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 $R=15\,\Omega$
$R_{eq}=\frac{15\times\frac{\pi}{4}\times\left(2\pi-\frac{\pi}{4}\right)}{4\pi^{2}}$
$=\frac{\frac{15\pi }{4}\left(\frac{7\pi}{4}\right)}{4\pi^{2}}$
$=1.64\,\Omega$