Question:
A cell $\mathrm{E}_{1}$ of emf $6 \mathrm{~V}$ and internal resistance $2 \Omega$ is connected with another cell $E_{2}$ of emf $4 \mathrm{~V}$ and internal resistance $8 \Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is :
Correct Option: , 3
Solution:
$I=\frac{6-4}{10}=\frac{1}{5} \mathrm{~A}$
$\mathrm{V}_{\mathrm{x}}+4+8 \times \frac{1}{5}-\mathrm{V}_{\mathrm{y}}=0$
$\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}}=-5.6 \Rightarrow|\mathrm{Vx}-\mathrm{Vy}|=5.6 \mathrm{~V}$
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