Q 12-03-256JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be
Answer: (B) $5\ \text{A}$
Let the potential of junction $C$ be $V$. With $S$ closed, the lower end of the $2\ \Omega$ resistor is at $0\ \text{V}$. Kirchhoff's current law at $C$ (current in = current out):
$$\frac{20 - V}{2} + \frac{10 - V}{4} = \frac{V - 0}{2}$$
Multiplying by 4: $40 - 2V + 10 - V = 2V \Rightarrow V = 10\ \text{V}$.
$$i = \frac{10 - 0}{2} = 5\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics