Question

In the circuit shown in figure-$3.349$, the potential difference across the capacitor is $10\, V$. Each resistance is of $3\, \Omega$. The cell is ideal. The EMF of the cell is :

• A. $14\, V$
• B. $16\, V$
• C. $18\, V$
• D. $24\, V$

Answer 1

Answer: (A)

In the circuit shown in figure if a current $i$ is supplied by the battery then it is divided in resistances $R$ and $3 R$ in parallel as shown.

In steady state no current will flow through the branch of capacitor so equivalent resistance across the battery is given as
$R_{ eq }=R+\frac{(3 R)(R)}{(3 R+R)}=\frac{7}{4} R$
Thus current through battery is given as
$i=\frac{E}{(7 / 4) R}=\left(\frac{4 E}{7 R}\right)$
In above circuit writing KVL equation from terminal $a$ to $b$ of capacitor gives
$ V_{a}-\frac{i}{4} R-i R=i R=V_{b} $
$\Rightarrow V_{a}-V_{b}=\frac{5}{4} i R=10$
Substituting value of current in above equation, we get
$\left(\frac{5}{4}\right)\left(\frac{4 E}{7 R}\right) =10$
$\Rightarrow E =14\, V$