Question

Four capacitors are connected as shown in figure to a $30V$ battery. The potential difference between points a and b is :

• A. $5V$
• B. $9V$
• C. $20V$
• D. $13V$

Answer 1

The correct option is D $13V$

As the capacitors 1 and 1.5 are iin series so

potential across $1\mu F$ is $V_p-V_a=\frac{1.5}{1+1.5}{V}_{pq}=\frac{1.5}{2.5}\left(30\right)$

or $V_p-V_a=18...\left(1\right)$

similarly for bottom branch , potential across $2.5\mu F$ is

$V_p-V_b=\frac{0.5}{2.5+0.5}{V}_{pq}=\frac{0.5}{3}\left(30\right)=5..\left(2\right)$

$\left(1\right)-\left(2\right),V_b-V_a=18-5=13V$