Question
A capacitor of 4 ĀµF is charged to 100 V. It is then
connected across an uncharged 2 ĀµF capacitor. Final common voltage is:Solution
Initial charge Q = CV = 4 Ć 100 = 400Ī¼C After connection, total capacitance = 4 + 2 = 6 ĀµF. 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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