The circuit is closed for long time so that the capacitors are fully charged. Switch is opened at a certain instant. The heat developed in resistor $R_{2}$ after that instant is $K \times 1 0^{- 4} J$ . Then find the value of K.

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TestHub · Accepted Answer

Steady state: PD across R 1 ​ = 30 + 50 100 ​ × 30 = 20 V Energy stored: U = 2 1 ​ ( 1 + 2 ) × 1 0 − 6 × ( 20 ) 2 = 6 × 1 0 − 4 J After opening the switch, heat divides as H 2 ​ H 1 ​ ​ = R 1 ​ R 2 ​ ​ H 2 ​ = U ⋅ R 1 ​ + R 2 ​ R 2 ​ ​ = 6 × 1 0 − 4 × 300 200 ​ = 4 × 1 0 − 4 J ⇒ K = 4.00

Physics - Capacitor Question with Solution | TestHub

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