Question

A and B are two points on a uniform ring of resistance 15$6.5 \times 10^{- 6}m^{2}V^{- 1}s^{- 1}$. The <AOB = $2.5 \times 10^{6}m^{2}V^{- 1}S^{- 1}$.The equivalent resistance between A and B is

• A. $v_{d} \propto E$
• B. $v_{d} \propto E^{2}$
• C. $v_{d} \propto \sqrt{E}$
• D. $v_{d} \propto \frac{1}{E}$

Answer 1

Answer: (A)

$v_{d} \propto E$

Answer 2

: Resistance per unit length of ring, $\rho = \frac{R}{2\pi r}$

Length of sections ADB and ACB are $r\theta$and

$r(2\pi - \theta)$

$\therefore$ 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, $R_{2} = \rho r(2\pi - \theta)$

Now, $R_{1}$and $R_{2}$are connected in parallel between A and B then

$R_{eq} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} = \frac{\frac{R\theta}{2\pi} \times \frac{R(2\pi - \theta)}{2\pi}}{\frac{R\theta}{R\pi} + \frac{R(2\pi - \theta)}{2\pi}} = \frac{\frac{R\theta \times R(2\pi - \theta)}{(2\pi)^{2}}}{\frac{2\theta + R(2\pi - \theta)}{2\pi}}$Putting $\theta = 45^{o} = \frac{\pi}{4}$ rad and $R = 15\Omega$

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

$= \frac{105}{64}\Omega = 1.64\Omega$

$R_{eq} = 5\Omega$