Galvanometer conversion

A moving-coil galvanometer is a delicate instrument: push more than a few milliamps through its coil and the pointer slams past full scale or the coil itself burns out. Yet the very same galvanometer, with one extra resistor placed correctly, becomes a working ammeter reading amperes or a working voltmeter reading volts — instruments an engineer can trust on a real circuit board. The trick is always the same two-parameter story: a galvanometer is fully described by its coil resistance $G$ and its full-scale current $I_g$ . Add a small resistance in parallel and it becomes a low-resistance ammeter; add a large resistance in series and it becomes a high-resistance voltmeter. This article derives both conversion formulas, works fully through the two conversions of a single sample galvanometer, and pins down the reasoning examiners look for when they ask "why must an ammeter have low resistance?"

The core insight

A galvanometer becomes an ammeter by shunting most of the current around it through a parallel low resistance $S = I_gG/(I-I_g)$ , and becomes a voltmeter by blocking most of the voltage in front of it through a series high resistance $R = V/I_g - G$ .

The core idea

Every galvanometer carries two fixed numbers on its spec sheet: the coil resistance $G$ (also written $R_g$ ) and the full-scale deflection current $I_g$ , the largest current the coil can safely carry before the pointer is driven off the end of the scale. Conversion means adding one resistor so that the combination reads a chosen larger range of current or voltage while the coil itself never sees more than $I_g$ .

Ammeter: shunt in parallel. To read currents up to a larger value $I$ , connect a small resistance $S$ — the shunt — in parallel with the galvanometer. The two branches share the same potential difference, so when the total current is exactly $I$ , the coil carries $I_g$ and the shunt carries the remainder $I - I_g$ :

I_g G = (I - I_g)S \quad\Rightarrow\quad S = \frac{I_g G}{I - I_g}

where: $I_g$ is the full-scale current (A), $G$ is the coil resistance ( $\Omega$ ), $I$ is the new ammeter range (A), and $S$ is the shunt resistance ( $\Omega$ ) needed.

Voltmeter: high resistance in series. To read voltages up to a larger value $V$ , connect a large resistance $R$ in series with the galvanometer. The series combination must draw exactly $I_g$ when the voltage across it reaches $V$ :

V = I_g(G + R) \quad\Rightarrow\quad R = \frac{V}{I_g} - G

where: $V$ is the new voltmeter range (volts), $I_g$ and $G$ are the same coil constants, and $R$ is the series resistance ( $\Omega$ ) needed.

Why the placement matters. A shunt in parallel gives the combined ammeter a very low net resistance — lower than either $G$ or $S$ alone — because it opens an easy bypass path. That is exactly what an ammeter needs: it sits in series in the main circuit, so if it added significant resistance it would throttle the very current it is trying to measure. A series resistor, by contrast, gives the voltmeter a very high net resistance, $G + R$ . That is what a voltmeter needs: it is connected in parallel across a component, so if it drew significant current it would steal current from that component and change the very voltage it is trying to measure. The two conversions are mirror images — parallel for low resistance, series for high resistance — and the placement is not a convention, it follows directly from Kirchhoff's laws applied to each configuration.

Figure

Two circuit sketches, one showing a shunt resistor in parallel with a galvanometer forming a low resistance ammeter, the other showing a resistor in series with a galvanometer forming a high resistance voltmeter

Derivation of the shunt resistance formula

Derivation AShunt resistance for converting a galvanometer to an ammeter

The goal is to find, from first principles using only the parallel-resistor rule that equal branches share equal voltage, exactly how small the shunt resistance $S$ must be so that a galvanometer of resistance $G$ and full-scale current $I_g$ reads correctly up to a larger total current $I$ , without ever letting more than $I_g$ pass through the delicate coil.

Step 1. Set up the circuit. The shunt $S$ is connected directly across the galvanometer's two terminals, so the pair forms a single parallel combination with two terminals of its own; the full current $I$ that we want to measure enters this combination and splits between the two branches.

Step 2. Apply the full-scale condition. At the top of the new ammeter's range, the coil must be carrying exactly its safe maximum, $I_g$ , no more. By Kirchhoff's current law the rest of the current, $I - I_g$ , must be flowing through the shunt.

Step 3. Use the parallel-voltage condition. Because the coil and the shunt are the same two nodes, the potential difference across them is identical: $V_G = V_S$ . Writing each with Ohm's law, $I_g G = (I - I_g)S$ .

Step 4. Solve for $S$ . Dividing both sides by $(I - I_g)$ gives $S = \dfrac{I_g G}{I - I_g}$ . Because $I \gg I_g$ for any useful ammeter, the denominator is close to $I$ and $S$ comes out much smaller than $G$ — confirming that the shunt is, as its name suggests, a small resistor that diverts nearly all of the current away from the coil. Check: as $I \to I_g$ (no real conversion needed), the denominator $\to 0$ and $S \to \infty$ — an infinite shunt is the same as no shunt at all, which is consistent, since then all the current is already allowed through the coil directly.

