Question

    A capacitor of 4 µF is charged to 100 V. It is then

    connected across an uncharged 2 µF capacitor. Final common voltage is:
    A 33.3 V Correct Answer Incorrect Answer
    B 66.6 V Correct Answer Incorrect Answer
    C 50 V Correct Answer Incorrect Answer
    D 25 V Correct Answer Incorrect Answer

    Solution

    Initial charge Q = CV = 4 × 100 = 400μC After connection, total capacitance = 4 + 2 = 6 µF. Charge is conserved. Final voltage V = Q/Ceq = 400/6 = 66.6 V Hence, the final common voltage is 66.6 V.

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