In the circuit, the battery emf is 20 V, the resistance is 150 ohm, the inductan

Question

In the circuit, the battery emf is 20 V, the resistance is 150 ohm, the inductance is 700 mH and the capacitance is 0.01 mu F. The switch S is closed for a long time. After the switch is opened, what is the value of maximum voltage across the capacitor? Epsilon = 20 mho, R = 150 ohm, L = 700 mH, C = 01 mu F I_max = epsilon/R I_max = (20 mho)/(150 ohm) I_max = 133 A - when switch is opened, the circuit is reduced to an LC circuit. u = 1/2 LI^2 u = 1/2 (700 mH)(.133 A)^2 u = 0062 J u = 1/2 C epsilon^2 0062 J = 1/2 (.01 mu F) epsilon^2 0124 J = (.01 mu F) epsilon^2 squareroot 1244444.5 V = squareroot epsilon^2 1115.55 mho = epsilon

Explanation / Answer

The maximum current in the circuit is:

I(max) = V(bat)/R = 20/150 = 0.133 A

from the conservation of energy, stored energy is inductor is equal to that in capacitor

1/2 C V(max)^2 = 1/2 L I(max)^2

V = I sqrt (L/C) = 0.133 sqrt(700 x 10^-3/0.01 x 10^-6) = 1112.76 V

Hence, V = 1112.76 Volts

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