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Voltage and Current in AC Circuits
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 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 account 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 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 I_R(t) through the resistor as a function of time?
Hint A.1
Ohm's law
Ohm's law I=\frac{V}{R} is still true at any moment in time.
Express your answer in terms of V_0, R, omega, and t.
ANSWER:
  I_R(t)  =  \frac{V_{0}}{R}{\cos}\left({\omega}{\cdot}t\right) 
Note that the voltage and the current are in phase; that is, in the expressions for V_R(t) and 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 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 V_L(t) that must be driving the inductor.
Part B.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 V_L(t) for the voltage from the source, and describe the voltage drop across the inductor in terms of L and dI_L(t)/dt
ANSWER:
  \Sigma V_i  =

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Part B.2
The derivative of \cos(\omega t)

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Hint B.3
The phase relationship between sine and cosine

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Express your answer in terms of I_0, L, omega, and t. Use the cosine function, not the sine function, in your answer.
ANSWER:
  V_L(t)  =  I_{0}{\cdot}{\omega}{\cdot}L{\cdot}{\cos}\left({\omega}{\cdot}t+\left(\frac{{\pi}}{2}\right)\right) 
Graphs of V_L(t) and I(t) are shown below. As you can see, for an inductor, the voltage leads (i.e., reaches its maximum before) the current by \frac{\pi}{2}; in other words, the current lags the voltage by \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 \frac{\pi}{2} is called the phase angle.

Part C
Again consider an inductor with inductance 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 I_L(t) through the inductor as a function of time?
Hint C.1
Using Part B
You can obtain the answer almost immediately if you consider the results of Part B: The amplitude of the voltage ( V_0) is I_0 \omega L; the frequency is the same as in Part B, and the phase difference is \frac{\pi}{2}. Do you remember what leads and what lags?
Express your answer in terms of V_0, L, omega, and t. Use the cosine function, not the sine function, in your answer.
ANSWER:
  I_L(t)  =  (V_0/(omega*L))*cos(omega*t-(pi/2)) 
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 X_L (sometimes 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 \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 D.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 I(t) must be positive (clockwise).
The current I(t) must be directed counterclockwise.
The derivative of the current \frac{dI(t)}{dt} must be negative.
The derivative of the current \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 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 t_m, which of the following statements is true?
Hint E.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 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 I_C(t) through the capacitor as a function of time?
Hint F.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 F.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 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 I(t)=\frac{dq(t)}{dt}, where q(t) is the charge on the capacitor plate b.
Hint F.3
Mathematical details
To obtain the current, obtain q(t) from the voltage and then obtain the current from the time derivative of q(t). Finally, rewrite the expression you obtain in terms of the cosine (instead of sine) function.
Express your answer in terms of V_0, C, omega and t. Use the cosine function with a phase shift, not the sine function, in your answer.
ANSWER:
  I_C(t)  =  V_{0}{\cdot}C{\cdot}{\omega}{\cdot}{\cos}\left({\omega}t+\frac{{\pi}}{2}\right) 
For the amplitude values of voltage and current, one can write 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 \frac{1}{\omega C}, measured in ohms, is called capacitive reactance; it is denoted by X_C (sometimes 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 \frac{\pi}{2}, while in the case of an inductor, the current lags the voltage by the same quantity \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 G.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 G.2
Graphs of V_C(t) and I(t)

ANSWER:
The current must be directed clockwise.
The current must be directed counterclockwise.
The current must be at 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 H.1  

Hint not displayed

Hint H.2  

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ANSWER:
The current must be directed clockwise.
The current must be directed counterclockwise.
The current must be at 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 I.1
How to approach the problem

Hint not displayed

Hint I.2  

Hint not displayed

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 J.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 currents in the capacitor and inductor.
It is always less than the sum 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 pass easily, but will tend to short those at other freqeuncies to ground. This is the basic mechanism for selecting a radio station.
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