the current lags the voltage by π . 2 For a capacitor, the current leads the voltage by π . 2
Part A Phasors are helpful in determining the values of current and voltage in complex AC circuits. Consider this phasor diagram: The diagram describes a circuit that contains two elements connected in parallel to an AC source. The vector labeled V0 corresponds to the voltage across both elements of the circuit. Based on the diagram, what elements can the circuit contain?
ANSWER:
two resistors two inductors two capacitors a capacitor and a resistor an inductor and a resistor a capacitor and an inductor
Part B The phasor diagram from Part A describes a circuit that looks like the one in the figure: What are the respective amplitudes of the currents in the capacitor and the resistor in the diagram for Part A?
ANSWER:
I 1 ;I 2 I 2 ;I 1
Part C Find the amplitude of the current I 0 through the voltage source. Express your answer in of the magnitudes of the individual currents I 1 and I 2 .
Hint 1. The parallel connection In a parallel circuit, the current through the voltage source is the sum of the individual currents. That rule, learned first for DC circuits, holds true for AC circuits, too: The law of conservation of charge is still valid! Therefore, at any instant, the current through the voltage source is the sum of the currents through the resistor and the capacitor. However, you have to be careful in adding these currents.
Hint 2. The current through the voltage source as a vector sum The current through the voltage source may also be represented by the component of a rotating vector whose length represents its amplitude I 0 : You should find the value of I 0 as the magnitude of the vector sum of the individual currents. Note that the phase angle between I 1 and I 2 is π . 2
ANSWER:
I0 =
Part D What is the tangent of the phase angle ϕ between the voltage and the current through the voltage source? Express tan(ϕ) in of I 1 and I 2 .
Hint 1. The current through the voltage source as a vector sum The current through the voltage source may also be represented by the component of a rotating vector whose length represents its amplitude I 0 : You should find the value of I 0 as the magnitude of the vector sum of the individual currents. Note that the phase angle between I 1 and I 2 is π . 2
ANSWER:
tan(ϕ) =
As you can see, phasors can be helpful in visualizing the processes in AC circuits.
Problem 31.14 Description: You have a 0.400 H inductor and a 6.00 mu F capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 (rad)/s. (a) What is... You have a 0.400 H inductor and a 6.00 µF capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s .
Part A What is the impedance of the curcuit? ANSWER: 567
Ω
Part B What is the current amplitude? ANSWER:
5.29×10−2 A
Part C What is the voltage amplitude across the capacitor. ANSWER: 35.3 V
Part D What is the voltage amplitude across the inductor. ANSWER: 5.29 V
Part E What is the phase angle ϕ of the source voltage with respect to the current? ANSWER: -90.0 ∘
Exercise 31.19 Description: The power of a certain CD player operating at ## V rms is ## W. Assume that the CD player behaves like a pure resistance. (a) Find the maximum instantaneous power. (b) Find the rms current. (c) Find the resistance of this player. The power of a certain CD player operating at 120V rms is 20.0 W . Assume that the CD player behaves like a pure resistance.
Part A Find the maximum instantaneous power. ANSWER:
p max = 40.0 {\rm W}
Part B Find the rms current. ANSWER: I_{\rm rms} = 0.167 {\rm A}
Part C Find the resistance of this player. ANSWER: R = 720 {\rm {}}\Omega
Problem 31.25 Description: An L-R-C series circuit with an inductance of L, a resistance of R, and a capacitance of C carries an rms current of I with a frequency of f.
(a) What is the phase angle? (b) What is the power factor for this circuit? (c) What is the impedance of... An L-R-C series circuit with an inductance of 0.118{\rm H} , a resistance of 241{\rm \Omega} , and a capacitance of 7.26{\rm \mu F} carries an rms current of 0.455{\rm A} with a frequency of 392{\rm Hz} .
