Question

Find the equivalent capacitance between $A$ and $B$ in the circuit shown in the figure. Given $C=5\mu \text{F}$

• A. $25\mu \text{C}$
• B. $10\mu \text{C}$
• C. $7.5\mu \text{C}$
• D. $7\mu \text{C}$

Answer 1

The correct option is D $7\mu \text{C}$

Let us imagine a battery of EMF $E$ connected between $A$ and $B$. Let's suppose that the battery supplies a charge $Q$ which moves from teminal $A$ to $B$.


From the symmetry of circuit, the charges appearing on diagonally opposite capacitors will be equal.

Applying the charge conservation at junction $P$ we get, the charge$\left(Q_2\right)$ on capacitor ${}^{\prime }{C}^{\prime }$ connected between $P\&R$ as,
$-Q_1+(Q-Q_1)+Q_2=0$
$\because$The three plates of the capacitor taken together forms an isolated system
$\Rightarrow Q_2=2Q_1-Q$
The distribution of charges on each capacitor is shown below;


The potential difference between $A$ and $B$ can be written as,
$V_A-V_B=(V_A-V_P)+(V_P-V_B)$

Substituting for the potential difference across capacitors in terms of charge on them we get,
$E=\frac{Q_1}{C}+\frac{Q-Q_1}{2C}$

$\text{Or,}2CE=2Q_1+(Q-Q_1)=Q_1+Q....\left(1\right)$

Also, $V_A-V_B=(V_A-V_P)+(V_P-V_R)+(V_R-V_B)$
$\Rightarrow E=\frac{Q_1}{C}+\frac{2Q_1-Q}{C}+\frac{Q_1}{C}$
$\Rightarrow CE=Q_1+(2Q_1-Q)+Q_1=4Q_1-Q....\left(2\right)$

From Eq.$\left(1\right)$ and $\left(2\right)$,

$Q_1=\frac{3CE}{5}$

$\Rightarrow Q=2CE-Q_1=\frac{7CE}{5}$

Now replacing the whole circuit by an equivalent capacitance we get,

$Q=C_{eq}\times E$

$C_{eq}=\frac{Q}{E}=\frac{7CE}{5E}=\frac{7\times C}{5}$

$\Rightarrow C_{eq}=\frac{7\times 5}{5}=7\mu \text{C}$

Why this question? $\underset{–}{\text{Tip}}$: This problem presents the situation of an unbalanced wheatstone bridge. In such problems always stick to the application of fundmental concepts like symmetric distribution of charge, charge conservation & KVL.