A cell has internal resistance $r$ . It sends current through two resistances $R_{1}$ and $R_{2}$ separately. then, the heat produced in resistance $R_{1}$ and $R_{2}$ will be in the ratio

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

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

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\frac{R_{1}}{R_{2}} {(\frac{R_{2} + r}{R_{1} + r})}^{2}

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

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Solution

The correct option is D $\frac{R_{1}}{R_{2}} {(\frac{R_{2} + r}{R_{1} + r})}^{2}$
Let, E be the emf of the cell, then currents through two resistances $R_{1}$ and $R_{2}$ are
$I_{1} = \frac{E}{R_{1} + r}$ and $I_{2} = \frac{E}{R_{2} + r}$
The heat produced in resistor $R_{1}$ is
$H_{1} = (I_{1})^{2} R_{1}$
$H_{1} = {(\frac{E}{R_{1} + r})}^{2} R_{1}$ .........................................(1)
similarly, the heat produced in resistor $R_{2}$ is
$H_{2} = (I_{2})^{2} R_{2}$
$H_{2} = {(\frac{E}{R_{2} + r})}^{2} R_{2}$ ............................................(2)
Hence, from (1) and (2) we get
$\frac{H_{1}}{H_{2}} = \frac{{(\frac{E}{R_{1} + r})}^{2} R_{1}}{{(\frac{E}{R_{2} + r})}^{2} R_{2}}$
$\frac{H_{1}}{H_{2}} = \frac{{(\frac{1}{R_{1} + r})}^{2} R_{1}}{{(\frac{1}{R_{2} + r})}^{2} R_{2}}$
$\frac{H_{1}}{H_{2}} = \frac{(R_{2} + r)^{2} R_{1}}{(R_{1} + r)^{2} R_{2}}$

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A cell has internal resistance r. It sends current through two resistances R1 and R2 separately. then, the heat produced in resistance R1 and R2 will be in the ratio

A cell has internal resistance $r$ . It sends current through two resistances $R_{1}$ and $R_{2}$ separately. then, the heat produced in resistance $R_{1}$ and $R_{2}$ will be in the ratio