Question

A $10m$ long wire of resistance $15$ ohm is connected in series with a battery of emf $2V$ (no internal resistance) and a resistance of $5$ ohm. The potential gradient along the wire is:

• A. $0.15V{m}^{-1}$
• B. $0.45V{m}^{-1}$
• C. $1.5V{m}^{-1}$
• D. $4.5V{m}^{-1}$

Answer 1

Answer: (A)

$0.15V{m}^{-1}$

$E=IR$

$2=I[5+15]$

$I=\frac{2}{20}=0.1$

$\Rightarrow$ P.D across potentiometer $=$$I\times 15$

$=$ $1.5$

Potential Gradient $=\frac{P.D}{l}=\frac{1.5}{10}=0.15V/m$