Question

The ratio of potential differences between $1\mu F$ and $5\mu F$ capacitors is

• A. $1:2$
• B. $3:1$
• C. $1:5$
• D. $10:1$

Answer 1

Answer: (C)

$3\mu F$ & $1\mu F$ are in parallel hence $C_{net}=C_1+C_2$ hence is equal to $4\mu F$ .
Now $2\mu F$ & $2\mu F$ are in series with this equivalent capacitor $\left(4\mu F\right)$ .Solving $2\mu F$ and $2\mu F$ are in series, we get $1\mu F$ .
$\therefore$ $V_{4\mu F}=\frac{V\times C_{1\mu F}}{\left(C_{1\mu F}+C_{4\mu F}\right)}$
$\Rightarrow V_{4\mu F}=\frac{10\times 1}{\left(1+4\right)}=2$
$\Rightarrow V_{4\mu F}=2V$
(potential difference across $4\mu F$ capacitor is calculated by voltage division rule)
$\therefore V_{5\mu F}=10V$
(Potential difference across $5\mu F$ is same as battery)
$\therefore \frac{V_{1\mu F}}{V_{5\mu F}}=\frac{2}{10}=\frac{1}{5}$