Question

When the generator emf in Sample Problem 31.07 is a maximum, what is the voltage across (a) the generator, (b) the resistance, (c) the capacitance, and (d) the inductance? (e) By summing these with appropriate signs, verify that the loop rule is satisfied. 

   When the generator emf in Sample Problem 31.07 is a maximum, what is the voltage across (a) the generator, (b) the resistance, (c) the capacitance, and (d) the inductance? (e) By summing these with appropriate signs, verify that the loop rule is satisfied.
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Fundamentals of Physics
Fundamentals of Physics
David Halliday,… 10th Edition
Chapter 31, Problem 93 ↓
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When the generator emf in Sample Problem 31.07 is a maximum, what is the voltage across (a) the generator, (b) the resistance, (c) the capacitance, and (d) the inductance? (e) By summing these with appropriate signs, verify that the loop rule is satisfied.
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Key Concepts

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Inductive Voltage Drop
An inductor resists changes in current due to the magnetic field that builds around it, which according to Faraday’s Law induces a voltage across the inductor that is proportional to the rate of change of current. This induced voltage, often out of phase with the current, must be accounted for when summing voltages around the loop.
Resistor Voltage Drop (Ohm’s Law)
The voltage drop across a resistor is determined by Ohm’s Law, which defines the relationship between voltage, current, and resistance (V = IR). This drop represents the energy dissipated as heat due to the flow of current through the resistor, and it is in phase with the current.
Capacitive Voltage Drop
A capacitor stores energy in the form of an electric field, and the voltage across it is related to the amount of charge stored relative to its capacitance. In an AC circuit, the capacitor's voltage can lag or lead the current, and its instantaneous value is key when assessing the circuit at the moment of maximum generator emf.
Electromotive Force (emf)
The emf of a generator represents the driving force that moves charge around the circuit. In AC circuits, this emf varies with time and reaches a maximum value at certain moments, setting the stage for analyzing instantaneous voltages across all circuit elements.
Kirchhoff’s Voltage Law (Loop Rule)
Kirchhoff’s Voltage Law states that the sum of all electrical potential differences around any closed loop in a circuit must equal zero. This principle is essential for verifying that the energy supplied by the emf is exactly balanced by the energy drops across the resistor, capacitor, and inductor in the loop.
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An ac generator produces emf $\mathscr{E}=\mathscr{E}_{m} \sin \left(\omega_{d} t-\pi / 4\right)$, where $\mathscr{E}_{m}=30.0 \mathrm{~V}$ and $\omega_{d}=350 \mathrm{rad} / \mathrm{s}$. The current in the circuit attached to the generator is $i(t)=I \sin \left(\omega_{d} t+\pi / 4\right)$, where $I=$ $620 \mathrm{~m}$ A. (a) At what time after $t=0$ does the generator emf first reach a maximum? (b) At what time after $t=0$ does the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as the case may be?

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Transcript

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00:01 For this problem on the topic of ac circuits, we are told that for the sample problem provided the generator's emf is a maximum, and we want to find the voltage across the generator, the resistance, the capacitance, and the inductance.
00:16 And by summing these voltages, we want to show rather that the loop rule is satisfied.
00:25 Now, we note that we obtain a maximum value when the time t is equal to.
00:32 To pi over 2 omega d, which is 1 over 4 times the frequency f, which is 1 over 4 times 60 hertz.
00:51 And this gives a time of 0 .00417 seconds, which is 4 .17 milliseconds.
01:07 And so therefore we have that the maximum emf times the sign of pi over 2 is equal to epsilon m times the sign of 90 degrees, which is 36 volts.
01:40 Now for part b, when t is equal to 4 .17 milliseconds, then the kind of current i is equal to i sine omega d times t minus phi which is i sign 90 degrees minus 24 .3 degrees which is 0 .164 ampiers times the cosine of 20 .5 times the cosine of 20.
02:31 4 .3 degrees...
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