Galvanometer, ammeter and voltmeter

How does a meter know the current?

Look at the pointer of an old meter on a workshop bench. The needle swings further when more current flows. How does the current move the needle?

It uses the torque of the last lesson. A coil with a pointer sits between two magnets. The current in the coil gets a torque. A spring pulls back. The needle stops where the two torques are equal. The instrument is a moving-coil galvanometer.

A radial field

In a uniform field, the torque $NIAB\sin\theta$ changes as the coil turns. The scale would be uneven. Worse, the torque would die away as the coil swings to the field.

The fix is to shape the field. Curved pole pieces surround a fixed cylinder of soft iron. In the narrow gap, the field points along the radius, like the spokes of a wheel. This is a radial field.

The coil turns in the gap. Wherever it turns, its plane contains the field, so the normal is at $90^\circ$ to the field. The torque is always

A spring (or a pair of hair-springs) pulls back with a torque $k\phi$ when the coil has turned through an angle $\phi$. Here $k$ is the torsion constant of the spring, in N m/rad. The needle rests when the two are equal:

Look at the result. The deflection is in direct proportion to the current. That is why the scale of a galvanometer has even divisions. In the graph, the magnetic torque is a flat line, the spring torque is a rising line, and a bigger current moves the meeting point out in proportion.

Sensitivity

A galvanometer is judged by how far the needle moves for a small current.

• Current sensitivity: $\dfrac{\phi}{I} = \dfrac{NAB}{k}$. It is the deflection for each unit of current.
• Voltage sensitivity: $\dfrac{\phi}{V} = \dfrac{NAB}{kG}$, where $G$ is the resistance of the coil. It is the deflection for each unit of voltage.

To raise the current sensitivity, use more turns, a larger area, a stronger magnet and a weaker spring. But take care with the voltage sensitivity. Doubling $N$ doubles $NAB/k$. It also doubles the length of wire, so it doubles $G$. The voltage sensitivity then stays the same.

A worked example. A coil has $100$ turns of area $5.0\ \text{cm}^2$ in a radial field of $0.16$ T. The spring has $k = 1.6 \times 10^{-6}\ \text{N m/rad}$. Find the deflection for $60\ \mu\text{A}$.

1. The factor. $NAB = 100 \times 5.0 \times 10^{-4}\ \text{m}^2 \times 0.16\ \text{T} = 8.0 \times 10^{-3}\ \text{N m/A}$.
2. The sensitivity. $\dfrac{\phi}{I} = \dfrac{8.0 \times 10^{-3}}{1.6 \times 10^{-6}\ \text{N m/rad}} = 5000\ \text{rad/A}$.
3. The deflection. $\phi = 5000\ \text{rad/A} \times 60 \times 10^{-6}\ \text{A} = 0.30\ \text{rad}$, which is $17^\circ$.

Too delicate for the real world

A galvanometer is made to be sensitive. Its full-scale current $I_g$ is only a fraction of a milliampere or a few milliampere. Its coil has a resistance $G$ of tens of ohms. Send a whole ampere through it and the coil burns. A voltage of a few volt across it does the same.

So we cannot use it as it is to measure amperes or volts. We add a resistor to it. The resistor makes the meter fit the job.

An ammeter: add a shunt

An ammeter measures a current, so it is joined in series with the part of the circuit that carries that current. It must not change the current. So it needs a very small resistance.

We join a small resistor, the shunt $S$, in parallel with the coil. Most of the current goes through the shunt. The coil gets only $I_g$ when the circuit current is $I$, the range we want.

The coil and the shunt have the same voltage across them:

We want $S$, so

The ammeter is the coil and the shunt together, in parallel. At full scale the voltage across it is $I_gG$ and the current into it is $I$. So its resistance is $R_A = \dfrac{I_gG}{I}$, which is smaller than both $G$ and $S$.

A worked example. The coil has $G = 80\ \Omega$ and $I_g = 1.5$ mA. We want an ammeter for $0$ to $3.0$ A.

