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JEE Main 2020 (4 Sep, Shift 2) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2020 (4 Sep, Shift 2)

Q143JEE Main 2020 · 4 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

Four resistances 40 $\Omega$ , 60 $\Omega$ , 90 $\Omega$ and 110 $\Omega$ make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Path ABC: $40 + 60 = 100\ \Omega$ , current $0.4$ A; path ADC: $90 + 110 = 200\ \Omega$ , current $0.2$ A.
With $V_A = 40$ V: $V_B = 40 - 0.4 \times 40 = 24$ V.
$V_D = 40 - 0.2 \times 90 = 22$ V
$V_{BD} = 24 - 22 = 2$ V, so the answer is 2 V.

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