Q.A galvanometer of resistance $10\ \Omega$ that gives maximum (full-scale) deflection for a current of $1\ \text{mA}$ is to be converted into a multirange voltmeter reading $2\ \text{V}$, $20\ \text{V}$ and $200\ \text{V}$. Three resistors $R_1$, $R_2$ and $R_3$ are joined in series with the galvanometer, one after another. The $2\ \text{V}$ terminal is tapped just after $R_1$, the $20\ \text{V}$ terminal after the series pair $R_1+R_2$, and the $200\ \text{V}$ terminal after $R_1+R_2+R_3$; each range terminal together with the common galvanometer terminal forms the two leads of the voltmeter for that range. Find $R_1$, $R_2$ and $R_3$.

Galvanometer to Voltmeter Conversion

A moving-coil galvanometer deflects fully only at a small current $I_g$ (its full-scale deflection current) and has internal resistance $G$. To use it as a voltmeter — reading much larger voltages $V$ — a large resistance $R_s$ (the multiplier) is connected in series with the galvanometer coil. This series resistance limits the current to exactly $I_g$ when the full-scale voltage $V$ is applied, so the needle deflects fully and the scale is recalibrated to read volts instead of amperes.

The Conversion Formula

The galvanometer and $R_s$ together form a series circuit of total resistance $R_s + G$. Applying $V$ across this series combination drives a current

We want this current to equal $I_g$ exactly when $V$ is the maximum (full-scale) voltage, so $I_g = \dfrac{V}{R_s+G}$, which rearranges to:

Since the coil's deflection is proportional to the current through it, and that current is proportional to the applied voltage (Ohm's law), the resulting scale is linear in $V$: equal voltage steps give equal angular deflections.

Why series, not parallel?

A voltmeter must be connected across the component whose voltage is being measured, without diverting current away from it — so it should draw as little current as possible, meaning its own resistance must be as large as possible. Adding $R_s$ in series does exactly that: it raises the meter's total resistance to $R_s + G$, which is deliberately made large. (This is the opposite requirement to an ammeter, which sits in the current path and needs the smallest possible resistance — achieved there with a small shunt in parallel, not a large resistor in series.)

Worked Example

For $I_g = 1\ \text{mA}$, $G = 50\ \Omega$, converting to a $0$–$10\ \text{V}$ voltmeter:

A $9.95\ \text{k}\Omega$ resistor in series gives full-scale deflection at exactly $10\ \text{V}$.

Key Properties

• High input resistance: $R_v = R_s + G$ is large (kΩ to MΩ), so the voltmeter draws minimal current and barely disturbs the circuit it measures.
• Linear scale: deflection ∝ current ∝ voltage. …