Part A What is the phase angle? ANSWER:
= 0.772 radians
Part B What is the power factor for this circuit? ANSWER:
= 0.716
Part C What is the impedance of the circuit? ANSWER:
= 336 \Omega
Part D What is the rms voltage of the source? ANSWER:
= 153 V
Part E What average power is delivered by the source? ANSWER:
= 49.9 W
Part F What is the average rate at which electrical energy is converted to thermal energy in the resistor?
ANSWER:
= 49.9 W
Part G What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? ANSWER: 0 W
Part H What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor? ANSWER: 0 W
Problem 31.33 Description: A Step-Down Transformer. A transformer connected to an AC line with an rms voltage of V_source is to supply an rms voltage of V_rms to a portable electronic device. The load resistance in the secondary is R. (a) What should the ratio of primary to ... A Step-Down Transformer. A transformer connected to an AC line with an rms voltage of 120{\rm V} is to supply an rms voltage of 12.0{\rm V} to a portable electronic device. The load resistance in the secondary is \texttip{R}{R} = 4.50{\rm \Omega} .
Part A What should the ratio of primary to secondary turns of the transformer be? ANSWER: = 10.0
Part B What rms current must the secondary supply? ANSWER: = 2.67 A
Part C What average power is delivered to the load? ANSWER: = 32.0 W
Part D What resistance connected directly across the source line (which has a voltage of 120{\rm V} ) would draw the same power as the transformer? ANSWER: = 450 {\rm \Omega}
A Low- Filter Description: In this problem, the student will derive an expression for the ratio of the voltage out to the voltage in (from source) as a function of source frequency for an L-R-C circuit where the output is taken over the capacitor. The figure shows a low- filter; the output voltage is taken across the capacitor in an L-R-C series circuit.
Part A Derive an expression for V_{\rm out}/V_{\rm s}, the ratio of the output and source voltage amplitudes, as a function of the angular frequency \texttip{\omega }{omega} of the source. Express your answer in of the capacitance \texttip{C}{C}, resistance \texttip{R}{R}, inductance \texttip{L}{L}, and angular frequency \texttip{\omega }{omega} of the circuit.
Hint 1. Voltage over a capacitor The voltage over a capacitor in an L-R-C circuit is given by the equation V_C = I X_C, where \texttip{V_{\mit C}}{V_C} is the voltage across the capacitor (also equal to \texttip{V_{\rm out}}{V_out} in this case), \texttip{I}{I} is the current flowing through the capacitor, and \large{X_C = \frac{1}{\omega C}} is the capacitive reactance.
Hint 2. Current through the circuit Recall that the current through an L-R-C circuit is given by the equation \large{I = \frac{V}{Z}}, where \texttip{I}{I} is the current, \texttip{V}{V} is the source voltage, and \texttip{Z}{Z} is the circuit impedance.
Hint 3. What is an expression for impedance? Choose the correct expression for the impedance ( \texttip{Z}{Z}) for a series L-R-C circuit. ANSWER:
\large{Z = \sqrt{R^2+\left(L\omega-\frac{1}{C\omega}\right)^2}} \large{Z = \sqrt{R^2+\left(L\omega+\frac{1}{C\omega}\right)^2}} \large{Z = \sqrt{R^2+\left(\frac{1}{L\omega}+C\omega\right)^2}} \large{Z = \sqrt{R^2+\left(\frac{1}{L\omega}-C\omega\right)^2}}
ANSWER:
\large{\frac{V_{\rm out}}{V_{\rm s}}} =
You can see from the shape of the graph (see the figure) of V_{\rm out}/V_{\rm s} why the circuit in this problem is called a low- filter: Low frequencies through (i.e., have output voltage) at a much higher multple of their source voltage than high frequencies. No units are shown, since this graph is simply meant to give a general idea of how the graph is shaped. The exact details depend on the inductance, resistance, and
capacitance of the components of the circuit.
Problem 31.35 Description: The internal resistance of an ac source is R_1. (a) What should the ratio of primary to secondary turns of a transformer be to match the source to a load with a resistance of R_2? ("Matching" means that the effective load resistance equals the... The internal resistance of an ac source is 12.3{\rm k \Omega} .
