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: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.
Answer the following question based on the direction given below.
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Answer the questions based on the information given below:
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P # Q means P is brother of Q
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