electromagnetism - 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?

Answer

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.

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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! ;)