Kirchhoff's Laws & Measuring Instruments question from JEE Main 2023 (30 Jan, Shift 1)

Answer: 1.5

1. Take the bottom node at $0$; the $2$ V cell fixes the top node at $2$ V. Let the middle node be $V_M$ and the left wire $V_L$.
2. KCL at M: $\frac{2 - V_M}{1} - \frac{V_M}{1} = (V_M + 5 - V_L) \Rightarrow V_L = 3V_M + 3$
3. KCL at L: $V_M + 5 - V_L = \frac{V_L - 2}{2} + \frac{V_L}{2} \Rightarrow V_M = 0$, $V_L = 3$ V.
4. Top node: $I_3 = \frac{3 - 2}{2} = 0.5$ A in, $\frac{2 - 0}{1} = 2$ A out, so $I_1 = 2 - 0.5 = 1.5$ A.
5. So the answer is 1.5 A.