Q 12-03-259JEE MainJEE Main 2019 (8 Apr, Shift 1)Medium
For the circuit shown, with $R_1 = 1.0\ \Omega$, $R_2 = 2.0\ \Omega$, $E_1 = 2\ \text{V}$ and $E_2 = E_3 = 4\ \text{V}$, the potential difference between the points $a$ and $b$ is approximately (in V)
Answer: (A) $3.3$
Between $a$ and $b$ there are three branches, each a cell with its positive terminal towards $a$:
- left: $E_1 = 2\ \text{V}$ with $R_1 + R_1 = 2\ \Omega$
- middle: $E_2 = 4\ \text{V}$ with $R_2 = 2\ \Omega$
- right: $E_3 = 4\ \text{V}$ with $R_1 + R_1 = 2\ \Omega$
Let $V = V_a - V_b$. The currents into node $a$ must add to zero:
$$\frac{2 - V}{2} + \frac{4 - V}{2} + \frac{4 - V}{2} = 0 \Rightarrow 3V = 10$$
$$V \approx 3.3\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics