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When two identical batteries of internal resistance $1\Omega $ each are connected in series across a resistor R, the rate of heat produced in R is ${{J}_{1}}$. When the same batteries are connected in parallel across R, the rate is ${{J}_{2}}$. If ${{J}_{1}}=2.25{{J}_{2}}$ then the value of R in $\Omega $ is,

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Last updated date: 25th Jul 2024
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Answer
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Hint: When we connect two identical batteries in series the effective resistance in series combination will be the sum of those two internal resistances and in while the batteries are connected in parallel the reciprocal of those two internal resistances are added together to find the effective internal resistances. Then by calculating ${{I}_{1}}$and ${{I}_{2}}$ , we can find the heat in those two cases. Hence by substituting in the given equation we will get the value of R.

Complete step by step answer:
Given that the internal resistance $1\Omega $ is connected in series with a resistor R.
Then the total resistance in series combination becomes 2r+R.
Hence the current flow through ${{I}_{1}}$ is given by,
Current ${{I}_{1}}=\left[ \frac{2E}{2r+R} \right]$
Heat in the first case can be calculated by using the equation,
${{J}_{1}}=I_{1}^{2}Rt$
$\Rightarrow {{J}_{1}}={{\left( \frac{2E}{2r+R} \right)}^{2}}Rt$
Similarly we can calculate the current ${{I}_{2}}$. Here the same battery is connected in parallel across R. Thus,
${{I}_{2}}=\left[ \frac{E}{\frac{r}{2}+R} \right]$
Hence heat produced in the second case,
${{J}_{2}}=I_{2}^{2}Rt$
$\Rightarrow {{J}_{2}}={{\left( \frac{E}{\frac{r}{2}+R} \right)}^{2}}Rt$
Given that,
${{J}_{1}}=2.25{{J}_{2}}$
Substituting the values of ${{J}_{1}}$and ${{J}_{2}}$ in the above equation we get, ${{\left( \frac{2E}{2r+R} \right)}^{2}}Rt=2.25{{\left( \frac{E}{\frac{r}{2}+R} \right)}^{2}}Rt$
${{\left( \frac{2E}{2r+R} \right)}^{2}}Rt=2.25{{\left( \frac{2E}{r+2R} \right)}^{2}}Rt$
Rearranging the equation and cancelling the common terms we get,
${{\left( r+2R \right)}^{2}}=2.25{{\left( 2r+R \right)}^{2}}$
${{\left( r+2R \right)}^{2}}=\frac{9}{4}\times {{\left( 2r+R \right)}^{2}}$
$4{{\left( r+2R \right)}^{2}}=9{{\left( 2r+R \right)}^{2}}$
Taking the square root the above equation becomes,
$2\left( r+2R \right)=3\left( 2r+R \right)$
$R=4r$
Given that,
$r=1\Omega $
Then the value of R in $\Omega $ is,
$R=4\Omega $

Note: In series the effective resistance in series combination will be the sum of those two resistances and parallel the reciprocal of those two internal resistances are added together to find the effective internal resistances.