Moving coil galvanometer — conversion to ammeter and voltmeter

A galvanometer is a sensitive current detector. To convert it:

• To ammeter: Connect a low resistance (shunt $S$) in parallel — most current bypasses through the shunt.
• To voltmeter: Connect a high resistance (multiplier $R$) in series — the high resistance limits current to the galvanometer’s safe range.

We want the ammeter to read up to $I = 5$ A, but the galvanometer can handle only $I_g = 2$ mA $= 0.002$ A.

The remaining current $(I - I_g)$ must flow through the shunt. Since galvanometer and shunt are in parallel:

$S = \frac{I_g \cdot G}{I - I_g} = \frac{0.002 \times 50}{5 - 0.002} = \frac{0.1}{4.998} \approx \mathbf{0.02 \text{ }\Omega}$

This tiny resistance ensures almost all current flows through the shunt.

We want the voltmeter to read up to $V = 100$ V. At full deflection, current through the galvanometer is $I_g = 0.002$ A.

Total resistance needed: $R_{total} = V / I_g$

$R_{total} = \frac{100}{0.002} = 50,000 \text{ }\Omega$

The multiplier resistance $R$:

$R = R_{total} - G = 50,000 - 50 = \mathbf{49,950 \text{ }\Omega \approx 50 \text{ k}\Omega}$