Part A What should the ratio of primary to secondary turns of a transformer be to match the source to a load with a resistance of 8.05{\rm \Omega} ? ("Matching" means that the effective load resistance equals the internal resistance of the source.) ANSWER: = 39.1
Part B If the voltage amplitude of the source is 60.0{\rm V} , what is the voltage amplitude in the secondary circuit under open circuit conditions? ANSWER:
= 1.53 V
Voltage and Current in AC Circuits
Description: Find the current when an AC voltage is applied to a resistor, a capacitor, and an inductor in turn. Conceptual questions about reactance. Learning Goal: To understand the relationship between AC voltage and current in resistors, inductors, and capacitors, especially the phase shift between the voltage and the current. In this problem, we consider the behavior of resistors, inductors, and capacitors driven individually by a sinusoidally alternating voltage source, for which the voltage is given as a function of time by V(t)=V_0 \cos (\omega t). The main challenge is to apply your knowledge of the basic properties of resistors, inductors, and capacitors to these "single-element" AC circuits to find the current \texttip{I\left(t\right)}{I(t)} through each. The key is to understand the phase difference, also known as the phase angle, between the voltage and the current. It is important to take into the sign of the current, which will be called positive when it flows clockwise from the b terminal (which has positive voltage relative to the a terminal) to the a terminal (see figure). The sign is critical in the analysis of circuits containing combinations of resistors, capacitors, and inductors.
Part A First, let us consider a resistor with resistance \texttip{R}{R} connected to an AC source (diagram 1). If the AC source provides a voltage V_R(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit R}\left(t\right)}{I_R(t)} through the resistor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{R}{R}, \texttip{\omega }{omega}, and \texttip{t}{t}.
Hint 1. Ohm's law Ohm's law \large{I=\frac{V}{R}} is still true at any moment in time.
ANSWER: \texttip{I_{\mit R}\left(t\right)}{I_R(t)} =
Note that the voltage and the current are in phase; that is, in the expressions for \texttip{V_{\mit R}\left(t\right)}{V_R(t)} and \texttip{I_{\mit R}\left(t\right)}{I_R(t)}, the arguments of the cosine functions are the same at any moment of time. This will not be the case for the capacitor and inductor.
Part B Now consider an inductor with inductance \texttip{L}{L} in an AC circuit (diagram 2). Assuming that the current in the inductor varies as I_L(t) = I_0 \cos(\omega t), find the voltage \texttip{V_{\mit L}\left(t\right)}{V_L(t)} that must be driving the inductor. Express your answer in of \texttip{I_{\rm 0}}{I_0}, \texttip{L}{L}, \texttip{\omega }{omega}, and \texttip{t}{t}. Use the cosine function, not the sine function, in your answer.
Hint 1. Kirchhoff's loop rule Apply Kirchhoff's loop rule to this circuit, going clockwise around the circuit (the same direction as the current arrow). Write down the sum of the voltage drops across each of the circuit elements.
Use \texttip{V_{\mit L}\left(t\right)}{V_L(t)} for the voltage from the source, and describe the voltage drop across the inductor in of \texttip{L}{L} and \texttip{dI_{\rm L(t)/dt}}{dI_L(t)/dt} ANSWER: \Sigma V_i =
Hint 2. The derivative of \cos(\omega t) What is the derivative of \texttip{\cos(\omega t)}{cos(omega*t)}, with respect to time? Express your answer in of any or all of \texttip{\sin(\omega t)}{sin(omega*t)}, \texttip{\omega }{omega}, and \texttip{t}{t}. ANSWER: \large{\frac{d\cos(\omega t)}{dt}} =
Hint 3. The phase relationship between sine and cosine Note that \large{\sin(x) = - \cos\left(x + \frac{\pi}{2}\right)}
ANSWER: \texttip{V_{\mit L}\left(t\right)}{V_L(t)} =
Graphs of \texttip{V_{\mit L}\left(t\right)}{V_L(t)} and \texttip{I\left(t\right)}{I(t)} are shown below. As you can see, for an inductor, the voltage leads (i.e., reaches its maximum before) the current by \large{\frac{\pi}{2}}; in other words, the current lags the voltage by \large{\frac{\pi}{2}}. This can be conceptually understood by thinking of inductance as giving the current inertia: The voltage "tries" to push current through the inductor, but some sort of inertia resists the change in current. This is another manifestation of Lenz's law. The difference \large{\frac{\pi}{2}} is called the phase angle.
