Question

A 18.0−μF capacitor is placed across a 22.5−V battery for a few seconds and is then connected across a 12.0−mH inductor that has no appreciable resistance.A) After connecting the capacitor and inductor together, find the maximum current in the circuit.B) When the current is a maximum, what is the charge on the capacitor?C) How long after connecting the capacitor and inductor together does it take for the capacitor to be completely discharged for the first time?D) How long after connecting the capacitor and inductor together does it take for the capacitor to be completely discharged for the second time?

Answer 1

Answer:

Part a)

$i = 10.4 mA$

Part b)

in this case the charge on the capacitor will become zero

Part c)

$t_1 = 0.73 ms$

Part d)

$t = 2.2 ms$

Explanation:

As we know that first capacitor is charged with the battery and then it is connected to the inductor

So here we will have

$Q = CV$

$Q = (18\mu F)(22.5 V)$

$Q = 405 \mu C$

Part a)

now since the total energy of capacitor is converted into the energy of inductor

so by energy conservation we can say

$\frac{Q^2}{2C} = \frac{1}{2}Li^2$

so maximum current is given as

$i = \sqrt{\frac{L}{C}}Q$

$i = \sqrt{\frac{12\times 10^{-3}}{18\times 10^{-6}}}(405\times 10^{-6})$

$i = 10.4 mA$

Part b)

When current is maximum then whole energy of capacitor is converted into magnetic energy of inductor

So in this case the charge on the capacitor will become zero

Part c)

Time period of oscillation of charge between the plates and inductor is given as

$T = 2\pi\sqrt{LC}$

$T = 2\pi\sqrt{(18\mu F)(12 mH)}$

$T = 2.92 ms$

now capacitor gets discharged first time after 1/4 of total time period

$t_1 = 0.73 ms$

Part d)

Since time period is T and capacitor gets discharged two times in one complete time period of the motion

so first it will discharges in T/4 time

then next T/4 it will get charged again

then next T/4 time it will again discharged

so total time taken

$t = 3T/4$

$t = 2.2 ms$