Q1 = C1V1.
V1 = Q1 / C1. = 135 / 170 = 0.794 Volts.
V2 = 25-0.794 = 24.206 Volts.
Q2 = Q1 = 135pC.
Q2 = C2V2.
C2 = Q2/V2 = 135 / 24.206 = 5.577 pF.
A 170 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are connected to a battery with an emf of 25.0 V. If the 170 -pF capacitor stores 135 pC of charge on its plates, what is the unknown capacitance?
1 answer
1 answer