A 25 V batteryis connected across capacitors of capacitances C1 = C6 = 5.0 μF and C3 = C5 = 2.0C2 = 2.0C4 = 5.0 μF. What are theAnswers
Figure out total equivalent capacitance of the circuit by combining various capacitor combinations:
C35 = 1/[(1/2)+(1/2)] = 1uF
C235 = 1 + 2 = 3uF
C2345 = 3 + 6 = 9uF
C16 = 3 + 3 = 6uF
C123456 = 1/[(1/9)+(1/6)] = 3.6uF
total Q of the circuit = CV = 3.6*15 = 54uC
Charge Q over the capacitor combinations C2345 and C16 is going to be 54uC over each combo, since they are in series.
V2345 = Q2345/C2345 = 54/9 = 6V = V4 = V35 = V2 (voltage over the capacitors C4, C35, and C2 are all going to be 6V since the combos are all inparallel)
Since C3 and C5 are equal, they split the voltage equally and so V3 = 3V.
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