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MHT CET 2026 (11 Apr, Shift 2) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from MHT CET 2026 (11 Apr, Shift 2)

Q8MHT CET 2026 · 11 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

In the following network, the current flowing through $15\ \Omega$ resistance is

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Answer: (c)

$\frac{15}{3} = \frac{20}{4} = 5$ , so the bridge is balanced and no current flows through G.
Path ABC: $15 + 3 = 18\ \Omega$ ; path ADC: $20 + 4 = 24\ \Omega$ .
$I_{15} = 2.1 \times \frac{24}{18 + 24} = 1.2$ A
So the answer is (c) $1.2$ A.

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