Q.Discuss the conversion of a galvanometer into an ammeter and also a voltmeter.

Step 1. Ammeter: a shunt resistance $S$ is connected in parallel with the galvanometer, so most of the circuit current $I-I_g$ bypasses the coil (which carries only $I_g$, the full-scale deflection current). Equating the potential difference across the galvanometer and the shunt, $I_gR_g=(I-I_g)S$, gives $I=I_g\left(1+\dfrac{R_g}{S}\right)$. The combined (ammeter) resistance is $R_a=SR_g/(S+R_g)$, small (ideal: 0); to extend the range $n$ times, $S=R_g/(n-1)$.

Step 2. Voltmeter: a high resistance $R_h$ is connected in series, so the same current $I_g$ flows through both; $V=I_g(R_g+R_h)$. The combined (voltmeter) resistance is $R_v=R_g+R_h$, large (ideal: infinite); to extend the range $n$ times, $R_h=(n-1)R_g$. …

Derive both conversions from the current/potential-difference matching …

• Swapping the series/parallel placement between the ammeter …