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 (in ohms) should the series resistance have so that the current in the circuit is $0.44$? (round off the value to the nearest integer)

Answer 1

Voltage drop across the Rod$\Rightarrow I=0.2(V{)}^{\frac{5}{2}}$

$0.44=0.2(V{)}^{\frac{5}{2}}$

$2.2=(V{)}^{\frac{5}{2}}$

$V=\left({2.2}^{\frac{2}{5}}\right)$

$logV=\frac{2}{5}\times log\left(2.2\right)$

$logV=0.4\times 0.34=0.14V\Rightarrow V=1.15volt$

Now voltage at point X$\Rightarrow 6-1.15=4.85volt$

apply V=IR on the resistance $R\Rightarrow (4.85-0)=0.44R\Rightarrow R=11.02\Omega$

Nearest intrger value=$11\Omega$