Question

A 24.0@mF capacitor is charged by a 170.0-V power supply, then disconnected from the power and connected in series with a 0.210-mH inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time t = 0 ms (the moment of connection with the inductor); (c) the energy stored in the inductor at t = 1.35 ms.

Answer 1

Answer:

Part a)

$f = 70.9 Hz$

Part b)

U = 111 J

Explanation:

As we know that the capacitor is of capacitance

$C = 24 mF$

$V = 170 Volts$

now the maximum charge on it

$Q = CV$

$Q = (24 mF)(170 V)$

$Q = 4.08 C$

Part a)

Oscillation frequency of the charge is given as

$f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}$

$f = 70.9 Hz$

Part b)

Now the equation of charge oscillation is given as

$q = Q cos(\omega t)$

$q = 4.08 cos(2\pi(70.9) t)$

now current in the circuit is given as

$i = \frac{dq}{dt}$

$i = (4.08)(2\pi(70.9)) sin(2\pi(70.9) t)$

now at t = 1.35 ms we have

$i = 1028.36 A$

So the energy stored in inductor is given as

$U = \frac{1}{2}Li^2$

$U = \frac{1}{2}(0.210 \times 10^{-3})(1028.36)^2$

$U = 111 J$