Question

A galvanometer has a current range of 15mA and a voltage range of 750mV. To convert this galvanometer into an ammeter of range 25A, What is the required shunt?

Answer 1

Step1: Given data

It is given that the current range of the galvanometer is $I_G=15mA=15\times {10}^{-3}A(\because 1mA={10}^{-3}A)$
It is also given that the voltage range of the galvanometer is $V_G=750mV=750\times {10}^{-3}V(\because 1mV={10}^{-3}V)$

Step2: Formula used

$V=IR[whereV=voltage,I=current,R=resis\tan ce]$

Step3: Calculating the value of the required shunt

Now, according to Ohm’s law, the current $I$ passing through a resistor of resistance $R$ is related to the voltage V across it by $V=IR...........\left(i\right)$
Therefore, using (i), we get, $V_G=I_{G}G$
$\therefore G=\frac{V_G}{I_G}=\frac{750\times {10}^{-3}}{15\times {10}^3}=50\Omega$
Hence, the resistance of the galvanometer is $50\Omega$
Now, to convert it into an ammeter, a very small shunt resistance has to be connected in parallel. This increases the current range of this setup. This is because, in absence of the shunt, a maximum current of $IG$ could enter the galvanometer.

However, when a shunt is connected in parallel to form an ammeter setup, a large current will enter the setup, $IG$ current will flow in the branch of the galvanometer while the rest of the current can flow through the shunt.

Hence, let us connect a shunt of resistance $S$ in parallel.
Let the total current entering the ammeter setup be $I$

Now, the current entering the branch of the galvanometer is given by $I_G=\frac{S}{G+S}I(\because FromOhm’slaw,weseeI\propto \frac{1}{R})$
Now according to the question$I=25A$
Therefore putting this value, we get,
$$\begin{gathered} 15\times {10}^{-3}=\frac{S}{50+S}25 \\ 0.015(50+S)=25S \\ (0.015\times 50)+0.015S=25S \\ 0.75=25S-0.015S \\ 0.75=24.985S \\ S=\frac{0.75}{24.985} \\ S=0.03\Omega \end{gathered}$$
Hence, the required resistance of the shunt is $0.03\Omega$.