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A capacitor of charge $100 μC, 2 μF$ is connected another uncharged capacitor of capacity $6 μ F$ then common potential of both the capacitors is

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detailed solution

Correct option is C

100−Q2=Q6⇒300−3Q=Q⇒4Q=300⇒Q=75∴V=756 = 12.5 V

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A capacitor of charge 100μC, 2μF is connected another uncharged capacitor of capacity 6 μF then common potential of both the capacitors is

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Ready to Test Your Skills?

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A capacitor of charge $100 μC, 2 μF$ is connected another uncharged capacitor of capacity $6 μ F$ then common potential of both the capacitors is