Question

Three capacitors are connected in the arms of a triangle $ABC$ as shown in figure, $5V$ 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
$Q=\frac{16}{5}\mu F\times 5V=16\mu C$
$Q_1=(2\mu F)\times 5V=10\mu C$
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
$V_{BC}=\frac{Q_2}{3\mu F}=\frac{6\mu C}{3\mu F}=2V$