Nine uncharged capacitors with equal capacitances are combined in series. The combination is connected to a 5.39 V battery, which charges the capacitors. The charging process involves 0.000291 C of charge moving through the battery. Find the capacitance of each capacitor.
The equivalent capacitance of this circuit which will be given by -
Ceq = Q / V
where, Q = charge moving through a battery = 0.000291 C
V = voltage of a battery = 5.39 V
then, we get
Ceq = [(0.000291 C) / (5.39 V)]
Ceq = 5.39 x 10-5 F
We know that, nine uncharged capacitors with equal capacitances are combined in series.
Then, the capacitance of each capacitor which will be given as -
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5 + 1 / C6 + 1 / C7 + 1 / C8 + 1 / C9
1 / Ceq = 9 / C
C = 9 Ceq
C = [9 (5.39 x 10-5 F)]
C = 4.85 x 10-4 F
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