Question

An ideal cell of $emf10V$ is connected in circuit shown in figure. Each resistance is $2\Omega$. The potential difference $\left(inV\right)$ across the capacitor when it is fully charged is _______.

Answer 1

Step1: Given data and assumptions.

Supply $emf$$=10V$

Each resistance, $R=2\Omega$

Step2: Finding potential differences across the capacitor.

We know that,

When the capacitor becomes fully charged then it behaves as an open circuit.

Then by applying Kirchhoff's voltage law for loop DEFHCD and FGAHF in the above figure, we get.

$2i_1+2i=10$

$\Rightarrow$ $i_1+i=5$

$\Rightarrow$$2i_1+i_2=5$ $\left[\because i=i_1+i_2\right]$ …$\left(i\right)$

Again

$2i_2+2i_2-2i_1=0$

$\Rightarrow$ $i_1=2i_2$ ……$\left(ii\right)$

Substitute equation $\left(i\right)$ and $\left(ii\right)$ we get.

$4i_2+i_2=5$

$\Rightarrow$ $i_2=1$

Then from the equation, $\left(ii\right)$ we get.

$\Rightarrow$ $i_1=2$

Now, after a fully charged capacitor behaves as an open circuit then, from the figure right side resistor is negligible.

Therefore, The potential difference $V_{AB}$ across capacitors by applying Kirchhoff's voltage law between AB.

$V_{AB}=2i_2+2\left(i_1+i_2\right)$

$\Rightarrow$$V_{AB}=8V$

Hence, The potential difference (in V) across the capacitor when it is fully charged is $\mathbf{8}\mathbf{}\mathit{V}$.