Refer to the figure given below. The values of $\mathrm{I}_{1}$, $\mathrm{I}_{2}$ and $\mathrm{I}_{3}$ are ______.

(A) $\mathrm{I}_{1}=2.5$ A, $\mathrm{I}_{2}=1.875$ A, $\mathrm{I}_{3}=1.875$ A

(B) $\mathrm{I}_{1}=1.875$ A, $\mathrm{I}_{2}=2.5$ A, $\mathrm{I}_{3}=1.875$ A

(C) $\mathrm{I}_{1}=1.875$ A, $\mathrm{I}_{2}=1.875$ A, $\mathrm{I}_{3}=2.5$ A

(D) $\mathrm{I}_{1}=2.5$ A, $\mathrm{I}_{2}=2.5$ A, $\mathrm{I}_{3}=1.875$ A

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Current Electricity

Refer to the figure given below. The values of

I_{1}

I_{2}

and

I_{3}

are ______.

Answer: (A)

Take the bottom junction as

0

V; the

5

V cell on the right then fixes the top junction at

5

V. Let the left junction be at

V_{L}

and the central junction (between the two

2 Ω

resistors) at

V_{M}

. The branch joining the central and left junctions contains the

10

V cell and the

1 Ω

resistor. Kirchhoff's current law gives, at the central junction,

\frac{V _{M} - 5}{2} + \frac{V _{M}}{2} + (V_{M} - V_{L} + 10) = 0

, and at the left junction,

\frac{V _{L} - 5}{4} + \frac{V _{L}}{4} + (V_{L} - V_{M} - 10) = 0

. Solving these,

V_{M} = 0

V and

V_{L} = 7.5

V. The current through the

10

V–

1 Ω

branch is

I_{1} = V_{M} - V_{L} + 10 = 2.5

A, the current through the lower-left

4 Ω

resistor is

I_{3} = \frac{V _{L}}{4} = 1.875

A, and the current delivered by the

5

V cell is

I_{2} = \frac{5 - V _{L}}{4} + \frac{5 - V _{M}}{2} = 1.875

A. Hence

I_{1} = 2.5

I_{2} = 1.875

A and

I_{3} = 1.875

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option