In the circuit of Figure P32.48, the battery emf is 80 V, the resistance R is...

uploaded image In the circuit of Figure P32.48, the battery emf is 80 V, the resistance R is 240 , and the capacitance C is 0.500 µF. The switch S is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of 150 V. What is the value of the inductance L?

Expert Solution

Imax = 80/240 = 0.33 A

also, .5*C*Vmax^2 = .5*L*Imax^2
so, L = .5*10^-6*150^2/(.33^2) = 0.1H

genius_generous answered 2 years ago

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