Question 456: Electrostats & Capacitance

Question

A $4 \mu \mathrm{~F}$ capacitor is charged to 10 V . The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is

Accepted Answer

We have an LC oscillation setup:
Capacitance $C = 4 \mu F = 4 	imes 10^{-6} F$
Initial capacitor voltage $V = 10\ 	ext{V}$
Inductance $L = 10\ 	ext{mH} = 10 	imes 10^{-3} H = 0.01\ H$
Step 1: Initial energy stored in the capacitor
$ E_C = \frac{1}{2} C V^2 $
$ E_C = 	frac{1}{2} \cdot 4 	imes 10^{-6} \cdot (10^2) $
$ E_C = 2 	imes 10^{-6} \cdot 100 = 2 	imes 10^{-4} \ 	ext{J} $
Step 2: Maximum energy in the inductor
At peak current,
all energy resides in the inductor
:
$ E_L = \frac{1}{2} L I_{	ext{max}}^2 $
Equating:
$ \frac{1}{2} L I_{	ext{max}}^2 = E_C $
$ \frac{1}{2} \cdot 0.01 \cdot I_{	ext{max}}^2 = 2 	imes 10^{-4} $
$ 0.005 \, I_{	ext{max}}^2 = 2 	imes 10^{-4} $
$ I_{	ext{max}}^2 = \frac{2 	imes 10^{-4}}{0.005} = \frac{2 	imes 10^{-4}}{5 	imes 10^{-3}} = 0.04 $
$ I_{	ext{max}} = 0.2\ 	ext{A} $
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Final Answer:
$ I_{	ext{max}} = 0.2\ 	ext{A} $
Correct Option: A (0.2 A)

Answer

0.2 A

Answer

0.1 A

Answer

0.4 A

Answer

0.25 A

Electrostats & Capacitance

A 4 \mu \mathrm{~F} capacitor is charged to 10 V . The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is

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