Question

A cell $E_1$of emf $6\text{V}$ and internal resistance $2\Omega$ is connected with another cell $E_2$ of emf $4\text{V}$ and internal resistance $6\Omega$ as shown in the figure below. The potential difference between the points $X$ and $Y$ is

• A. $5.5\text{V}$
• B. $2.5\text{V}$
• C. $-5.5\text{V}$
• D. $-2.5\text{V}$

Answer 1

The correct option is C $-5.5\text{V}$


$E_1=6\text{V};E_2=4\text{V}$ and $r_1=2\Omega ;r_2=6\Omega$

$E_{eff}=E_1-E_2=6-4=2\text{V}$

$r_{eff}=r_1+r_2=2+6=8\Omega$

$I=\frac{E_{eff}}{r_{eff}}=\frac{2}{8}=\frac{1}{4}=0.25\text{V}$

The Potential difference between $X$ and $Y$

$V_x+I_{r2}+E_2=V_y$

$V_x-V_y=-I_{r2}-E_2$

$=-(0.25\times 6)-4$

$=-1.5-4$

$=-5.5V$

Hence, option (c) is the correct answer.