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Q.
The ratio of potential differences between $1 \, μF$ and $5 \, μF$ capacitors is
NTA AbhyasNTA Abhyas 2020Electrostatic Potential and Capacitance
Solution:
$3 \, μF$ & $1 \, μF$ are in parallel hence $C_{net}=C_{1}+C_{2}$ hence is equal to $4 \, μF$ .
Now $2 \, μF$ & $2 \, μF$ are in series with this equivalent capacitor $\left(4 \, μF\right)$ .Solving $2 \, μF$ and $2 \, μF$ are in series, we get $1 \, μF$ .
$\therefore $ $V_{4 μF}=\frac{V \times C_{1 μF}}{\left(C_{1 μF} + C_{4 μF}\right)}$
$\Rightarrow V_{4 μF}=\frac{10 \times 1}{\left(1 + 4\right)}=2$
$\Rightarrow V_{4 μF}=2 \, V$
(potential difference across $4 \, μF$ capacitor is calculated by voltage division rule)
$\therefore \, V_{5 μF}=10 \, V$
(Potential difference across $5 \, μF$ is same as battery)
$\therefore \, \, \frac{V_{1 μF}}{V_{5 μF}}=\frac{2}{10}=\frac{1}{5}$
