Q 12-03-286JEE MainJEE Main 2017 (2 Apr)Medium
In the circuit shown, the current in each resistance is:
Answer: (A) $0$ A
Apply Kirchhoff's loop law to the first loop: left wire, top $2$ V cell, first $1\ \Omega$ resistor and bottom $2$ V cell. The top and bottom cells face the same way, so going round the loop one is crossed from $-$ to $+$ and the other from $+$ to $-$: their emfs cancel. Hence
$$I_1(1\ \Omega) = 2 - 2 = 0 \;\Rightarrow\; I_1 = 0$$
The same argument applies to the loop through the first and second resistors, and then to the second and third: the cell emfs cancel in every loop and the resistors start with equal potential differences of zero. So the current in every resistor is $0$ A.
Solution by Sreeraj P, M.Sc Physics