A capacitor of capacity 10μF is charged to 40V and a second capacitor of capacity 15μF is charged to 30V. 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 C60 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 Q′1=C1V=10×34=340μC and Q′2=C2V=15×34=510μC