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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 J
Let 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(V30)+3(V30)+5(V0)+9(V0)=0

21V12090=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×(3010)=80μ C

Charge of 3μ F capacitor is Q3=3μ×(3010)=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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