A cell E₁ of emf 6V and internal resistance $2\Omega$ is connected with another cell E₂ of emf 4V and internal resistance $8\Omega$  (as shown in the figure). The potential difference across points X and Y is:

 

Correct Option - 3


Detailed Solution:

$i=\frac{6-4}{2+8}=0.2A$

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

$\Rightarrow V_Y-V_X=5.6V$

 

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