Part C Again consider an inductor with inductance \texttip{L}{L} connected to an AC source. If the AC source provides a voltage V_L(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit L}\left(t\right)}{I_L(t)} through the inductor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{L}{L}, \texttip{\omega }{omega}, and \texttip{t}{t}. Use the cosine function, not the sine function, in your answer.
Hint 1. Using Part B You can obtain the answer almost immediately if you consider the results of Part B: The amplitude of the voltage ( \texttip{V_{\rm 0}}{V_0}) is I_0 \omega L; the frequency is the same as in Part B, and the phase difference is \large{\frac{\pi}{2}}. Do you what leads and what
lags? ANSWER: \texttip{I_{\mit L}\left(t\right)}{I_L(t)} =
For the amplitudes (magnitudes) of voltage and current, one can write V_0=I_0 R (for the resistor) and V_0=I_0 \omega L (for the inductor). If one compares these expressions, it should not come as a surprise that the quantity \omega L, measured in ohms, is called inductive reactance; it is denoted by \texttip{X_{\mit L}}{X_L} (sometimes \texttip{\chi _{\mit L}}{chi_L}). It is called reactance rather than resistance to emphasize that there is no dissipation of energy. Using this notation, we can write V_0=I_0 R (for a resistor) and V_0=I_0 X_L (for an inductor). Also, notice that the current is in phase with voltage when a resistor is connected to an AC source; in the case of an inductor, the current lags the voltage by \large{\frac{\pi}{2}}. What will happen if we replace the inductor with a capacitor? We will soon see.
Part D Consider the potentials of points a and b on the inductor in diagram 2. If the voltage at point b is greater than that at point a, which of the following statements is true?
Hint 1. How to approach the problem Try drawing graphs of the current through the inductor and voltage across the inductor as functions of time. ANSWER: The current \texttip{I\left(t\right)}{I(t)} must be positive (clockwise). The current \texttip{I\left(t\right)}{I(t)} must be directed counterclockwise. The derivative of the current \large{\frac{dI(t)}{dt}} must be negative. The derivative of the current \large{\frac{dI(t)}{dt}} must be positive.
It may help to think of the current as having inertia and the voltage as exerting a force that overcomes this inertia. This viewpoint also explains the lag of the current relative to the voltage.
Part E Assume that at time \texttip{t_{\rm m}}{t_m} , the current in the inductor is at a maximum; at that time, the current flows from point b to point a. At time \texttip{t_{\rm m}}{t_m}, which of the following statements is true?
Hint 1. How to approach the problem Try drawing graphs of the current through the inductor and voltage across the inductor as functions of time. ANSWER: The voltage across the inductor must be zero and increasing. The voltage across the inductor must be zero and decreasing. The voltage across the inductor must be positive and momentarily constant. The voltage across the inductor must be negative and momentarily constant.
Part F Now consider a capacitor with capacitance \texttip{C}{C} connected to an AC source (diagram 3). If the AC source provides a voltage V_C(t)=V_0 \cos (\omega t), what is the current \texttip{I_{\mit C}\left(t\right)}{I_C(t)} through the capacitor as a function of time? Express your answer in of \texttip{V_{\rm 0}}{V_0} , \texttip{C}{C}, \texttip{\omega }{omega} and \texttip{t}{t}. Use the cosine function
with a phase shift, not the sine function, in your answer.
