You visited us 2 times! Enjoying our articles? Unlock Full Access!

Power Delivered and Heat Dissipated in a Circuit

A cell of con...

Question

A cell of constant emf first connected to a resistance $R_{1}$ and then connected to a resistance $R_{2}$ . If power delivered in both cases is same, then the internal resistance of the cell is

\sqrt{R_{1} R_{2}}

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

\sqrt{\frac{R_{1}}{R_{2}}}

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

\frac{R_{1} - R_{2}}{2}

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

\frac{R_{1} + R_{2}}{2}

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

Open in App

Solution

The correct option is A $\sqrt{R_{1} R_{2}}$

Given, same power is delivered for both load resistances $R_{1}$ and $R_{2}$

$∴ i_{1}^{2} R_{1} = i_{2}^{2} R_{2}$

$\Rightarrow {(\frac{E}{R_{1} + r})}^{2} R_{1} = {(\frac{E}{R_{2} + r})}^{2} R_{2}$

$R_{1} {(R_{2} + r)}^{2} = R_{2} {(R_{1} + r)}^{2}$

$R_{1} (R_{2}^{2} + r^{2} + 2 R_{2} r) = R_{2} (R_{1}^{2} + r^{2} + 2 R_{1} r)$

$R_{1} {R_{2}}^{2} + R_{1} r^{2} + 2 R_{1} R_{2} r = R_{2} {R_{1}}^{2} + R_{2} r^{2} + 2 R_{1} R_{2} r$

$R_{1} R_{2}^{2} - R_{2} R_{1}^{2} = r^{2} (R_{2} - R_{1})$

$R_{1} R_{2} (R_{2} - R_{1}) = r^{2} (R_{2} - R_{1})$

$\Rightarrow r^{2} = R_{1} R_{2}$

$∴ r = \sqrt{R_{1} R_{2}}$

Hence, option $(a)$ is correct.

Key concept
power = ${(current)}^{2} \times Resistance$

Suggest Corrections

Similar questions

E

R_{1}

t

H

R_{1}

R_{2}

t

R_{2}

4 H

E

r_{1}

r_{2}

R

R

r_{1}

r_{2}

r_{1} > r_{2}

n

(r)

(R)

R

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program