Obtain an expression for equivalent capacitance C, when three capacitors having capacitance $C_{1}, C_{2}$ and $C_{3}$ are connected m series.

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Solution

Let three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected in series. We know charge remains same in series combination.
$V_{1} = \frac{q}{C_{1}}$
$V_{2} = \frac{q}{C_{2}}$
$V_{3} = \frac{q}{C_{3}}$
$V = V_{1} + V_{2} + V_{3} = q (\frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}})$
If the $3$ capacitors are replaced by a single capacitor such that on giving it charge q is capacitance is C then $V = \frac{q}{C}$
$∴ \frac{q}{C} = q (\frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}})$
or $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

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Obtain an expression for equivalent capacitance C, when three capacitors having capacitance C1,C2 and C3 are connected m series.

Obtain an expression for equivalent capacitance C, when three capacitors having capacitance $C_{1}, C_{2}$ and $C_{3}$ are connected m series.