question_answer

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 \[{{J}_{1}}\]. If \[{{J}_{1}}=2.25\,{{J}_{2}}\] then the value of R in \[\Omega \] is

A) \[4\,\Omega \]               

B)        \[6\,\Omega \]

C) \[0.5\,\,\Omega \]           

D)        \[8\,\,\Omega \]

Correct Answer: A

Solution :

Cells connected in series \[{{J}_{1}}={{I}^{2}}R={{\left( \frac{2E}{2r+R} \right)}^{2}}.\,R\]                               ? (1) Cells connected in parallel \[{{J}_{2}}={{I}^{2}}R={{\left( \frac{E}{R+\frac{r}{2}} \right)}^{2}}\times \,R\] Given \[{{J}_{1}}=2.25\,{{J}_{2}}\] \[\frac{{{(2E)}^{2}}}{(2r+R)}.\,R=2.25\frac{E}{{{\left( R+\frac{r}{2} \right)}^{2}}}.\,R\] \[\therefore \,\,\,\,\frac{4}{{{(2r+R)}^{2}}}\,\,=\,\,\frac{2.25}{{{\left( R+\frac{r}{2} \right)}^{2}}}\] \[\therefore \,\,\,4{{[R+0.5]}^{2}}=2.25{{[2+R]}^{2}}\,\,\,\,\,\,\,\,\,\,[\because \,\,\,r=1\,\Omega ]\] \[\therefore \,\,\,\,2\left( R+0.5 \right)= 1.5\left( 2+R \right)\] \[\therefore \,\,\,\,R=4\,\Omega \]