Question

Draw labelled diagrams and derive expressions for the resultant capacity when capacitors are connected in:
(a) Series Combination
(b) Parallel Combination.

Answer 1

(a) Series Combination: Let $C_1$, $C_2$, $C_3$ are the capacitance of the given capacitors and the potential of capacitance $V_1$, $V_2$ and $V_3$ respectively. Let $Q$ is the charge of each capacitor.
Now the net potential of the combination
$V=V_1+V_2+V_3$ ......(i)
for first capacitor
$C_1=\frac{Q}{V_1}$
Or $V_1=\frac{Q}{C_1}$ .......(ii)
Similarly for two capacitors
$V_2=\frac{Q}{C_2}$ .......(iii)
and $V_3=\frac{Q}{C_3}$ .......(iv)
Putting the values of $V_1$, $V_2$, $V_3$ in equation (i)
$V=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}$
$V=Q(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3})$ .......(v)
If the total capacitance is $C$ then
$C=\frac{Q}{V}$
Or $\frac{1}{C}=\frac{V}{Q}$
Putting the value in equation (v)
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
(b) Parallel Combination: Let $C_1$, $C_2$ and $C_3$ are the capacitance of the given capacitor and the charge of capacitors are $Q_1$, $Q_2$, $Q_3$ and $V$ is potential of each capacitor.
$Q=Q_1+Q_2+Q_3$ (i)
For first capacitor
$C_1=\frac{Q_1}{V}$
Or $Q_1=C_{1}V$ .......(ii)
Similarly for other two capacitor
$Q_2=C_{2}V$ .......(iii)
$Q_3=C_{3}V$ .......(iv)
Put the values from equation (ii), (iii) and (iv) in equation (i)
$Q=C_{1}V+C_{2}V+C_{3}V$
Or $Q=V(C_1+C_2+C_3)$
$\frac{Q}{V}=(C_1+C_2+C_3)$ ......(v)
If the net capacitance is $C$, Now
$C=\frac{Q}{V}$
Put it in equation (v)
$C=C_1+C_2+C_3$