A cell $E _{1}$ of $emf 6 V$ and internal resistance $2 \Omega$ is connected with another cell $E _{2}$ of $emf 4 V$ and internal resistance $8 \Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is............ $V$
JEE MAIN 2021, Diffcult
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$I =\frac{6-4}{10}=\frac{1}{5} A$
$V _{ x }+4+8 \times \frac{1}{5}- V _{ y }=0$
$V _{ x }- V _{ y }=-5.6 \Rightarrow| Vx - Vy |=5.6 V$
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