You visited us 1 times! Enjoying our articles? Unlock Full Access!

Resistance and Resistors

A cell of emf...

Question

A cell of emf $ε$ is connected across thick spherical shell of inner radius $R_{1}$ and outer radius $R_{2}$ as shown in figure. If resitivity of sphere is $ρ$ , current through the cell is :

\frac{π ε (R_{1} + R_{2})^{2}}{R_{2} - R_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 π ε R_{1} R_{2}}{ρ (R_{2} - R_{1})}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π ε (R_{1} + R_{2})^{2}}{(R_{2} - R_{1})^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π ε R_{1} R_{2}}{ρ (R_{2} - R_{1})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{4 π ε R_{1} R_{2}}{ρ (R_{2} - R_{1})}$

If we consider a spherical shell of of radius $r$ and thickness $d r$ , the resistance of the shell will be :
$d R = \frac{ρ d r}{4 π r^{2}}$

Therefore, the resistance of whole sphere is given by :
$\Rightarrow R = \int d R = \frac{ρ}{4 π} \int_{R_{1}}^{R_{2}} \frac{1}{r^{2}} d r$

$\Rightarrow R = \frac{ρ}{4 π} {[\frac{1}{r}]}_{R_{2}}^{R_{1}} = \frac{ρ}{4 π} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\Rightarrow i = \frac{ε}{R} = \frac{4 π ε R_{1} R_{2}}{ρ (R_{2} - R_{1})}$

Suggest Corrections

Similar questions

ε

R_{1}

R_{2}

ρ

t

ρ

R_{1}

R_{2}

V_{0}

I

I

R_{1}

R_{2}

(R)

r_{1}

r_{2}

K

Join BYJU'S Learning Program