Question

A galvanometer has a capacity to carry a maximum current of $25\text{ }mA$ . How can it be used as an ammeter to read the current up to $0.1A$?

Answer 1

We can convert a galvanometer into an ammeter by connecting a shunt resistance parallel. This resistance increases the current range of the whole setup. A lot of current passes through the shunt. By using the formula for the distribution of current in two parallel branches, we can get the required equation for the resistance of the shunt.

Formula Used:
$V=IR$

Complete answer:
Given, Current range of galvanometer, ${{I}_{G}}=25mA$
Let the resistance of the galvanometer be $G$.
Now, according to ohm’s law, current passing through a resistor is related to voltage as,
$\begin{aligned} & V=IR \\\ & \Rightarrow {{V}_{G}}={{I}_{G}}G \\\ & \therefore G=\dfrac{{{V}_{G}}}{{{I}_{G}}} \\\ \end{aligned}$
Now, to convert a galvanometer into an ammeter by connecting a shunt resistance parallel. This resistance increases the current range of the whole setup. A lot of current passes through the shunt.

Hence, the shunt resistance connected parallel is $S$.
Now. The current entering the galvanometer branch is given by,
${{I}_{G}}=\left( \dfrac{S}{G+S} \right)I$
Where, $I$is the total current entering the ammeter setup.
Given in the question, $I=0.1A$
$\begin{aligned} & {{I}_{G}}=\left( \dfrac{S}{G+S} \right)I \\\ & \Rightarrow 25\times {{10}^{-3}}=\left( \dfrac{S}{G+S} \right)0.1 \\\ & \Rightarrow \left( \dfrac{S}{G+S} \right)=\dfrac{25\times {{10}^{-3}}}{0.1} \\\ & \Rightarrow S=\left( 0.25G+0.25S \right) \\\ & \Rightarrow S-0.25S=0.25G \\\ & \Rightarrow \dfrac{3}{4}S=\dfrac{3}{4}G \\\ & \therefore S=\dfrac{G}{3} \\\ \end{aligned}$
Hence, required shunt resistance will be $\dfrac{G}{3}\Omega$.

Note: Students must remember that to convert a galvanometer to ammeter, a very small shunt resistance parallel always. This resistance increases the current range. On the other hand, to convert an ammeter to a galvanometer, a large resistance has to be connected in series.