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Question

# If at t=0, switch S is closed, at steady state find heat generated (H) in the given circuit having a battery of 16 volts.

A
1280 μJ
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B
640 μJ
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C
320 μJ
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D
192 μJ
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Solution

## The correct option is B 640 μJLet us solve this problem using potential method, assigning potential to the known points, as shown in figure. Using Kirchoff`s law at junction P, (V0−16)×10+(V0−0)×4+(V0−0)×6=0 ⇒10V0−160+4V0+6V0=0 ⇒20V0=160 ∴V0=8 V Using V0 we can get potential across C1, C2 and C3 as follows, Potential difference across C1 is, ⇒V1=(16−8)=80 V Similarly V2 and V3 as, V2=(8−0)=8 V V3=(8−0)=8 V Now using V1, V2 and V3 we can calculate energy stored in each capacitor as follows, U1=12C1V12=12(10×10−6)(64)=320 μJ U2=12C2V22=12(4×10−6)(64)=128 μJ U3=12C3V32=12(6×10−6)(64)=192 μJ Work done by battery: WB=(Qthrough circuit)(VBattery) Qthrough circuit=Qacross C1=(10×10−6)(8)=80 μC ⇒WB=(80 μC)(16V)=1280 μJ As we know that, WB=Usystem + H ⇒H=WB−Usystem ⇒H=WB−(U1+U2+U3) ⇒H=1280−[320+128+192] ∴H=640 μJ Hence, option (b) is the correct answer.

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