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Parallel and Series Combination of Capacitors

Question

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

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Solution

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Consider that three capacitors having capacitance $C_{1}, C_{2}$ and $C_{3}$ are connected in series to each other shown in the figure above,
Consider that $+ q$ charge is given to first plat of first capacitor. By induction a charge $(- q)$ will produce at internal surface of second plate of this capacitor and so on.
Consider that potential between the plates of capacitor are $C_{1}, C_{2}$ and $C_{3}$ respectively so,
$V_{1} = \frac{q}{C_{1}}, V_{2} = \frac{q}{C_{2}}$ and $V_{3} = \frac{q}{C_{3}}$
Now $V = V_{1} + V_{2} + V_{3}$
$∴$ $V = \frac{q}{C_{1}} + \frac{q}{C_{2}} + \frac{q}{C_{3}}$ ...(i)
If equivalent capacitance of combination of capacitors is $C$ then
$V = \frac{q}{C}$ ...(ii)
By equations (i) and (ii), we get
$\frac{q}{C} = \frac{q}{C_{1}} + \frac{q}{C_{2}} + \frac{q}{C_{3}}$
$⟹$ $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

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