CUET · PHYSICS · PYQ PAPER 2025
A 16 \( \Omega \) 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 :
- A 6V
- B 4.5 V
- C 8V
- D 9V
Answer & Solution
Correct Answer
(A) 6V
Step-by-step Solution
Detailed explanation
\( R_{side} = \frac{16 \, \Omega}{4} = 4 \, \Omega \) \( R_{ADCB} = 3 \times 4 \, \Omega = 12 \, \Omega \) \( I_{ADCB} = \frac{9 \, V}{12 \, \Omega} = 0.75 \, A \) \( V_C = I_{ADCB} \times R_{CB} = 0.75 \, A \times 4 \, \Omega = 3 \, V \) (assuming \( V_B = 0 \, V \))…
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