Q. An ideal battery of $4\, V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1\, m$ and resistance $5\Omega$. The value of $R$, to give a potential difference of $5 \,mV$ across $10\, cm$ of potentiometer wire, is :
Solution:
Let current flowing in the wire is $i$.
$\therefore i = \left(\frac{4}{R +5}\right)A $
If resistance of 10 m length of wire is x
then $ x = 0.5 \Omega = 5 \times\frac{0.1}{1} \Omega $
$ \therefore \Delta V = P $ d. on wire = i. x
$ 5 \times10^{-3} = \left(\frac{4}{R +5}\right).\left(0.5\right) $
$\therefore \frac{4}{R+5} = 10^{-2} R +5 = 400 \Omega $
$ \therefore R = 395 \Omega $
