A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω and a standard cell of emf 4 V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is ____________ Ω.

Solution

Physics sir · 4 steps

Use the null point to find the voltage across the whole wire, then treat it as a voltage divider with 780 Ω.

1. Potential gradient. At the null point, the drop across 60 cm of wire equals the cell's emf. So the gradient is k = (20 mV)/(60 cm) = 1/3 mV cm⁻¹.
2. Drop across wire. The current is the same along the whole wire. So the drop across all 300 cm is 1/3 × 300 = 100 mV = 0.1 V.
3. Voltage divider. The 4 V cell drives one current I = 4/(780 + R) through the 780 Ω resistor and the wire in series. The wire's share of the voltage is IR, and this must equal 0.1 V.

   4R/(780 + R) = 0.1 ⇒ 4R = 78 + 0.1R ⇒ 3.9R = 78

4. So. R = 78/3.9 = 20 Ω. The resistance of the potentiometer wire is 20.

The idea to keep. In a potentiometer, the null length gives you the potential gradient. Gradient times full length gives the drop across the wire. Then the series circuit with the driver cell is just a voltage divider.

Answer from the NTA final answer key. Solution by PrepWiser.