Question

When two identical batteries of intemal 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 $J_1=2.25J_2$ then the value of $R$ in $\Omega$ is :

Answer 1

Current $I_1=\left(\frac{2E}{2r+R}\right)$

Heat in situation 1$\Rightarrow J_1={I_1}^{2}Rt\Rightarrow J_1={\left(\frac{2E}{2r+R}\right)}^{2}Rt$

Similarly $I_2=\left(\frac{E}{\frac{r}{2}+R}\right)$

Heat in situation 2$\Rightarrow J_2={I_2}^{2}Rt\Rightarrow J_2={\left(\frac{E}{\frac{r}{2}+R}\right)}^{2}Rt$

$\Rightarrow J_1=2.25J_2$ (Given)

${\left(\frac{2E}{2r+R}\right)}^2=\frac{9}{4}\times {\left(\frac{2E}{r+2R}\right)}^2$

$4(r+2R{)}^2=9(2r+R{)}^2$

$2(r+2R)=3(2r+R)$

$2r+4R=6r+3R$

$R=4r=4\Omega$