The current flowing through $\mathrm{R}_{2}$ is:

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Current Electricity

The current flowing through

R_{2}

is:

Answer: (D)

Take the negative terminal A as the zero of potential, so the positive terminal E sits at

8 V

. The resistors joining the nodes are

R_{5} = 2 Ω

(E–D),

R_{4} = 3 Ω

(D–C),

R_{6} = 4 Ω

(E–A),

R_{7} = 3 Ω

(D–A) and

R_{3} = 6 Ω

(A–C), while

R_{1} = 2 Ω

(A–B) and

R_{2} = 4 Ω

(B–C) together form a single series branch of

6 Ω

between A and C. Applying Kirchhoff's current law at nodes D, C and B gives

7 V_{D} - 2 V_{C} = 24

V_{D} = 2 V_{C}

and

V_{B} = \frac{1}{3} V_{C}

. Solving these,

V_{C} = 2 V

V_{D} = 4 V

and

V_{B} = \frac{2}{3} V

. The current in

R_{2}

I = \frac{V _{C} - V _{B}}{R _{2}} = \frac{2 - \frac{2}{3}}{4} = \frac{1}{3} A

. The current flowing through

R_{2}

\frac{1}{3} A

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option