Question

A potentiometer wire has length 10 m and resistance $10\Omega$. It is connected to a battery of EMF 11 volt and internal resistance $1\Omega$, then the potential gradient in the wire is

• A. 10 V/m
• B. 1 V/m
• C. 0.1 V/m
• D. none

Answer 1

Answer: (B)

1 V/m

R=resistance of potentiometer wire=$10\Omega$

L=length of potentiometer wire=$10m$

r=internal resistance of cell=$1\Omega$

V=emf of cell=$11V$

Effective resistance of circuit=$R_{eq}=R+r$

$⟹R_{eq}=10+1=11\Omega$

Current in circuit=$I=\frac{V}{R_{eq}}=\frac{11}{11}$

$⟹I=1A$

Potential difference across ends of potentiometer wire AB=$V_o=IR$

$⟹V_o=1\times 10=10V$

Potential gradient of potentiometer= $\frac{V_o}{L}=\frac{10}{10}$

$=1V/m$

Answer-(B)