In the circuit shown, emf of the battery is 15.1 volts, C=197 microFarads, R1=7.

Question

In the circuit shown, emf of the battery is 15.1 volts, C=197 microFarads, R1=7.2 ohms, R2=15.9 ohms. After the switch being closed for a long time, the switch is opened. Find the total energy U dissipated in the resistor R2 after the switch is opened. (Enter your answer in milliJoules)

QUESTION 4 5 points Save Answer In the circuit shown, emf of the battery is 15.1 volts C-197 microFarads, R1=72 ohms, R2=15.9 ohms. After the switch being closed for a long time, the switch Is opened. Find the total energy U dissipated in the resistor R2 after the switch is opened. (Enter your answer in milli Joules) R1 R2

Explanation / Answer

Here,

C = 197 uF = 0.197 mF

R1 = 7.2 Ohm

R2 = 15.9 Ohm

before opening the switch , potential across the capacitor is V

V = V * R2/(R1 + R2)

V = 15.1 * 15.9/(15.9 + 7.2)

V = 10.4 V

as the energy dissipated in the resistance R2 is energy stored in the capacitor before opening the switch

energy dissipated in the resistance R2 = 0.5 * 0.197 * 10.4^2

energy dissipated in the resistance R2 = 10.65 mJ

the energy dissipated in the resistance R2 is 10.65 mJ

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