Refer to the figure given below. The values of I₁, I₂ and I₃ are ______. (a) I₁ = 2.5 A, I₂ = 1.875 A, I₃ = 1.875 A (b) I₁ = 1.875 A, I₂ = 2.5 A, I₃ = 1.875 A (c) I₁ = 1.875 A, I₂ = 1.875 A, I₃ = 2.5 A (d) I₁ = 2.5 A, I₂ = 2.5 A, I₃ = 1.875 A

Answer: (a). Take left node L at 0; top T, bottom D, middle M. The 5 V cell gives VT - VD = 5; branch current I₁ = VM + 10 (1 Ω, 10 V). KCL at M: (VT - VM)/(2) + (VD - VM)/(2) = VM + 10; KCL for T+D: (VT + VD)/(4) = (2VM - VT - VD)/(2). Solving: VT + VD = -10, VM = -7.5, so VT = -2.5 V, VD = -7.5 V. I₁ = -7.5 + 10 = 2.5 A, I₃ = (0 - (-7.5))/(4) = 1.875 A At T: I₂ = (VT - VM)/(2) - (0 - VT)/(4) = 2.5 - 0.625 = 1.875 A So the answer is (a).

Refer to the figure given below. The values of I₁, I₂ and I₃ are… | JEE Main 2026 (5 Apr, Shift 1)

Q12JEE Main 2026 · 5 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are ______.

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Answer: (a)

Take left node $L$ at 0; top $T$ , bottom $D$ , middle $M$ . The 5 V cell gives $V_T - V_D = 5$ ; branch current $I_1 = V_M + 10$ (1 Ω, 10 V).
KCL at M: $\frac{V_T - V_M}{2} + \frac{V_D - V_M}{2} = V_M + 10$ ; KCL for T+D: $\frac{V_T + V_D}{4} = \frac{2V_M - V_T - V_D}{2}$ .
Solving: $V_T + V_D = -10$ , $V_M = -7.5$ , so $V_T = -2.5\text{ V}$ , $V_D = -7.5\text{ V}$ .
$I_1 = -7.5 + 10 = 2.5\text{ A}$ , $I_3 = \frac{0 - (-7.5)}{4} = 1.875\text{ A}$
At T: $I_2 = \frac{V_T - V_M}{2} - \frac{0 - V_T}{4} = 2.5 - 0.625 = 1.875\text{ A}$
So the answer is (a).

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