Hint 1. The relationship between charge and voltage for a capacitor Recall that by definition of capacitance, the relationship q(t)=CV(t) among charge, capacitance, and voltage is true at all times. The convention here is that the voltage is that on the plate with charge +q relative to the other plate (which has charge -q).
Hint 2. The relationship between charge and current No charged particles "flow" through the capacitor. Instead, the current deposits or drains charge from the plates. If the voltage \texttip{V_{\mit C}\left(t\right)}{V_C(t)} is the voltage on terminal b (appropriate for Kirchhoff's law in this circuit), then if the current flow is positive (clockwise here as usual), the charge will increase. Hence \large{I(t)=\frac{dq(t)}{dt}}, where \texttip{q\left(t\right)}{q(t)} is the charge on the capacitor plate b.
Hint 3. Mathematical details To obtain the current, obtain \texttip{q\left(t\right)}{q(t)} from the voltage and then obtain the current from the time derivative of \texttip{q\left(t\right)}{q(t)}. Finally, rewrite the expression you obtain in of the cosine (instead of sine) function.
ANSWER: \texttip{I_{\mit C}\left(t\right)}{I_C(t)} =
For the amplitude values of voltage and current, one can write \large{V_0=\frac{I_0} {\omega C}}. If one compares this expression with a similar one for the resistor, it should come as no surprise that the quantity \large{\frac{1}{\omega C}}, measured in ohms, is called capacitive reactance; it is denoted by \texttip{X_{\mit C}}{X_C} (sometimes \texttip{\chi _{\mit C}}{chi_C}). It is called reactance rather than resistance to emphasize that there is no dissipation of energy. Using this notation, we can write V_0=I_0 X_C, and voltage lags current by \pi/2 radians (or 90 degrees). The notation is analogous to V_0=I_0 R for a resistor, where voltage and current are in phase, and V_0=I_0 X_L for an inductor, where voltage leads current by \pi/2 radians (or 90 degrees). We see, then, that in a capacitor, the voltage lags the current by \large{\frac{\pi}{2}}, while in the case of an inductor, the current lags the voltage by the same quantity \large{\frac{\pi}{2}}. In a capacitor, where voltage lags the current, you may think of the current as driving the change in the voltage.
Part G Consider the capacitor in diagram 3. Which of the following statements is true at the moment the alternating voltage across the capacitor is zero?
Hint 1. How to approach the problem Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)}
ANSWER:
The current must be directed clockwise. The current must be directed counterclockwise. The magnitude of the current must be a maximum. The current must be zero.
Part H Consider the capacitor in diagram 3. Which of the following statements is true at the moment the charge of the capacitor is at a maximum?
Hint 1. How to approach this problem Since the voltage is directly proportional to the charge, when the charge is maximum, so is the voltage. Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time. Find the current when the voltage drop is maximum.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)}
ANSWER: The current must be directed clockwise. The current must be directed counterclockwise. The magnitude of the current must be a maximum. The current must be zero.
Part I Consider the capacitor in diagram 3. Which of the following statements is true if the voltage at point b is greater than that at point a?
Hint 1. How to approach the problem Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitor as functions of time.
Hint 2. Graphs of \texttip{V_{\mit C}\left(t\right)}{V_C(t)} and \texttip{I\left(t\right)}{I(t)} Look at the region where \texttip{V_{\mit C}\left(t\right)}{V_C(t)} is positive. Is the current positive or negative or both?
ANSWER: The current must be directed clockwise. The current must be directed counterclockwise. The current may be directed either clockwise or counterclockwise.
Part J Consider a circuit in which a capacitor and an inductor are connected in parallel to an AC source. Which of the following statements about the magnitude of the current through the voltage source is true?
Hint 1. Driven AC parallel circuits The voltage across each element is the same at every moment in time. However, the magnitudes of the currents in an AC circuit cannot be added without consideration of the phase angle between the currents. ANSWER: It is always larger than the sum of the magnitudes of the currents in the capacitor and inductor. It is always less than the sum of the magnitudes of the currents in the capacitor and inductor. At very high frequencies it will become small. At very low frequencies it will become small.
