Question

A wire is bent in the front of circle of radius $2m$. Resistance per unit length of wire is $1/\pi \Omega /m$. Battery of $6V$ is connected between $A$ and $B$. $\angle AOB={90}^o$. Find the current through the battery

• A. $4A$
• B. $8A$
• C. $3A$
• D. $9A$

Answer 1

Answer: (B)

$8A$

Resistance of complete wire $=2\pi r\times \frac{1}{\pi }$
$=2r=2\times 2=4\Omega$
$\therefore$ Resistance of each $\frac{1}{4}$ part $=\frac{1}{4}\times 4$
$=1\Omega$
Hence, equivalent resistance
$R_{AB}=\frac{3}{4}\times 1=\frac{3}{4}\Omega$
The current through the battery
$=\frac{V}{R_{AB}}=\frac{6}{3/4}=\frac{6\times 4}{3}=8A$