JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 Ω per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :
- (a) 5 V
- (b) 2.25 V
- (c) 6 V
- (d) 2.75 V
Answer and solution
Answer (a) 5 V
Solution
Physics sir · 4 steps
The most a potentiometer can measure is the full voltage across its wire, not the driver battery's emf.
- Wire resistance. The wire is 10 m = 1000 cm long at 0.1 Ω per cm. So R_(AB) = 1000 × 0.1 = 100 Ω.
- Driver current. The 6 V cell drives current through the wire and the 20 Ω resistor in series. That gives I = 6/(100 + 20) = 0.05 A.
V_(AB) = I R_(AB) = 0.05 × 100 = 5 V
- The upper limit. A balance point exists only where the potential difference along the wire equals the emf being measured. The largest potential difference available is across the whole wire, so any emf above 5 V has no balance point.
- So. The maximum emf this potentiometer can measure is 5 V, option (a).
The idea to keep. Find the current from the driver cell and every series resistance. The full wire's drop, I R_(wire), is the largest emf the potentiometer can measure.
Answer from the NTA final answer key. Solution by PrepWiser.