AP Physics C E and M 13.6 LC Circuits Overview

Describe capacitor-inductor circuits through energy exchange, maximum current, oscillatory charge, and simple harmonic motion.

[Maximum number: 7]

E\&M. 2.
The circuit represented above contains a 9.0 V battery, a 25 mF capacitor, a 5.0 H inductor, a $500 \Omega$ resistor, and a switch with two positions, $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ . Initially the capacitor is uncharged and the switch is open.

In experiment 2 the capacitor is again uncharged when the switch is closed to position $\mathrm{S}_{1}$ at time $t_{1}$ . The switch is then moved to position $\mathrm{S}_{2}$ at time $t_{2}$ when the magnitude of the charge on the capacitor plate is 105 mC, allowing electromagnetic oscillations in the LC circuit.

Calculate the energy stored in the capacitor at time $t_{2}$ .

Calculate the maximum current that will be present during the oscillations.

Calculate the time rate of change of the current when the charge on the capacitor plate is 50 mC.

Figure for Question 13.6 Circuits with Capacitors and Inductors (LC Circuits) question 1 — AP Physics C: Electricity & Magnetism

2 points

U_{C}=\frac{1}{2} Q V=\frac{1}{2} Q \frac{Q}{C}=\frac{1}{2} \frac{Q^{2}}{C}

For substituting correct values into a correct expression 1 point
For example, $U_{C}=\frac{1}{2} \frac{\left(105 \times 10^{-3} \mathrm{C}\right)^{2}}{\left(25 \times 10^{-3} \mathrm{~F}\right)}$
For a consistent answer with correct units 1 point
$U_{C}=0.22 \mathrm{~J}$

2 points
The maximum current occurs when there is no charge on the capacitor and all the energy is stored in the inductor.

U_{L}=\frac{1}{2} L I^{2}

The total energy is the energy that was stored in the capacitor at time $t_{2}$ .
For a correct expression of energy conservation

\begin{aligned} & \frac{1}{2} L I^{2}=U_{C} \\ & I=\sqrt{2 U_{C} / L} \end{aligned}

Substituting the given value for L and the value of $U_{C}$ determined in part (b) i

I=\sqrt{2(0.22 \mathrm{~J}) / 5.0 \mathrm{H}}

For an answer with units consistent with previous work

I=0.30 \mathrm{~A}

3 points
For a correct application of the loop rule

\begin{aligned} & L \frac{d I}{d t}+\frac{Q}{C}=0 \\ & \frac{d I}{d t}=-\frac{Q}{C L} \\ & \frac{d I}{d t}=-\frac{\left(50 \times 10^{-3} \mathrm{C}\right)}{\left(25 \times 10^{-3} \mathrm{~F}\right)(5.0 \mathrm{H})} \end{aligned}

For a correct numerical answer obtained from a correct procedure, with or without the negative sign
For the correct units on a calculated answer

\frac{d I}{d t}=-0.40 \mathrm{~A} / \mathrm{s}

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Covered syllabus points

13.6 Circuits with Capacitors and Inductors (LC Circuits) question 1