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Question

# The capacitors shown in the figure are in steady state. If at t=0 switch S is closed, then find the heat generated in the circuit.

A
1200μ J
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B
2100μ J
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C
1800μ J
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D
4200μ J
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Solution

## The correct option is B 2100μ JLet V be the potential at point P as shown in the figure. Applying the law conservation of charges on the given capacitor network we get, 4(V−30)+3(V−30)+5(V−0)+9(V−0)=0 ⇒21V−120−90=0 ⇒V=10 volts Therefore, Charge of 9μ F capacitor is Q9=9μ×10=90μ C Charge of 5μ F capacitor is Q5=5μ×10=50μ C Charge of 4μ F capacitor is Q4=4μ×(30−10)=80μ C Charge of 3μ F capacitor is Q3=3μ×(30−10)=60μ C Thus, charge flow in the circuit Q=Q9+Q5=Q3+Q4=140μ C Workdone by the battery W=Q×30=140μ×30=4200μ J Energy stored in 4μ F capacitor E1=12×4μ×(20)2=800μ J Energy stored in 3μ F capacitor E2=12×3μ×(20)2=600μ J Energy stored in 9μ F capacitor E3=12×9μ×(10)2=450μ J Energy stored in 5μ F capacitor E4=12×5μ×(10)2=250μ J So, H=W−(E1+E2+E3+E4) ⇒H=4200−(800+600+450+250)μ J ​​​​​​​⇒H=2100μ J Hence, option (b) is the correct answer.

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