CUET Preparation Today
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: |
6 V 4.5 V 8 V 9 V |
6 V |
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: $I = \frac{V}{R_{total}} = \frac{9\text{ V}}{4\Omega + 4\Omega + 4\Omega} = \mathbf{0.75\text{ A}}$
Calculating the voltage drop ($I \times R$) of 3 V across each side:
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): $\Delta V = V_C - V_A = 6\text{ V} - 0\text{ V} = \mathbf{6\text{ V}}$
Note: The given answer is as per NTA.
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