The circuit shown in the figure contains an inductor L, a capacitor C0,...
The circuit shown in the figure contains an inductor L, a capacitor C0, a resistor R0 and an ideal battery. The circuit also contains two keys K1 and K2. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K1 is closed and immediately after this the current in R0 is found to be I1. After a long time, the current attains a steady state value I2. Thereafter, K2 is closed and simultaneously K1 is opened and the voltage across C0 oscillates with amplitude V0 and angular frequency ω0.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

Solution:
(P) When K1 is closed current in R0 is I1 At t=0; circuit will be
I1=0P→(1)
(Q) After long time inductor behave as a wire so I2
I2=205=4 AQ→(3)
(R) When K2 is closed and K1 open
ω0=1√LCω0=1√25×10−3×10×10−6=15×10−4ω0=2×103rad/sω0=2 kilo-radian /sR→(2)
(S) Now K2 is closed and K1 open
12LI22=12CV2025×10−3×(4)2=10×10−6×V20 V20=2500×16 V0=50×4=200 V S→(5)

(P) When K1 is closed current in R0 is I1 At t=0; circuit will be

I1=0P→(1)
(Q) After long time inductor behave as a wire so I2

I2=205=4 AQ→(3)
(R) When K2 is closed and K1 open

ω0=1√LCω0=1√25×10−3×10×10−6=15×10−4ω0=2×103rad/sω0=2 kilo-radian /sR→(2)
(S) Now K2 is closed and K1 open

12LI22=12CV2025×10−3×(4)2=10×10−6×V20 V20=2500×16 V0=50×4=200 V S→(5)
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