Question

A 4 m long wire of resistance 8 $\Omega$ is connected in series with a battery of emf 2V and a resistor of 7 $\Omega$. The internal resistance of the battery is 1 $\Omega$. What is the potential gradient along the wire?

• A. $1.00V{m}^{-1}$
• B. $0.75V{m}^{-1}$
• C. $0.50V{m}^{-1}$
• D. $0.25V{m}^{-1}$

Answer 1

Answer: (D)

$0.25V{m}^{-1}$

Total resistance of the circuit $R=8+7+1\Omega$

Current in the potentiometer
$i=\frac{E}{R}=\frac{2}{8+7+1}=\frac{1}{8}A$
Voltage dro across potentiometer wire
$V_p=\frac{1}{8}\times 8=1V$

Length of the wire $l=4m$
$\therefore$ Potential gradient of potentiometer wire $=\frac{V_p}{l}=\frac{1}{4}=0.25V/m$