Question

In experiment of the potentiometer wire AB of length 100 cm has a resistance of 10 Ω. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A source emf 10mV is balanced against a length of 40cm of the potentiometer wire. What is the value of the external resistance? 

(A) 900Ω 

(B) 820Ω 

(C) 790Ω 

(D) 670Ω

Answer 1

The correct option is  (C) 790Ω.

Explanation:

Resistance per unit length for wire AB = [(100) / 10] = 0.1 Ω/cm 

∴ Resistance of 40 cm of length of wire = 40 × 0.1 = 4Ω 

Current I in the wire = [(10 mV) / (R of 40cm length)] 

= [(10 × 10^–3) / 4] = 2.5 × 10^–3 A 

Also I = [V / (R_Total)] = [2 / (R + 10)] 

∴ [2 / (R + 10)] = 2.5 × 10^–3 

∴ 2.5R + 25 = 2000 

∴ 2.5R = 1975 

R = [(1975) / (2.5)] = 790Ω