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Question

# Initially, the switch S is open in the circuit for a long time. After the switch is closed, the heat generated in the circuit will be:

A
600 μJ
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B
750 μJ
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C
350 μJ
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D
450 μJ
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Solution

## The correct option is D 450 μJWhen the switch is open: When the switch S is open, both the capacitors are in series. so, effective capacitance will be Ceq=6×36+3=189=2μF Net emf =20+10=30V And charge on each capacitor will be, q=CeqV ⇒q=2×(20+10)=60μC After the switch is closed: Let the higher emf battery dominates the circuit and redraw the circuit. From above diagram, we can find the charge on each capacitor. Therefore, charge on 6μF capacitor; q=20×6=120μC Thus, 20 V battery has supplied a charge of Δq=120−60=60μC Therefore, workdone by 20V battery is W1=60×20=1200μJ (∵W=qV) Charge on 3μF capacitor after switch S closes; q2=CV=3×10=30μC thus, final charge =+30μC Hence, charge supplied by 10V battery is −30μC(=30−60). So, workdone by battery will be W2=−30×10=−300 μJ And we know that energy stored by capacitor will be U=Q22C So, initially stored energy in capacitor is Ui=[(60)22×6+(60)22×3]=900 μJ And final energy stored in capacitors is Uf=12C1V21+12C2V22=12×6×(20)2+12×3×(10)2 ⇒Uf=1350μJ From energy conversation, Ui+Wbattery=Uf+HLost ⇒900+(W1+W2)=1350+HLost ⇒900+(1200−300)=1350+HLost ∴HLost=1800−1350=450μJ Hence, option (d) is correct. Why this question? Tip: After the switch is closed, we observe that charge on 3μF capacitor reduced. If represents that 10V battery connected in parallel has done a negative work or work has been done on the battery.

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