Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 Ω resistor will be, (a) 0.2 A (b) 0.6 A (c) 0.4 A (d) 1.0 A

Answer: (c). Let the junction potential be V. KCL: (30 - V)/(10) = (V - 12)/(20) + (V - 2)/(30) Multiply by 60: 180 - 6V = 3V - 36 + 2V - 4 ⇒ V = 20 V I₂₀ = (20 - 12)/(20) = 0.4 A So the answer is (c) 0.4 A.

Figure shows a part of an electric circuit. The potentials at points… | JEE Main 2023 (6 Apr, Shift 2)

Q86JEE Main 2023 · 6 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30\text{ V}, 12\text{ V}$ and $2\text{ V}$ respectively. The current through the $20\ \Omega$ resistor will be,

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Answer: (c)

Let the junction potential be $V$ . KCL: $\frac{30 - V}{10} = \frac{V - 12}{20} + \frac{V - 2}{30}$
Multiply by 60: $180 - 6V = 3V - 36 + 2V - 4 \Rightarrow V = 20\text{ V}$
$I_{20} = \frac{20 - 12}{20} = 0.4\text{ A}$
So the answer is (c) 0.4 A.

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Kirchhoff's Laws & Measuring Instruments question from JEE Main 2023 (6 Apr, Shift 2)

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