Worked example

A galvanometer has coil resistance $G = 100\ \Omega$ and full-scale deflection current $I_g = 1\ \text{mA}$ . Convert it (a) into an ammeter reading up to $I = 1\ \text{A}$ , and (b) into a voltmeter reading up to $V = 10\ \text{V}$ .

Step 1 — Convert units. $I_g = 1\times10^{-3}\ \text{A}$ , $I = 1\ \text{A}$ , $V = 10\ \text{V}$ , $G = 100\ \Omega$ . Keeping everything in SI base units (amps, volts, ohms) avoids the commonest slip in this calculation.

Step 2 — Shunt resistance. Using $S = \dfrac{I_g G}{I - I_g}$ :

S = \frac{(1\times10^{-3})(100)}{1 - 1\times10^{-3}} = \frac{0.1}{0.999} \approx 0.1001\ \Omega

A shunt of about $0.1\ \Omega$ in parallel gives a $0$ – $1\ \text{A}$ ammeter. Almost all the current ( $0.999\ \text{A}$ ) takes the shunt; only $1\ \text{mA}$ passes through the coil, exactly its safe limit.

Step 3 — Series resistance. Using $R = \dfrac{V}{I_g} - G$ :

R = \frac{10}{1\times10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega

A series resistance of $9900\ \Omega$ (about $9.9\ \text{k}\Omega$ ) gives a $0$ – $10\ \text{V}$ voltmeter. The total voltmeter resistance is $G + R = 10{,}000\ \Omega$ , one hundred times the bare coil resistance — the hallmark of a good voltmeter.

Step 4 — Compare the two conversions. The ammeter's net resistance (galvanometer parallel with $0.1001\ \Omega$ ) works out to roughly $0.1\ \Omega$ , far below $G$ ; the voltmeter's net resistance is $10{,}000\ \Omega$ , far above $G$ . The shunt required is about $0.1\ \Omega$ for the ammeter, and the series resistor required is $9900\ \Omega$ for the voltmeter — the same coil, pushed to opposite resistance extremes by the two different circuit placements.

Common traps

Wiring the shunt in series or the series resistor in parallel — this reverses the whole physics; a shunt must divert current around the coil (needs parallel), a series resistor must drop voltage ahead of the coil (needs series).
Using $I$ instead of $I - I_g$ in the shunt-formula denominator — the shunt only has to carry the excess current beyond what the coil itself draws, not the full current $I$ .
Assuming a bigger shunt makes a "better" ammeter — the opposite is true: a good ammeter needs the smallest possible net resistance so it does not choke the current it is inserted to measure.
Assuming a smaller series resistance makes a "better" voltmeter — the opposite is true: a good voltmeter needs the largest possible net resistance so it draws negligible current from the component it is measuring.

Ask the tutor: Want to try converting this same galvanometer to a different ammeter or voltmeter range, or see why the shunt and series formulas fall straight out of Kirchhoff's laws?

Why does the shunt formula use I minus Ig instead of just I in the denominator?

Because $I$ is the total current entering the ammeter, but the coil itself is already carrying $I_g$ of that current when the meter reads full scale. Kirchhoff's current law says the shunt only has to handle whatever is left over, $I - I_g$ . Writing $I_gG = IS$ instead would overstate how much current the shunt must carry, giving a shunt value that is slightly too small and would let the coil see more than $I_g$ at full-scale reading — risking damage.

Can the same galvanometer be converted into more than one ammeter or voltmeter range?

Yes — $G$ and $I_g$ are fixed properties of the coil, but you can choose a different shunt $S$ for each desired ammeter range $I$ , or a different series resistance $R$ for each desired voltmeter range $V$ . Multi-range meters actually do exactly this: a rotary switch selects between several pre-calculated shunt or series resistors, letting one galvanometer serve as an ammeter or voltmeter across several ranges without rebuilding the circuit.

Does the shunt or series resistor change the full-scale deflection current of the galvanometer itself?

No. $I_g$ is a fixed property of the coil — the maximum current that safely gives full-scale deflection — and it never changes. What the shunt or series resistor changes is how much of the external current or voltage corresponds to that same $I_g$ flowing through the coil. The conversion resistor is chosen precisely so that when the external ammeter or voltmeter reads its maximum ( $I$ or $V$ ), the coil is, not coincidentally, carrying exactly its own safe $I_g$ .

Galvanometer conversion

The core idea

Derivation of the shunt resistance formula

Worked example

Common traps

Before this

Where this leads