1. The shunt. $S = \dfrac{1.5 \times 10^{-3}\ \text{A} \times 80\ \Omega}{3.0\ \text{A} - 1.5 \times 10^{-3}\ \text{A}} = 0.0400\ \Omega = 40.0\ \text{m}\Omega$.
2. The meter. $R_A = \dfrac{I_gG}{I} = \dfrac{1.5 \times 10^{-3}\ \text{A} \times 80\ \Omega}{3.0\ \text{A}} = 40.0\ \text{m}\Omega$.

The scale is then marked again. The deflection that meant $1.5$ mA now means $3.0$ A.

Raise the range of the ammeter. Does the shunt get larger or smaller? Now raise the resistance of the coil. What happens to the shunt?

A voltmeter: add a resistance in series

A voltmeter measures a voltage, so it is joined in parallel across the part that is measured. It must not take much current from the circuit. So it needs a very large resistance.

We join a large resistor $R$ in series with the coil. At full scale, the voltage $V$ is across the resistor and the coil together:

We want $R$, so

The voltmeter is the coil and the resistor in series, so its resistance is $R_V = G + R = \dfrac{V}{I_g}$.

A worked example. The same coil, $G = 80\ \Omega$ and $I_g = 1.5$ mA, is to read $0$ to $9.0$ V.

1. The resistor. $R = \dfrac{9.0\ \text{V}}{1.5 \times 10^{-3}\ \text{A}} - 80\ \Omega = 6000\ \Omega - 80\ \Omega = 5920\ \Omega$.
2. The meter. $R_V = \dfrac{V}{I_g} = 6000\ \Omega$. Its rating is $\dfrac{1}{I_g} = 667\ \Omega$ for each volt of range. The higher the rating, the better the voltmeter.

Meters change the circuit

A meter is part of the circuit it measures, so it changes the circuit a little.

• An ammeter adds its resistance in series. The current falls a little. The reading is a little low.
• A voltmeter in parallel takes a current. The resistance of that part of the circuit falls, so the voltage across it falls. The reading is a little low.

An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance. No real meter is ideal. A real voltmeter should have a resistance far larger than the resistance it is joined across. The graph shows how the error falls as the resistance of the voltmeter grows.

Practise with new numbers

Try it in Python

The program uses one coil to make four ammeters and three voltmeters. It prints the shunt, the series resistor and the resistance of each meter.

Common mistakes

• Joining the shunt in series. A shunt is always in parallel with the coil.
• Using $I$ in place of $I - I_g$. The shunt carries what the coil does not take.
• Forgetting to subtract $G$. For a voltmeter, $R = V/I_g - G$.
• Joining an ammeter in parallel across a part of a circuit. Its tiny resistance takes a huge current.
• Giving the voltmeter a small resistance. A voltmeter wants a large resistance and an ammeter a small one.
• Mixing mA and A. Change to A before you divide.
• Thinking that more turns always give a better voltage sensitivity. The resistance of the coil grows with the turns too.

Remember these

• Radial field: the torque is $NIAB$ at every angle. Deflection: $\phi = \dfrac{NAB}{k}I$.
• Current sensitivity $\dfrac{NAB}{k}$. Voltage sensitivity $\dfrac{NAB}{kG}$.
• Ammeter: shunt in parallel, $S = \dfrac{I_gG}{I - I_g}$. Low resistance, joined in series.
• Voltmeter: resistance in series, $R = \dfrac{V}{I_g} - G$. High resistance, joined in parallel.
• An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A galvanometer has a full-scale current of $0.25$ mA and a coil resistance of $100\ \Omega$. What voltage is across the coil at full scale?

2. The same galvanometer is to read $0$ to $5.0$ V. What resistance must be joined in series?

3. Why must an ammeter have a small resistance and a voltmeter a large one?

4. A student joins an ammeter across a resistor by mistake, in parallel. What happens?