In the following circuit, the current I₃ is (a) 5 A (b) 3 A (c) - 3A (d) -5/6 A

Answer: (d). Take the bottom junction O at 0 V; all three currents flow into O, so I₁ + I₂ + I₃ = 0. The 6 V cell (+ at top) puts the top junction T at +6 V. Left branch: the 8 V cell (+ towards O) puts the left corner at -8 V, so I₁ = (6 - (-8))/(28) = 0.5 A. Right branch: the 12 V cell (+ towards O) puts the right corner at -12 V, so I₂ = (6 + 12)/(54) = 1/3 A. I₃ = -(0.5 + 1/3) = -5/6 A So the answer is (d) -5/6 A.

In the following circuit, the current I₃ is | MHT CET 2024 (2 May, Shift 2)

Q51MHT CET 2024 · 2 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

In the following circuit, the current $I_3$ is

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Answer: (d)

Take the bottom junction O at $0$ V; all three currents flow into O, so $I_1 + I_2 + I_3 = 0$ .
The $6$ V cell (+ at top) puts the top junction T at $+6$ V.
Left branch: the $8$ V cell (+ towards O) puts the left corner at $-8$ V, so $I_1 = \frac{6 - (-8)}{28} = 0.5$ A.
Right branch: the $12$ V cell (+ towards O) puts the right corner at $-12$ V, so $I_2 = \frac{6 + 12}{54} = \frac{1}{3}$ A.
$I_3 = -(0.5 + \frac{1}{3}) = -\frac{5}{6}$ A
So the answer is (d) $-\frac{5}{6}$ A.

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Kirchhoff's Laws & Measuring Instruments question from MHT CET 2024 (2 May, Shift 2)

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