A 16Ω wire is bent to form a square loop. A cell of 9 V is connected across one of its sides. The potential difference between the diagonals of the square loop is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6 V
Total resistance is 16 $\Omega$, so each side is 4 $\Omega$.
A 9 V battery is connected across side AB.
Let $V_A = 0\text{ V}$ and $V_B = 9\text{ V}$.
The current ($I$) through the series path $B \to C \to D \to A$ is:
Calculating the voltage drop ($I \times R$) of 3 V across each side:
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$V_B = 9\text{ V}$
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$V_C = 9\text{ V} - 3\text{ V} = \mathbf{6\text{ V}}$
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$V_D = 6\text{ V} - 3\text{ V} = \mathbf{3\text{ V}}$
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$V_A = 0\text{ V}$
The question is interpreted as asking for the potential difference across either diagonal of the square, i.e., between the opposite vertices (such as A and C):