Explain the solution to the difficulties that arise when using a galvanometer directly as an ammeter.

(N/A) Using a galvanometer directly as an ammeter presents two main difficulties: $(1)$ The galvanometer has a high resistance,which changes the total resistance of the circuit and thus alters the current being measured. $(2)$ The galvanometer is a sensitive device and can be damaged by high currents.
To overcome these difficulties,a small resistance,known as a shunt $(r_{s})$,is connected in parallel with the galvanometer.
$(i)$ Since the shunt resistance $(r_{s})$ is very small compared to the galvanometer resistance $(R_{G})$,most of the current passes through the shunt,protecting the galvanometer from damage.
$(ii)$ The equivalent resistance of the parallel combination is $R_{eq} = \frac{R_{G} r_{s}}{R_{G} + r_{s}}$. Since $R_{G} \gg r_{s}$,we can approximate $R_{G} + r_{s} \approx R_{G}$. Therefore,$R_{eq} \approx \frac{R_{G} r_{s}}{R_{G}} = r_{s}$.
Because $r_{s}$ is very low,the total resistance of the circuit remains nearly unchanged,ensuring that the original current in the circuit is not significantly altered by the insertion of the ammeter.