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Question
Derive the expression for resultant capacitance, when the capacitor is connected in series.
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Solution
Consider three capacitors of capacitance C1, C2 and C3 connected in series with a battery of voltage V as shown in figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from negative terminal to the right plate of C3which pushes the electrons of same amount -Q from left plate of C3 to the right plate of C2 due to electrostatic induction. Similarly, the left plate of C2 pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C1 At the same time, electrons of charge -Q are transferred from left plate of C1 to positive terminal of the battery.
(a)
Capacitors connected in series
(b)
Equivalent capacitors CS
By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different, so that the voltage across each capacitor is also different and are denoted as V1, V2 and V3 respectively.
The total voltage across each capacitor must be equal to the voltage of the battery.
V = V1 + V2 + V3 ….. (1)
Since Q = CV, we have V = `"Q"/"C"_1 + "Q"/"C"_2 + "Q"/"C"_3`
Q = `(1/"C"_1 + 1/"C"_2 + 1/"C"_3)` .....(2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = `"Q"/"C"_"s"`
Substituting this expression into equation (2) we get
V = `"Q"/"C"_"s" = "Q"(1/"C"_1 + 1/"C"_2 + 1/"C"_3)`
`1/"C"_"s" = 1/"C"_1 + 1/"C"_2 + 1/"C"_3` ...(3)
Thus, the inverse of the equivalent capacitance Cs of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance Cs is always less than the smallest individual capacitance in the series.
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