Question

Some capacitors each of capacitance $30\mu \text{F}$ are connected as shown in the figure. Calculate equivalent capacitance between terminals $\text{A}$ and $\text{B}$ $\left(\text{in}\mu \text{F}\right)$.

Answer 1

Clearly none of the capacitors are in series or parallel connection. Let us consider charge distribution.


Applying $\text{KVL}$ in loop $\text{ADC}$:

$\frac{Q_2}{C}+\frac{Q_3}{C}-\frac{Q_1}{C}=0$

$\Rightarrow Q_1=Q_2+Q_3...\left(1\right)$

Now, charge in FB = charge flowing in AC= $Q_1$ [By symmetry]

Similalrly, charge in CE= charge in DE = $Q_2-Q_3$

Applying KVL in loop DCFD:

$\frac{Q_3}{C}+\frac{Q_1-Q_2+2Q_3}{C}-\frac{(Q_2-Q_3)}{C}=0$

$\Rightarrow Q_1-Q_2+3Q_3=Q_2-Q_3$

$\Rightarrow Q_1-2Q_2+4Q_3=\mathrm{0...}\left(2\right)$

Using (1) in (2) we get:

$Q_1-2Q_2=4(Q_2-Q_1)$

$\Rightarrow 5Q_1=6Q_2...\left(3\right)$

Now, $Q=C_{eq}{V}_{AB}$

$V_{AB}=V_{AD}+V_{DF}+V_{FB}$

$\Rightarrow V_{AB}=\frac{Q_2}{C}+\frac{Q_2-Q_3}{C}+\frac{Q_1}{C}$

Using $\left(1\right)$ and $\left(3\right)$,

$\Rightarrow V_{AB}=\frac{2Q_2+Q_1-Q_3}{C}=\frac{3Q_2}{C}$

Now,
Q = $Q_1+Q_2$, so

$C_{eq}=\frac{Q_1+Q_2}{\frac{3Q_2}{C}}$

From equation $\left(1\right)$,

$\Rightarrow C_{eq}=\frac{\frac{6Q_2}{5}+Q_2}{\frac{3Q_2}{C}}$

$\Rightarrow C_{eq}=\frac{11C}{15}$

As, $C=30\mu \text{F}$,

$\Rightarrow C_{eq}=\frac{11\times 30}{15}=22\mu \text{F}$

Why this question? This question will help the students in developing a proper approach to deal with the questions where no series, parallel combination or bridges are observed.