Class 10 > Unit # 14: Electrostatics > Combination of Capacitors

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Electrostatics - Equivalent Capacitance

Q. Derive expressions for the equivalent capacitance when (i) Capacitors are connected in series (ii) Capacitors are connected parallel

(i) Capacitors in Series Combination

Figure: Three capacitors connected in series across a voltage source $V$.

Consider three capacitors having capacitances $C_1$, $C_2$, and $C_3$ connected in series with a battery of voltage $V$.

• Plate A of $C_1$ is connected to the positive terminal of the battery and acquires a positive charge $+q$.
• Plate F of $C_3$ is connected to the negative terminal and acquires a negative charge $-q$.
• Intermediate plates ($B, C, D,$ and $E$) do not draw direct charge from the battery; instead, they become charged via electrostatic induction. Consequently, every capacitor in the series chain acquires the same magnitude of charge $q$.

If $V_1$, $V_2$, and $V_3$ are the potential differences across $C_1$, $C_2$, and $C_3$ respectively, then:

$V_1 = \frac{q}{C_1}, \quad V_2 = \frac{q}{C_2}, \quad V_3 = \frac{q}{C_3}$

The total voltage $V$ supplied by the battery is the sum of the individual potential differences:

$V = V_1 + V_2 + V_3$

If these three capacitors are replaced by a single equivalent capacitor of capacitance $C_e$ connected to the same voltage $V$, it will also store charge $q$, such that:

$V = \frac{q}{C_e}$

Substituting the expressions for $V, V_1, V_2,$ and $V_3$ into the total voltage equation:

$\frac{q}{C_e} = \frac{q}{C_1} + \frac{q}{C_2} + \frac{q}{C_3}$

Factoring out $q$ from the right-hand side:

$\frac{q}{C_e} = q \left( \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right)$

Dividing both sides by $q$ yields the final series equivalent capacitance formula:

$\frac{1}{C_e} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Conclusion: The reciprocal of the equivalent capacitance in a series combination is equal to the sum of the reciprocals of individual capacitances. The equivalent capacitance is always less than the smallest individual capacitance in the combination.

(ii) Capacitors in Parallel Combination

Figure: Three capacitors connected in parallel across a voltage source $V$.

Consider three capacitors having capacitances $C_1$, $C_2$, and $C_3$ connected in parallel across a battery of voltage $V$.

Since all capacitors are connected across the same common terminals, the potential difference across each capacitor is identical and equal to $V$. However, each capacitor draws a distinct amount of charge from the battery depending on its capacitance.

Let $q_1$, $q_2$, and $q_3$ be the charges drawn by capacitors $C_1$, $C_2$, and $C_3$ respectively:

$q_1 = C_1 V, \quad q_2 = C_2 V, \quad q_3 = C_3 V$

The total charge $q$ supplied by the battery is the sum of the individual charges:

$q = q_1 + q_2 + q_3$

If the three capacitors are replaced by an equivalent capacitor of capacitance $C_e$, the total charge stored is:

$q = C_e V$

Substituting the expressions for $q, q_1, q_2,$ and $q_3$ into the total charge equation:

$C_e V = C_1 V + C_2 V + C_3 V$

Factoring out $V$ on the right side:

$C_e V = V (C_1 + C_2 + C_3)$

Dividing both sides by $V$ yields the final parallel equivalent capacitance formula:

$C_e = C_1 + C_2 + C_3$

Conclusion: The equivalent capacitance of a parallel combination is equal to the direct algebraic sum of the individual capacitances. The equivalent capacitance is greater than the largest individual capacitance in the combination.

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