Question

In the figure shown, the potential difference between points $A$ and $B$ is :

• A. $10V$
• B. $30V$
• C. $7.5V$
• D. None of the above

Answer 1

The correct option is A $10V$

The above circuit is simplified by replacing $C_2,C_3,C_4$ series capacitance by its equivalent $\frac{1}{C_{eq}}=\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$ $C_2=C_3=C_4=6\mu F$

$\therefore C_{eq}=2\mu F$

Next applying $KVL\left(KirchoffsVoltageLaw\right)$

To Loop $I$

$+V-V1=0$

$\therefore V-\frac{Q1}{C_1}=0$ ... 1

To Loop $II$

$-V1+V_{eq}=0$

$\therefore \frac{-Q1}{C_1}+\frac{Q}{C_{eq}}=0$

$V=30V,C_1=6\mu F,C_{eq}=2\mu F$

solving 1 and 2 we get

$\therefore \frac{Q}{C_{eq}}=\frac{Q1}{C_1}=V=30Volt$

$\therefore Q=30\times 2\mu F=60\mu C$

Consider the series combination of $C_2,C_3,C_4$ we have $Q2=Q3=Q4=Q$

Therefore between points A and B the voltage drop is given as $V_3=\frac{Q3}{C_3}=\frac{60\mu C}{6\mu F}=10V$

Thus the potential difference between points A and B is $10V$