JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Current Electricity

Taking the lower-right corner as zero potential, the bottom $10\Omega$ has $10$ V across it, so $I_3=\frac{10}{10}=1$ A. The two parallel $10\Omega$ resistors have their left ends at $10$ V and their right ends at $20$ V, a difference of $10$ V across each, so each carries $\frac{10}{10}=1$ A, giving $I_1=1$ A and $I_2=1$ A. Therefore $\left|\frac{I_1+I_3}{I_2}\right|=\left|\frac{1+1}{1}\right|=2$.