Q180 / 180 · COMEDK 2013 · Physics
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )
- A$\frac{3}{14} \mu \mathrm{F}$
- Correct answerB$\frac{14}{3} \mu \mathrm{F}$
- C$21 \mu \mathrm{F}$
- D$23 \mu \mathrm{F}$
Solution
As it is a balanced Wheatstone bridge, so $C$ and $D$ are same potential and the capacitor $2 \mu F$ is ineffective.
The $3 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ in upper portion are in series, so equivalent capacitance,
$$
C_{1}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}
$$
Similarly, $\quad C_{2}=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{F}$
$C_{1}$ and $C_{2}$ are in parallel, so their equivalent capacitance
$$
C=C_{1}+C_{2}=2+\frac{8}{3}=\frac{14}{3} \mu \mathrm{F}
$$
Hence, the correct answer is option: 'B'.