This surprising result occurs because the currents in inductor and capacitor are exactly out of phase with each other (i.e., one lags and the other leads the voltage), and hence they cancel to some extent. At a particular frequency, called the resonant frequency, the currents have exactly the same amplitude, and they cancel exactly; that is, no current flows from the voltage source to the circuit. (Lots of current flows around the loop made by the inductor and capacitor, however.) If an L-C parallel circuit like this one connects the wire between amplifier stages in a radio, it will allow frequencies near the resonance frequency to easily, but will tend to short those at other freqeuncies to ground. This is the basic mechanism for selecting a radio station.
Alternating Voltages and Currents Description: Find the rms voltage from the graph of the instantaneous voltage; then find the rms current and average power from a given value of R. The voltage supplied by a wall socket varies with time, reversing its polarity with a constant frequency, as shown in the graph.
Part A What is the rms value \texttip{V_{\rm rms}}{V_rms} of the voltage plotted in the graph? Express your answer in volts.
Hint 1. RMS value of a quantity with sinusoidal time dependence A quantity that varies with time as x=x_{\max}\sin \omega t (or x=x_{\max}\cos \omega t) has a maximum value equal to x_{\max} and an rms value given by \large{x_{\rm rms}=\frac{x_{\max}}{\sqrt{2}}}.
Hint 2. Find the maximum value of the voltage What is the maximum value \texttip{V_{\rm max}}{V_max} of the voltage plotted in the graph? Express your answer in volts. ANSWER: \texttip{V_{\rm max}}{V_max} = 170 \rm V
ANSWER: \texttip{V_{\rm rms}}{V_rms} = 120 \rm V
This is the standard rms voltage supplied to a typical household in North America.
Part B When a lamp is connected to a wall plug, the resulting circuit can be represented by a simplified AC circuit, as shown in the figure. Here the lamp has been replaced by a resistor with an equivalent resistance \texttip{R}{R} = 120{\rm \Omega} . What is the rms value \texttip{I_{\rm rms}}{I_rms} of the current flowing through the circuit? Express your answer in amperes.
Hint 1. Ohm's law in AC circuits Ohm's law can still be applied to an AC circuit, provided the values used to describe all physical quantities are consistent. For example, Ohm's law can be written using maximum values of voltage and current, or alternatively using rms quantities. ANSWER:
\texttip{I_{\rm rms}}{I_rms} =
= 1.00 \rm A
Part C What is the average power \texttip{P_{\rm avg}}{P_avg} dissipated in the resistor? Express your answer in watts.
Hint 1. Average power in an AC circuit The average power \texttip{P_{\rm avg}}{P_avg} dissipated in a resistor with resistance \texttip{R}{R} is given by P_{\rm avg}=I_{\rm rms}^2R, where \texttip{I_{\rm rms}}{I_rms} is the rms value of the current flowing through the resistor. Note that the average power can also be expressed as \large{P_{\rm avg}=\frac{V_{\rm rms}^2}{R}}, where \texttip{V_{\rm rms}}{V_rms} is the rms voltage supplied to the resistor .
ANSWER:
\texttip{P_{\rm avg}}{P_avg} =
= 120 \rm W
The instantaneous power dissipated in the resistor can be substantially higher than the average power. However, since the voltage supplied to the resistor varies in time, so does the instantaneous power. Therefore, a better estimate of the energy dissipated in an AC circuit is given by the average power. For example, the power rating on light bulbs is in fact the average power dissipated in the bulb.
The Resonance Peak Description: Short quantitative problem on resonance in a series RLC circuit. Based on Young/Geller Quantitative Analysis 22.5. In this problem we consider the resonance curve for a circuit with electrical components \texttip{L}{L}, \texttip{R}{R}, and \texttip{C}{C} and resonant frequency \texttip{\omega _{\rm 0}}{omega_0}.
