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 $8\Omega$ as shown in the figure. The potential difference across points $X$ and $Y$ is

• A. $3.6\text{V}$
• B. $10.0\text{V}$
• C. $5.6\text{V}$
• D. $2.0\text{V}$

Answer 1

The correct option is C $5.6\text{V}$

Given:
$E_1=6\text{V}$
$r_1=2\Omega$

$E_2=4\text{V}$
$r_2=8\Omega$

Now,
$E_{\text{eff}}=6-4=2\text{V}$
$R_{\text{eq}}=2+8=10\Omega$

So, current in the circuit will be
$I=\frac{E_{\text{eff}}}{R_{\text{eq}}}$
$\Rightarrow I=\frac{2}{10}=0.2\text{A}$

On going from point $X$ to $Y$ in clockwise, we get,

$V_X+4+0.2\times 8=V_Y$

$\Rightarrow V_X-V_Y=-5.6\text{V}$

Therefore, the potential difference across points $X$ and $Y$ is $5.6\text{V}$.