JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 20)
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Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p=1 \Omega$ as shown in the figure. An external resistance of $R_e=2 \Omega$ is connected via the sliding contact. The electric current in the circuit is :
1.35 A
0.9 A
1.0 A
0.3 A
Explanation
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We know, $$R = \int {{l \over A} \Rightarrow R \propto l} $$
As sliding contact of a potentiometer is in the middle of the potentiometer wire.
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$\therefore$ $${R_{eq}} = 0.5 + {{0.5 \times 2} \over {0.5 + 2}} = 0.5 + {1 \over {2.5}}$$
$$ = {1 \over 2} + {2 \over 5} = {{5 + 4} \over {10}} = {9 \over {10}} = 0.9\Omega $$
Hence, $$I = {v \over {{R_{eq}}}} = {{0.9v} \over {0.9\Omega }} \Rightarrow I = 1A$$
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