In the circuit of figure, the battery emf is 50 V, the resistance is 200 Ω, and the capacitance is 1.0 μ F. The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor.

Question

In the circuit of figure, the battery emf is 50 V, the resistance is 200 Ω, and the capacitance is 1.0 μ F. The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

When the switch has been closed for a long time, battery, resistor and coil carry constant current

I_{i} = \frac{ε}{R}

. When the switch is opened, current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the

L C

loop. We interpret the problem to mean that the voltage amplitude of these oscillations is

Δ V

, in

\frac{1}{2} C (Δ V)^{2} = \frac{1}{2} L I_{i}^{2}

.
Then,

L = \frac{C ( Δ V ) ^{2}}{I _{i}^{2} .} = \frac{C ( Δ V ) ^{2} R ^{2}}{ε ^{2}} = \frac{( 1.0 \times 1 0 ^{- 6} ) ( 100 ) ^{2} ( 200 ) ^{2}}{( 50 ) ^{2}}

= 0.16 H = 160 m H

Q. In the circuit of figure, the battery emf is 50V, the resistance is 200Ω, and the capacitance is 1.0μF. The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor.

Q. In the circuit of figure, the battery emf is $50 V$ , the resistance is $200Ω$ , and the capacitance is $1.0 μ F$ . The switch $S$ is closed for a long time interval, and zero potential difference is measured across the capacitor.