A current of 9A enters point P of an equilateral triangle PQR having three wires of 3 Ω each and leaves by point R. The currents I₁ and I₂ are respectively (a) 2 A, 7 A (b) 3 A, 6 A (c) 5 A, 4 A (d) 6 A, 3 A

Answer: (b). From P to R: direct path PR ( 3 Ω) carries I₂; path PQR ( 6 Ω) carries I₁. Current divides inversely as resistance: I₁ = 9 × 3/9 = 3 A, I₂ = 9 × 6/9 = 6 A So the answer is (b) 3 A, 6 A.

A current of 9A enters point P of an equilateral triangle PQR having… | MHT CET 2026 (17 Apr, Shift 2)

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Combination of Resistors

A current of 9A enters point P of an equilateral triangle PQR having three wires of $3\ \Omega$ each and leaves by point R. The currents $I_1$ and $I_2$ are respectively

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Answer: (b)

From P to R: direct path PR ( $3\ \Omega$ ) carries $I_2$ ; path PQR ( $6\ \Omega$ ) carries $I_1$ .
Current divides inversely as resistance: $I_1 = 9 \times \dfrac{3}{9} = 3$ A, $I_2 = 9 \times \dfrac{6}{9} = 6$ A
So the answer is (b) 3 A, 6 A.

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