A 2.47micro-F and a 6.38 micro-F capacitor are connected in series across a 40.0-V battery. A 9.80 micro-F capacitor is then connected in parallel across the 2.47micro-F capacitor. Determine the voltage across the 9.80 micro-F capacitor.
let
C1 = 2.47 micro F
C2 = 6.38 micro F
C3 = 9.8 micro F
V_battery = 40 volts
C13 = C1 + C3
= 2.47 + 9.8
= 12.27 micro F
voltage across C3, V3 = V_battery*C2/(C13 + C2)
= 40*6.38/(6.38 + 12.27)
= 13.7 volts <<<<<<<<------------Answer
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