Talha's Physics Academy
Equivalent Capacitance: Series and Parallel Combinations
Video Lecture
Watch the complete video lecture below to understand the derivations for equivalent capacitance in series and parallel circuits.
I) Capacitors in Series
Consider three capacitors having capacitances $C_1$, $C_2$, and $C_3$ connected in series with a battery of voltage $V$, as shown in the figure. Plate $A$ of $C_1$ is connected to the positive terminal of the battery; therefore, it draws positive charge $+q$ from the battery. Plate $F$ of $C_3$ is connected to the negative terminal of the battery; therefore, it acquires a charge $-q$.
The intermediate plates $B$, $C$, $D$, and $E$ do not draw any charge directly from the battery; instead, they get charged due to electrostatic induction. Consequently, each capacitor plate 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 drops across each capacitor:
$$V = V_1 + V_2 + V_3$$
If these three capacitors are replaced by a single equivalent capacitor of capacitance $C_e$ connected across the same voltage $V$, it will also store charge $q$. The potential difference across $C_e$ is given by:
$$V = \frac{q}{C_e}$$
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 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 combination formula:
$$\frac{1}{C_e} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$
This shows that the reciprocal of the equivalent capacitance in a series combination is equal to the sum of the reciprocals of the individual capacitances. Consequently, the equivalent capacitance is always less than the smallest individual capacitance in the combination.
II) Capacitors in Parallel
Consider three capacitors having capacitances $C_1$, $C_2$, and $C_3$ connected in parallel across a battery of voltage $V$, as shown in the figure. Since the capacitors are connected in parallel, the potential difference across each individual capacitor is the same and equal to $V$, while each capacitor draws a separate charge from the battery.
If $q_1$, $q_2$, and $q_3$ are the charges drawn by capacitors $C_1$, $C_2$, and $C_3$ respectively, then:
$$q_1 = C_1 V$$
$$q_2 = C_2 V$$
$$q_3 = C_3 V$$
If the three capacitors are replaced by an equivalent capacitor of capacitance $C_e$, the total charge $q$ supplied by the battery is the sum of the individual charges:
$$q = q_1 + q_2 + q_3$$
$$q = C_e V$$
Substituting the expressions for $q$, $q_1$, $q_2$, and $q_3$ into the total charge equation:
$$C_e V = C_1 V + C_2 V + C_3 V$$
$$C_e V = V (C_1 + C_2 + C_3)$$
Dividing both sides by $V$ gives the final parallel combination formula:
$$C_e = C_1 + C_2 + C_3$$
This shows that the equivalent capacitance of capacitors connected in parallel is equal to the direct sum of the individual capacitances. The equivalent capacitance is always greater than the largest individual capacitance in the combination.
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