Part A For given values of \texttip{R}{R} and \texttip{C}{C}, if you double the value of \texttip{L}{L}, how does the new resonance curve differ from the original one?
Hint 1. How to approach the problem In a series R-L-C circuit, the resonance frequency is determined by \texttip{L}{L} and \texttip{C}{C}, whereas the peak value of the current in the circuit depends only on \texttip{R}{R}. Since the value of \texttip{R}{R} is being kept constant, the peak of the resonance curve is unchanged. To solve this problem, use proportional reasoning to find a relation between the resonance frequency \texttip{f_{\rm 0}}{f_0} and inductance \texttip{L}{L}. Find the simplest equation that contains these variables and other known quantities from the problem. Write this equation twice, once to describe the original circuit and again for the circuit with greater inductance. You then need to write each equation with all the constants on one side and the variables on the other. Since the variable is \texttip{f_{\rm 0}}{f_0} in this problem, you will write the equations in the form f_0=\cdots. To finish the problem you need to compare the two cases presented. For this question you should find the ratio of the resonance frequency of the original circuit to that of the circuit with greater inductance.
Hint 2. Resonance frequency In a series R-L-C circuit, the resonance frequency \texttip{f_{\rm 0}}{f_0} is given by \large{f_0=\frac{1}{2\pi\sqrt{LC}}}.
Hint 3. Find an expression for the new resonance frequency Write an expression for the resonance frequency \texttip{f_{\rm 1}}{f_1} of the circuit when the inductance \texttip{L}{L} is doubled? Express your answer in of some or all of the variables \texttip{R}{R}, \texttip{L}{L}, and \texttip{C}{C}. ANSWER: \texttip{f_{\rm 1}}{f_1} =
ANSWER: Both the peak height and peak frequency double. The peak height will be half as great, and the peak frequency will double. The peak height won't change, and the peak frequency will be 1/\sqrt 2 times as great. The peak height won't change, and the peak frequency will be half as great.
In an R-L-C circuit, the resonance peak depends only on \texttip{R}{R}, while the resonant frequency is determined by both \texttip{L}{L} and \texttip{C}{C}. In particular, for a given value of \texttip{C}{C}, the resonance frequency is inversely proportional to the square root of \texttip{L}{L}. Similarly, for a given value of \texttip{L}{L}, the resonance frequency is inversely proportional to the square root of \texttip{C}{C}.
Part B For given values of \texttip{R}{R} and \texttip{C}{C}, if you double the value of \texttip{L}{L}, how does the new rms current at resonance \texttip{I_{\rm rms}}{I_rms} differ from its original value? Assume that the voltage amplitude of the ac source is the same.
Hint 1. How to approach the problem Recall that in an ac circuit, for a given voltage, the rms current is always inversely proportional to the impedance of the circuit. This holds also at resonance. To solve this problem, use proportional reasoning and find a relation between the rms current \texttip{I_{\rm rms}}{I_rms} and the impedance \texttip{Z}{Z}.
Find the simplest equation that contains these variables and other known quantities from the problem. Write this equation twice, once to describe the original circuit and again for the circuit with greater inductance. You need to write each equation so that all the constants are on one side and the variables are on the other. Since the variable is \texttip{I_{\rm rms}}{I_rms} in this problem, you will write the equations in the form I_{\rm rms}=\cdots. To finish the problem you need compare the two cases presented. For this question you should find the ratio of the rms current in the original circuit to that in the circuit with greater inductance.
Hint 2. \texttip{V_{\rm rms}}{V_rms} across an ac circuit In a series R-L-C circuit, the rms voltage \texttip{V_{\rm rms}}{V_rms} across the circuit is always proportional to the rms current \texttip{I_{\rm rms}}{I_rms} in it. If \texttip{Z}{Z} is the impedance of the circuit, then V_{\rm rms}=I_{\rm rms}Z.
Hint 3. Find the impedance of the circuit What is the impedance \texttip{Z}{Z} of a series R-L-C circuit at resonance? Express your answer in of some or all of the variables \texttip{R}{R}, \texttip{L}{L}, \texttip{C}{C}, and \texttip{\omega _{\rm 0}}{omega_0}.
