A 190 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are c?

If the 190 -pF capacitor stores 125 pC of charge on its plates, what is the unknown capacitance?

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  • 9 years ago
    Best Answer

    You are missing a critical piece of information.

    You need the Voltage of the Source.

    The Formula is Q = CV

    So Q1 = (C1) x (V1)

    For the first capacitor Q1 = 125 pC = (190pF) (V)

    So the voltage across the 190 pF capacitor is 0.658 Volts

    Since this is a series circuit, the Charge Q is the same across each capactitor.

    So the charge Q across the 2nd unknown capacitor is the same --- 125 pC

    So Q2 = (C2) x (V2)

    Therefore, Q2 = 125 pC = (C2) ((V2 which is the Source Voltage minus 0.658 Volts)

    Voltage (source) = V1 + V2

    Therefore --- you need the source voltage to calculate C2

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