Question

The current $I$ through a rod of a certain metallic oxide is given by $I=0.2V^{5/2}$, where $V$ is the potential difference across it. The rod is connected in series with a resistance to a $6V$ battery of negligible internal resistance. What value should the series resistance have so that the power dissipated in the rod is twice that dissipated in the resistance?

Answer 1

$\text{Let the resistance is series be R. As they are in series the current}$

$\text{in the rod ans the resistance will be the same.}$
$\text{For the rod}I=0.2V^{5/2}\Rightarrow \text{resistance of rod}=r=\frac{V}{I}=\frac{1}{0.2V^{3/2}}=\frac{5}{V^{3/2}}$
$\text{For the power dissipated in the rod be twice that dissipated in the resistance,}$
$I^{2}r=2I^{2}R\Rightarrow r=2R\Rightarrow \frac{5}{V^{3/2}}=2R...........\left(1\right)$
$V=\text{voltage across the rod}=\frac{r}{r+R}6=\frac{\frac{5}{V^{3/2}}\times 6}{\frac{5}{V^{3/2}}+R}..................\left(2\right)$
$\text{Using (1) in (2) ,}$
${\left(\frac{5}{2R}\right)}^{2/3}=\frac{2R}{2R+R}6=\frac{2}{3}6=4$
${\left(\frac{5}{2R}\right)}^2=4^3\Rightarrow R=0.3125\Omega$