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

Non-Ideal Battery

The heat deve...

Question

The heat developed in a resistor $R_{1}$ connected to a cell is equal to the heat developed when it is connected to another resistance, $R_{2}$ replacing $R_{1}$ for the same time. The internal resistance of the cell will be

\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 (R_{1} + R_{2})}

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

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

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

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

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

Open in App

Solution

The correct option is C $\sqrt{R_{1} R_{2}}$
Initially $R_{1}$ connected to the cell and the heat developed in a resistor $R_{1}$ hence, current through resistor $R_{1}$ is $I_{1} = \frac{E}{R_{1} + r}$
then $R_{2}$ connected to the cell and the heat developed in a resistor $R_{2}$ hence, current through resistor $R_{2}$ is
$I_{2} = \frac{E}{R_{2} + r}$
Given that the heat developed in a resistor $R_{1}$ = the heat developed in a resistor $R_{2}$
$(I_{1})^{2} R_{1} = (I_{2})^{2} R_{2}$
${(\frac{E}{R_{1} + r})}^{2} R_{1} = {(\frac{E}{R_{2} + r})}^{2} R_{2}$
${(\frac{R_{1}}{R_{1} + r})}^{2} = {(\frac{R_{2}}{R_{2} + r})}^{2}$
$R_{1} (R_{2} + r)^{2} = R_{2} (R_{1} + r)^{2}$
$R_{1} ((R_{2})^{2} + 2 r R_{2} + r^{2}) = R_{2} ((R_{1})^{2} + 2 r R_{1} + r^{2})$
$R_{1} R_{2} (R_{1} + R_{2}) = r^{2} (R_{1} + R_{2})$
$r^{2} = R_{1} R_{2}$
$r = \sqrt{R_{1} R_{2}}$

Suggest Corrections

Similar questions

R_{1}

R_{2}

R_{p}

R_{1} + R_{2}

R_{1} - R_{2}

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

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

R_{1}

R_{1}

R_{2}

R_{2}

r_{1}

r_{2}

R_{1} = 400 Ω

R_{2} = 20 Ω

R_{1}

R_{2}

When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

Join BYJU'S Learning Program