Hint 1. Impedance of a series R-L-C circuit The impedance \texttip{Z}{Z} of a series R-L-C circuit is a function of \texttip{R}{R}, \texttip{L}{L}, \texttip{C}{C}, and the angular frequency \texttip{\omega }{omega}. It can be expressed as Z=\sqrt{R^2 + (X_L-X_C)^2}, where \texttip{X_{\mit L}}{X_L} and \texttip{X_{\mit C}}{X_C} are the respective inductive and capacitive reactances of the circuit.
Hint 2. Condition for resonance At resonance, the inductive and capacitive reactances must be equal. That is, X_L=X_C .
ANSWER: \texttip{Z}{Z} =
At resonance, the impedance of the circuit does not depend on \texttip{L}{L}. Thus, for a given voltage, will \texttip{I_{\rm rms}}{I_rms} change when \texttip{L}{L} is increased?
ANSWER: \texttip{I_{\rm rms}}{I_rms} is twice as great. \texttip{I_{\rm rms}}{I_rms} is half as great. \texttip{I_{\rm rms}}{I_rms} is 1/\sqrt 2 times as great. \texttip{I_{\rm rms}}{I_rms} is unchanged.
In a series R-L-C circuit, for a given voltage, the rms current is always inversely proportional to the circuit impedance. Since at resonance the impedance depends only on \texttip{R}{R}, the rms current in the circuit remains constant when either \texttip{L}{L} or \texttip{C}{C} is changed. Note that this is true only at resonance.
Secondary Voltage and Current in a Transformer Ranking Task Description: Rank the current and voltage in the secondary coil of different transformers. (ranking task)
Six transformers have the rms primary voltages (\texttip{V_{\rm p}}{V_p}), number of primary turns (\texttip{N_{\rm p}}{N_p}), and number of secondary turns (\texttip{N_{\rm s}}{N_s}) listed below.
Part A Which of the transformers are step-up transformers? Which of the transformers are step-down transformers? Place the appropriate transformers into the two categories listed below.
Hint 1. Step-up and step-down transformers A transformer is referred to as step-up if the secondary voltage is larger than the primary voltage. Similarly, a transformer is referred to as stepdown if the secondary voltage is smaller than the primary voltage.
Hint 2. Turns ratio The turns ratio, N_{\rm s}/N_{\rm p}, is the factor that determines the characteristics of a transformer. If the ratio is greater than one, the transformer "steps up" the voltage by this factor. If it is less than one, it "steps down" the voltage. ANSWER:
Part B Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Voltage and number of turns The proportionality constant between voltage and number of turns is the same on both sides of the transformer.
Hint 2. Secondary voltage Since the proportionality constant between voltage and number of turns is the same on both sides of the transformer, we can set the ratios V/N on both sides equal to each other: \large{\frac {V_{\rm p}}{N_{\rm p}} = \frac {V_{\rm s}}{N_{\rm s}}}, or
\large{V_{\rm s} = \left(\frac {N_{\rm s}}{N_{\rm p}}\right)V_{\rm p}}. Therefore, the secondary voltage is just the product of the primary voltage and the turns ratio. ANSWER:
Part C 100 \rm A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Energy conservation Since energy flow (power) in a circuit is given by P = IV, energy conservation requires that any increase in voltage be accompanied by a corresponding decrease in current, and vice versa. Thus, a step-up transformer will step down current.
Hint 2. Stepping down current in a step-up transformer Since voltage is stepped up by the turns ratio, \large{V_{\rm s} = \left(\frac {N_{\rm s}}{N_{\rm p}}\right)V_{\rm p}}, conservation of energy ensures that current is stepped down by the same ratio. This can be written as \large{I_{\rm s} = \left(\frac {N_{\rm p}}{N_{\rm s}}\right)I_{\rm p}}.
ANSWER:
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