Two capacitors of capacitances 3μF and 6μF are charged to a potential of 12V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The

Question

Two capacitors of capacitances 3μF and 6μF are charged to a potential of 12V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be-

Accepted Answer

Answer: 4V — C₁ = 3μF, C₂ = 6μF & V = 12V each ∴ Q = CV ⇒ Q₁ = 36μC & Q₂ = 72μC After +ve & −ve plates of both capacitor are joined together, then Qnet = 72 − 36 = 36μC & Cnet = 3 + 6 = 9μF ∴ V = Q/C = 36/9 = 4V

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