Question

A $2.0V$ potentiometer used for the determination of internal resistance of a $1.5V$ cell. The balance point of the cell in open circuit is $76cm$. When a resistor of $9.5\Omega$ is used in the external circuit of the cell, the balance point shifts to $57cm$ length of the potentiometer wire. Determine the internal resistance of the cell

• A. $1.7\Omega$
• B. $2.5\Omega$
• C. $1.5\Omega$
• D. $0.5\Omega$

Answer 1

The correct option is A $1.7\Omega$

Emf of potentiometer wire, $E=2V$

Potential of cell, $V=1.5V$

Balance point obtained when the cell is in open circuit, $I_1=76.3cm$

New balance point obtained when a resistor is used in the external circuit, $I_2=64.8cm$

External resistor, $R=9.5\Omega$

Now,

$r=\frac{R(E-V)}{V}=\frac{R(I_1-I_2)}{I_2}[\frac{E}{V}=\frac{I_1}{I_2}]=\frac{9.5(76.3-64.8)}{64.8}\Omega =\frac{9.5\times 11.5}{64.8}\Omega =1.686\Omega \approx 1.7\Omega$