# Conversion of Moving coil galvanometer to ammeter

A galvanometer can be converted into an ammeter by connecting a low resistance (called shunt resistance) in parallel to the galvanometer.

Firstly, why do we need to connect the resistance?

If a resistance is connected in parallel, then some of the current will flow through resistance. So how can the current in the circuit be determined? Won't it be inaccurate, as some of the current flows through resistance?

First things first, the resistance is connected in the circuit parallel to the galvanometer to minimise the total resistance of the ammeter. Whenever 2 resistances are connected in parallel, the equivalent resistance is always less than the lowest resistance.

You know that:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

$$\implies R_{eq} = \frac{R_1R_2}{R_1 + R_2}$$

eg: Try plugging in values. If $R_1 = 5\Omega$ and $R_2 = 0.2\Omega$:
$$R = \frac{5\times0.2}{5 + 0.2} = \frac{1}{5.2} = 0.19$$ which is less than 0.2, thus reducing the equivalent resistance.

This is done so that the ammeter has the least resistance possible and it doesn't interfere with the potential drop across the circuit.

Now, the ammeter is callibrated in such a way that it gives the correct reading, even if the current through G isn't what it says.
It's like calculating your weight while holding a 20kg bag. You can still get the correct weight by shifting the scale of your weighing machine down by 20kg! ;)

• But why we need to connect resistance?? – vaibhav Feb 22 '14 at 8:52
• @vaibhav As I said, connecting a low-resistance brings down the total resistance of the ammeter, so that it doesn't interfere with the amount of current in the circuit. Ideally, an ammeter should have 0 resistance, but obviously, that isn't possible, so we try to bring the resistance as low as possible. – mikhailcazi Feb 22 '14 at 9:02
• @mikhailcazi Since the majority of current will flow through the resistance then how will the ammeter give full scale reading – Hydrous Caperilla Apr 16 '18 at 2:04

The galvanometer has it's own resistance $R$ (of course it will unless it is ideal). If entire current flows through it, it will have a lot of contribution to potential difference ($i*R$) across it and affect circuit working.
To decrease the potential difference across it a low resistance ($r << R$) is connected across it so that the potential difference becomes much less than it was previously ( $r*i_s << R*i$) and has a less influence of circuit working.