Question

A galvanometer of resistance $100\Omega$ which can measure a maximum current of $1$ mA is converted into an ohmmeter by connecting a battery of emf $1$V and a fixed resistance of $900\Omega$ in series. When an external resistance is measured the current reading is $0.1$mA. Calculate the value of the resistance.

Answer 1

Given: a galvanometer of resistance $100\Omega$ which can measure a maximum current of $1mA$ is converted into an ohmmeter by connecting a battery of emf $1V$ and a fixed resistance of $900\Omega$ in series. When an external resistance is measured the current reading is $0.1mA$.

To find the value of resistance

Solution:

According to the given criteria,

$I_g=1mA,V=1V,G=100\Omega ,R=900\Omega ,G+R=1000\Omega$

Hence,

$I=\frac{V}{R_{ex}}=\frac{1}{1000}={10}^{-3}A$

And, $I^{\prime }=\frac{V}{R_{ex}}⟹R_{ex}=\frac{V}{I^{\prime }}⟹\frac{1}{{10}^{-4}}⟹R_{ex}={10}^4\Omega$

Therefore, $G+R+R^{\prime }=10,000\Omega$

Therefore, $R^{\prime }=10000-1000=9000\Omega$

is the required value of the resistance.