A 16 Ω wire is bent to form a square loop. A 9V supply having… | JEE Main 2021 (25 Jul, Shift 2)

Q120JEE Main 2021 · 25 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

A 16 $\Omega$ wire is bent to form a square loop. A 9V supply having internal resistance of 1 $\Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _______________ $\times$ 10 $-$ 1 V

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

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Answer: 45

Each side $= 4\ \Omega$ . Across side AB: $4 \parallel 12 = 3\ \Omega$ .
$I = \frac{9}{3 + 1} = \frac{9}{4}$ A, so $V_{AB} = \frac{9}{4} \times 3 = 6.75$ V.
Current in path A–D–C–B: $\frac{6.75}{12} = 0.5625$ A.
Diagonal AC: $V_{AC} = 0.5625 \times 8 = 4.5$ V (same for BD).
So the answer is $45 \times 10^{-1}$ V.

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