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

Heating Effect of Current

A battery of ...

Question

A battery of emf E and internal resistance r is connected to a resistor of resistance $r_{1}$ and Q joules of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance $r_{2}$ the same quantity of heat is produced in the same time t the value of r is

\frac{r_{1}^{2}}{r_{2}}

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

\frac{r_{2}^{2}}{r_{1}}

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

\frac{1}{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 D $\sqrt{r_{1} r_{2}}$
In the first case the current in the circuit is
$l_{1} = \frac{E}{r_{1} + r}$
$Q_{1} = l_{2} r_{1} t = (\frac{E}{r_{1} + r})^{2} \times r_{1} \times t$
In the second case
$Q_{2} = (\frac{E}{r_{1} + r})^{2} \times r_{2} \times t$
equating above equations we get
$\frac{r_{1}}{(r_{1} + r)^{2}} = \frac{r_{2}}{(r_{2} + r)^{2}}$
$r_{1} (r_{2} + r)^{2} = r_{2} (r_{1} + r)^{2}$
$r_{1} (r_{2}^{2} + 2 r r_{2} + r^{2}) = r_{2} (r_{1}^{2} + 2 r r_{1} + r^{2})$
$r^{2} (r_{1} - r_{2}) = r_{1} r_{2} (r_{1} - r_{2})$
$r = \sqrt{r_{1} r_{2}}$

Suggest Corrections

Similar questions

R_{1}

R_{2}

R_{2}

E

R_{1}

t

H

R_{1}

R_{2}

t

R_{2}

4 H

R_{1}

R_{2}

R_{1}

ε_{1}

ε_{2}

r_{1}

r_{2}

Join BYJU'S Learning Program