Question

Three capacitors are connected in the arms of a triangle $A B C$ as shown in figure, $5 \,V$ is applied between $A$ and $B$. The voltage between $B$ and $C$ is

• A. 2 V
• B. 1 V
• C. 3 V
• D. 1.5 V

Answer 1

Answer: (A)

The equivalent circuit diagram as shown in the figure.

The equivalent capacitance between $A$ and $B$ is
$C_{ eq }=\frac{2\, \mu F \times 3 \,\mu F }{2\, \mu F +3\, \mu F }+2 \,\mu F$
$ =\frac{6}{5} \mu F +2\, \mu F =\frac{16}{5} \mu F$
Total charge on the given circuit is
$Q=\frac{16}{5} \mu F \times 5\, V =16\, \mu C$
$Q_{1}=(2\, \mu F ) \times 5 \,V =10 \,\mu C$
As $Q=Q_{1}+Q_{2}$
$\therefore Q_{2}=Q-Q_{1}=16\, \mu C -10 \,\mu C =6\, \mu C$
$\therefore $ Voltage between $B$ and $C$ is
$V_{B C}=\frac{Q_{2}}{3 \,\mu F }=\frac{6\, \mu C }{3\, \mu F }=2 \,V$