Question

A $6V$ 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 $50cm$ will be

• A. 2 V
• B. 3 V
• C. 1 V
• D. 1.5 V

Answer 1

Answer: (C)

Total current drawn from the battery
$i=\frac{E}{R+r}=\frac{6}{100+0}=0.06A$
Resistance of $50cm$ wire is $R^{\prime }=\frac{\rho l^{\prime }}{A}=\left(\frac{\rho }{A}\right)l^{\prime }$
$=\left(\frac{R}{l}\right)l^{\prime }\left(\because R=\frac{\rho l}{A}\right)$
$=\frac{100}{300}\times 50$
So, $R^{\prime }=\frac{50}{30}\Omega$
Hence, the potential between two point on the wire spareted by a distance $l$ is
$V=iR=0.06\times \frac{50}{3}=1V$