Today, I read in my textbook that in order to convert a Galvanometer (more precisely a Moving Coil Galvanometer) into an Ammeter we need to attach a small Resistor with it in parallel (of Resistance of the order of 10^-2 ohm or less) for obvious reasons. I understood completely that why we need to do that. So there's no problem with that.

The thing which I'm not getting is how the original scale of the Galvanometer is modified (or as they say "Calibrated" ) in order to use it as a current meter? Kindly explain this with an example.

I'm attaching some screenshots of my book in which this line is highlighted in green

Theory part The problem

PS: I'm just a high school student so pardon me if you find any mishaps in my description.


1 Answer 1


A galvanometer is an ammeter which is designed to detect and measure small currents.
Like any other ammeter to extend the range of a galvanometer a shunt resistor is connected in parallel with the galvanometer.

Since the potential difference across the galvanometer and the shunt is the same and using the symbols defined in the text

$I_{\rm G}R_{\rm G} = I_{\rm s}r_{\rm s}$

where $I_{\rm G}$ and $I_{\rm s}$ are the currents through the galvanometer and shunt.

The current $I$ that you wish to measure is the sum of the currents through the galvanometer and the shunt resistor.

$I = I_{\rm G}+I_{\rm s}$

Combining these two equations gives $I = I_{\rm G} \left (1+ \dfrac {R_{\rm G}}{r_{\rm s}} \right)$.

If $R_{\rm G}$ and $r_{\rm s}$ are known and the current through the galvanometer $I_{\rm G}$ is read off its scale the current $I$ can be found.

Another way would be to recalibrate the galvanometer scale so that one could directly read the current $I$ off the scale.

Another method of calibrating the galvanometer scale would be to pass known currents measured using an already calibrated ammeter through the galvanometer and shunt resistor arrangement.


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