Question

A $6\, V$ battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of $100\,\Omega$ . The difference of potential between two points on the wire separated by a distance of $50 \,cm$ will be

• A. 2 V
• B. 3 V
• C. 1 V
• D. 1.5 V

Answer 1

Answer: (C)

1 V

Answer 2

Total current drawn from the battery $i=\frac{E}{R+r}=\frac{6}{100+0}=0.06\,A$ Resistance of $50\, cm$ wire is $R=\frac{\rho l}{A}=\left( \frac{\rho }{A} \right)l$ $=\left( \frac{R}{l} \right)l$ $\left( \because R=\frac{\rho l}{A} \right)$ $=\frac{100}{300}\times 50$ So, $R=\frac{50}{3}\Omega$ Hence, the potential difference between two points on the wire separated by a distance $l$ is $V=iR=0.06\times \frac{50}{3}=1\,V$