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Question

A capacitor of capacity 10μF is charged to 40 V and a second capacitor of capacity 15μF is charged to 30 V. If the capacitors are connected in parallel, the amount of charge that flows from the smaller capacitor to higher capacitor in μC is:

A
30
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B
60
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C
200
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D
250
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Solution

The correct option is C 60
Initially, the charge of first capacitance, Q1=C1V1=10×40=400μC and the charge of second one, Q2=C2V2=15×30=450μC

When the capacitors are connected in parallel, potential across them will be same. Thus, the common potential, V=Q1+Q1C1+C2=400+45010+15=34V

Now charges on capacitors are Q1=C1V=10×34=340μC and Q2=C2V=15×34=510μC

Charge flow =510450=60μC

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