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10 August, 20:45

A 1.03 µF capacitor is connected in series with a 1.98 µF capacitor. The 1.03 µF capacitor carries a charge of + 11.1 µC on one plate, which is at a potential of 49.5 V. (a) Find the potential on the negative plate of the 1.03 µF capacitor. (b) Find the equivalent capacitance of the two capacitors.

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  1. 10 August, 22:36
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    Answer: a) 38.73 V; b) 0.67 μ F

    Explanation: In order to solve this problem we have to take into account the expression used for the capacitor and its conection in a circuit.

    We know that C=Q/V then Q = 11,1 μC and C1=1.03 µF

    V1=Q/C1=11.1 μC/1.03 * µF = 10.77 V so as the positive plate is at 49.5 V then negative plate is 38.73 V.

    The capaciro equivalent for a seri connection is given by;

    1/Cequi=1/c1+1/c2 = (1 / 1.03 µF) + (1 / 1,98 µF) = 0.67 µF
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