Question

$A$ and $B$ are two points on a uniform ring of resistance $15\Omega$. The $\angle AOB=45^{\circ }$. The equivalent resistance between $A$ and $B$ is

• A. $1.64\Omega$
• B. $2.84\Omega$
• C. $4.57\Omega$
• D. $2.64\Omega$

Answer 1

Answer: (A)

Resistance per unit length of ring, $\rho =\frac{R}{2\pi r}$
Length of sections $ADB$ and $ACB$ are $r\theta$ and $r\left(2\pi -\theta \right)$
$\therefore$ Resistance of section $ADB$,
$R_1=\rho r\theta =\frac{R}{2\pi r}r\theta =\frac{R\theta }{2\pi }$
and resistance of section $ACB$,
$R_2=\rho r\left(2\pi -\theta \right)$
$=\frac{R}{2\pi r}r\left(2\pi -\theta \right)=\frac{R\left(2\pi -\theta \right)}{2\pi }$
Now, $R$ and $R_2$ are connected in parallel between $A$ and $B$ then
$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{\frac{R\theta }{2\pi }\times \frac{R\left(2\pi -\theta \right)}{2\pi }}{\frac{R\theta }{2\pi }+\frac{R\left(2\pi -\theta \right)}{2\pi }}$
$=\frac{R\theta \left(2\pi -\theta \right)}{4{\pi }^2}$
Putting $\theta =45^{\circ }=\frac{\pi }{4}rad$ and $R=15\Omega$
$R_{eq}=\frac{15\times \frac{\pi }{4}\times \left(2\pi -\frac{\pi }{4}\right)}{4{\pi }^2}$
$=\frac{\frac{15\pi }{4}\left(\frac{7\pi }{4}\right)}{4{\pi }^2}$
$=1.64\Omega$