Question:
An ideal cell of emf $10 \mathrm{~V}$ is connected in circuit shown in figure. Each resistance is $2 \Omega$. The potential difference (in V) across the capacitor when it is fully charged is _______
Solution:
As capacitor is fully charged no current will flow through it.
We have the current distribution as shown in the figure.
Equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=\left(\frac{4 \times 2}{4+2}\right)+2$
Net current, $i=\frac{10}{\frac{4}{3}+2}=\frac{10 \times 3}{10}=3$ Amp
$i_{1}=2 \mathrm{~A}$ and $i_{2}=1 \mathrm{~A}$
$V_{A E B}=1 \times 2+3 \times 2=8 \mathrm{~V}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.