Question

Part 1

An LC circuit is shown in the figure below. The 32 pF capacitor is initially charged by the 10 V battery when S is at position a. Then S is thrown to position b so that the capacitor is shorted across the 12 mH inductor

12 mH 32 pF s b 10 V

What is the maximum value for the oscillating current assuming no resistance in the circuit? Answer in units of A.

Part 2

What is the maximum energy stored in the magnetic field of the inductor? Answer in units of J.

Answer 1

energy of capacitor = C V ^2 /2

= (32 x 10^-12) (10^2)/ 2

= 1.6 x 10^-9 J

there is no energy loss in LC circuit.

so 1.6 x 10^-9 = (12 x 10^-3) I_max^2 /2

I_max = 5.16 x 10^-4 A ....Ans

2. maximum energy stored = 1.6 x 10^-9 J