A cell of emf $E$ is connected to a resistance $R_{1}$ for time $t$ and the amount of heat generated in it is $H$ . If the resistance $R_{1}$ is replaced by another resistance $R_{2}$ and is connected to the cell for the same time $t$ , the amount of heat generated in $R_{2}$ is $4 H$ . Then internal resistance of the cell is:

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

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

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

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

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

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

\sqrt{R_{1} R_{2}} \times \frac{\sqrt{R_{2}} - \sqrt{R_{1}}}{\sqrt{R_{2}} + \sqrt{R_{1}}}

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

Open in App

Solution

The correct option is C $\sqrt{R_{1} R_{2}} \times \frac{\sqrt{R_{2}} - 2 \sqrt{R_{1}}}{2 \sqrt{R_{2}} - \sqrt{R_{1}}}$
Heat generated is given as, $H = I^{2} R \times t$
According to given condition:
$\Rightarrow I_{1}^{2} R_{1} \times t = H$ and $I_{2}^{2} R_{2} \times t = 4 H$
$\Rightarrow \frac{E^{2}}{{(R_{1} + r)}^{2}} R_{1} \times t = H$ and $\frac{E^{2}}{{(R_{2} + r)}^{2}} R_{2} \times t = 4 H$

$\Rightarrow \frac{R_{2}}{{(R_{2} + r)}^{2}} = 4 \frac{R_{1}}{{(R_{1} + r)}^{2}}$

$\Rightarrow \sqrt{(R_{2} + r)} (R_{1} + r) = 2 \sqrt{R_{1}} (R_{2} + r)$

On solving, $r = \sqrt{R_{1} R_{2}} \times \frac{\sqrt{R_{2}} - 2 \sqrt{R_{1}}}{2 \sqrt{R_{2}} - \sqrt{R_{1}}}$

Suggest Corrections

Similar questions

Q. Two cells of same emf are in series with an external resistance R. The internal resistance

r_{1}

r_{2}

. The p.d across the

1^{s t}

cell is found to be zero the value of R is

Q. For the following reactions, the correct reaction rates

r_{1}, r_{2}, r_{1}^{'}

r_{2}^{'}

is:

Q. The emf (V) and internal resistance (

r

) of a single battery which is equivalent to a parallel combination of two batteries of emf's

V_{1}

and

V_{2}

and internal resistance

r_{1}

and

r_{2}

, respectively, with polarities as shown in the figure are given by

\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{1} + r_{2}}, \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Two cells with the same emf E and different internal resistance $r_{1} a n d r_{2}$ , are connected in series to an external resistance R. The value of R so that the pd across the ﬁrst cell is zero, is:

Q. Two resistances

R_{1}

and

R_{2}

provide series to parallel equivalents as

\frac{n}{1}

. Then the correct relationship is

Join BYJU'S Learning Program

A cell of emf E is connected to a resistance R1 for time t and the amount of heat generated in it is H. If the resistance R1 is replaced by another resistance R2 and is connected to the cell for the same time t, the amount of heat generated in R2 is 4H. Then internal resistance of the cell is: