The circuit shown in the figure contains three resistors ${\mathrm{R}}_1=100 \mathrm{\Omega }$, ${\mathrm{R}}_2=50 \mathrm{\Omega }$ and ${\mathrm{R}}_3=20 \mathrm{\Omega }$ and cells of E.M.F.s ${\mathrm{E}}_1=2 \mathrm{V}$ and ${\mathrm{E}}_2$. The ammeter indicates a current of $50 \mathrm{mA}$. Find the E.M.F (in $\mathrm{V}$) of the second cell. (The internal resistance of the ammeter and of the cell should be neglected)

Solution

Using KVL :

$100\left(\text{i}+\text{0.05}\right)+50\text{i}-2=0$

$150\text{i}+3=0$

$\text{i}=\frac{-3}{150}=-\text{0.02}A$

$=-20mA$

So, the direction of i is opposite to what we have assumed. If we redraw the circuit.



Using KVL :

$20\times \text{0.05}+100\times \text{0.03}-{\text{V}}_2=0$

$1+3-{\text{V}}_2=0$

${\text{V}}_2=4\text{V}$

Hence, the correct answer is 4.