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Question

# Initially switch S is in open state. How much charge flows through switch S when it is closed? Assume steady state condition to be achieved after S closed.

A
64 μC
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B
34 μC
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C
68 μC
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D
54 μC
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Solution

## The correct option is D −54 μCInitially switch ′S′ is open, then at the steady state the current will not flow through capacitor branch. Let the charge on both capacitors will be equal to (q) q=Cnet×V ⇒q=(6×36+3)×(18−0) ∴q=189×18=36 μC When switch ′S′ is closed for long time again at the steady state condition current will not flows through capacitors. Current through the circuit, i=VReq ⇒ i=186+3=2 A Potential difference across capacitance C1 V1=iR6Ω ⇒V1=2×6=12V ⇒q1=C1V1=6×12=72 μC Similarly potential diffreence across capacitor C2 V2=iR3Ω ⇒V2=2×3=6 V Charge on capacitor C2; q2=C2V2 ⇒q2=3×6=18 μC Charge flown through switch S =(Final charge on plates of 1 and 2)-(Initial charge on plates of 1 and 2) ⇒ qflown=(−72+18) μC−(−36+36) μC ∴ qflown=−54 μC <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (d) is correct.

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