Question

A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series to to a
resistance R and a cell of emf 2V and negligible internal resistance. A source of emf of 10mV is
balanced by a length of 40cm of the potentiometer wire. What is the value of the resistance R?
A.526.67 ohm
B.70 ohm
C.1580 ohm
D.zero ohm

Answer 1

Length of potentiometer wire=100cm

Total resistance of wire(r)=10$\Omega$

Resistance of 40 cm wire(r₁)=4Ω

EMF of driver cell(E)=$2V$

External resistance of circuit(R)=R

Current(I) in the circuit is given by $\frac{E}{r+R}$ =$\frac{2}{10+R}$ A

Voltage drop across 40 cm potentiometer wire=$Ir_1$ =

$\frac{2}{10+R}\times4$ =$\frac{8}{R+10}V$

This voltage drop is equal to the balancing EMF of $10mv$ or $10\times10^{-3}$ or $10^{-2}V$

so,

$$10^{-2}=\frac{8}{10+R} \implies 10+R=800 \implies R=790\Omega$$

It is the right answer.