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JEE Main 2019 (12 Jan, Shift 2) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2019 (12 Jan, Shift 2)

Q157JEE Main 2019 · 12 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

In the given circuit diagram, the currents, I1 = – 0.3 A, I4 = 0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are:

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Answer: (d)

Path S–P–Q has no junction at P: $I_6 = I_5 = 0.4$ A.
Junction R: $I_1 + I_2 = I_4 \Rightarrow I_2 = 0.8 - (-0.3) = 1.1$ A.
Junction S: $I_4 = I_5 + I_3 \Rightarrow I_3 = 0.8 - 0.4 = 0.4$ A.
So the answer is (d) 1.1 A, 0.4 A, 0.4 A.

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