Question

In the figure, R = 122, C =8 pF, and L = 3 mH, and the ideal battery has emf = 32 V. The switch is kept in position a for a long time and then thrown to position b. What are www HE R E a 수 b C L mo (a) the maximum charge in the capacitor plates? Select one: a. 256.00 C b. 96.00 C c. 32.00 LC d. 24.00 C e. 0.20 C

4 part question?

Answer 1

Part (a): When a capacitor is fully charged, there will not be any current pass through it. So, the potential drop across the capacitor will be equal to the potential drop of battery. And as the switch is in position ‘a’, the inductor will not be a part of the circuit. Therefore, the maximum charge on the capacitor = Qmax = CE = 8x106x32 = 256 MIC Part (b): Now the switch is thrown to position 'b' then there will be LC oscillation. 1 1 So, the frequency of oscillation = w= = 6454.97 rad/sec VLC 3x10-3x8x106 Part (c): When whole energy of capacitor will be transferred to the inductor then the current in the circuit will be maximum. 1 1 LI ==x8x10-6x32 X3 x 1 Imax = 0.1549A = 0.15A 2 2 2 Part (d): Maximum electromagnetic energy -CE2 ==x8x10x32 = 364J 2 max 1 ce? 1 -LI 2 = max 2 2

Thankyou

Part (a): When a capacitor is fully charged, there will not be any current pass through it. So, the potential drop across the capacitor will be equal to the potential drop of battery. And as the switch is in position ‘a’, the inductor will not be a part of the circuit. Therefore, the maximum charge on the capacitor = Qmax = CE = 8x106x32 = 256 MIC Part (b): Now the switch is thrown to position 'b' then there will be LC oscillation. 1 1 So, the frequency of oscillation = w= = 6454.97 rad/sec VLC 3x10-3x8x106 Part (c): When whole energy of capacitor will be transferred to the inductor then the current in the circuit will be maximum. 1 1 LI ==x8x10-6x32 X3 x 1 Imax = 0.1549A = 0.15A 2 2 2 Part (d): Maximum electromagnetic energy -CE2 ==x8x10x32 = 364J 2 max 1 ce? 1 -LI 2 = max 2 2