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March 26, 2017

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Capacitor A has a capacitance CA =29.4 uF and us initially charged to a voltage of V1= 10.70 V. Capacitor B is initially uncharged . When the switches are closed, connecting the two capacitors, the voltage on capacitor A drops to V2 = 6.40 V. What is the capacitance CB (in uF) of capacitor B?

  • physics - ,

    The voltage on A dropped from 10.70 to 6.40 V, or by 4.30 Volts. Capacitor A lost C(A)*deltaV = 1.264*10^-4 Coulombs of charge.

    That charge flowed to capacitor B, leaving it with 1.264*10^-4 Coulombs and a charge of 6.40 V

    C(B) = Q/V = 1.264*10^-4/6.4 = 19.8 uF

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