Multimeters and moving-iron instruments

Open an analogue multimeter and, more likely than not, you will find a single spring-restrained coil doing every job — reading current, voltage, and resistance depending on which resistor network the range switch has wired around it. Sitting alongside it in many panel meters is a different movement entirely: the moving-iron (MI) instrument, which reads current and voltage on both AC and DC because its deflecting torque depends on the square of the current, not its sign. This article picks up right after PMMC movements, shunts and multipliers to explain how a soft-iron vane converts current into torque, why that makes the scale non-uniform and RMS-reading, and how shunts, multipliers and a small internal battery let one sensitive movement masquerade as an ammeter, a voltmeter, and an ohmmeter in turn.

The core insight

A moving-iron instrument deflects in proportion to $I^2$ (so it works on AC and DC alike and reads RMS on a cramped, non-linear scale), while a multimeter is one such sensitive movement reconfigured by switched shunts, multipliers, and an internal battery into an ammeter, voltmeter, and ohmmeter.

The core idea

How the moving-iron movement produces torque. An MI instrument has a fixed coil that carries the current being measured (or a current proportional to the voltage under test) and one or two vanes of soft iron near it — one fixed, one free to move and coupled to the pointer. Current in the coil magnetises the vane(s): in the attraction type, a single moving vane is pulled into the coil's field; in the repulsion type, two similarly magnetised vanes push apart. Either way the force comes from the interaction of an induced magnetisation with the very field that induced it, so it is proportional to the square of that field, and hence to the square of the coil current:

T_d \propto I^2

where: $T_d$ is the deflecting torque and $I$ is the instantaneous coil current. Because $I^2$ is never negative, reversing the current direction does not reverse the torque — unlike a PMMC coil, which pins itself against the stop on reversed current, an MI vane is pulled in (or pushed apart) exactly the same way whichever half-cycle of an AC waveform is flowing.

Why the scale is non-uniform. A control spring supplies a restoring torque $T_c \propto \theta$ , and the pointer settles where $T_d = T_c$ . Since $T_d$ grows as the square of current while $T_c$ grows only linearly with angle, equal steps in current do not produce equal steps in deflection: the scale is cramped near zero and opens out at higher readings, in visible contrast to the uniform PMMC dial.

Why it reads RMS on AC. On AC the torque follows $i^2(t)$ instant by instant, but the pointer's inertia is far too sluggish to track that; it settles at the position set by the time-averaged torque over a cycle, which is proportional to the mean-square current. Since RMS is defined as the square root of the mean-square value, calibrating the dial in terms of $\sqrt{\overline{i^2}}$ makes the pointer read RMS current directly — and on DC, where $i$ is already constant, the "mean-square" is just $I^2$ , so the same calibration reads the true DC value.

Figure

Attraction type moving iron instrument with fixed coil soft iron vane pointer and control spring

How a multimeter reuses this movement. A multimeter wraps one sensitive movement — commonly a PMMC coil with a rectifier for AC ranges, or an MI coil for AC-only use — with a range switch bringing in different shunts for each current range, different multipliers (series resistors) for each voltage range, and an internal battery with a series resistor for the ohmmeter function, where deflection is calibrated against the unknown resistance across the test leads. A zero-ohms adjustment compensates for battery ageing.

Derivation AWhy a moving-iron ammeter's dial reads RMS current

The goal is to show, starting only from the torque law $T_d \propto I^2$ and the spring's linear restoring torque, that a moving-iron pointer driven by an alternating current settles at a deflection proportional to the mean-square value of that current, so a scale calibrated in the square root of that quantity reads the RMS current directly — the same reasoning that lets the same dial read a DC current correctly, since a steady current's mean square is just its own square.

Step 1. Write the instantaneous torque for a sinusoidal current $i(t) = I_m\sin(\omega t)$ flowing in the coil. Since $T_d \propto I^2$ at every instant, $T_d(t) = k\,i^2(t) = k\,I_m^2\sin^2(\omega t)$ for some constant $k$ fixed by the coil and vane geometry.

Step 2. Recognise that the pointer's mass and the damping in the movement give it far too much mechanical inertia to follow the rapid oscillation of $T_d(t)$ , which swings through a full cycle every $1/f$ seconds — typically $1/50$ or $1/60$ of a second — while the pointer needs many such cycles to settle. The pointer therefore responds only to the time-average of $T_d(t)$ over one cycle, not its instantaneous value.

Step 3. Average $\sin^2(\omega t)$ over a full cycle. A standard identity gives $\overline{\sin^2(\omega t)} = \tfrac12$ , so the average deflecting torque is $\overline{T_d} = k I_m^2 \times \tfrac12 = k\left(\dfrac{I_m}{\sqrt2}\right)^2$ . Since the RMS value of a sinusoid is by definition $I_{rms} = I_m/\sqrt2$ , this is exactly $\overline{T_d} = k\,I_{rms}^2$ .

Step 4. Balance against the spring. At the settled deflection $\theta$ , the control spring's restoring torque $K\theta$ equals the average deflecting torque, so $K\theta = k\,I_{rms}^2$ , giving

\theta = \frac{k}{K}\,I_{rms}^2

The pointer position depends only on $I_{rms}^2$ , whatever the waveform, so a dial marked in $I_{rms}$ reads the true RMS value directly — and for steady DC, where $i(t)$ never varies, the "average of the square" is simply $I^2$ , so the same calibration reads the true DC current too. Verified: $\overline{\sin^2(\omega t)} = 1/2$ over one full period, so $\overline{T_d} = k I_m^2/2 = k(I_m/\sqrt2)^2 = kI_{rms}^2$ , matching the RMS definition used throughout.

Worked example

A moving-iron ammeter marked to read RMS current directly is connected in a circuit carrying a sinusoidal current $i(t) = 10\sin(\omega t)$ A. Find the meter reading, then find what it would read if the same peak-value waveform were instead a steady DC current of 10 A.

Step 1. Identify the peak value of the given sinusoid: $I_m = 10$ A.

Step 2. Convert to RMS using $I_{rms} = I_m/\sqrt2$ from the derivation above: $I_{rms} = 10/\sqrt2 = 7.07$ A. This is what the meter shows on the AC waveform — not the peak value of 10 A.

Step 3. For the DC case, the current never varies, so its "mean square" is just $10^2$ and its RMS-equivalent value is exactly 10 A; the same $T_d \propto I^2$ torque law and the same calibrated dial read this correctly. The meter reads 7.07 A on the given AC sinusoid and 10 A on the equal-peak-value DC current — the same instrument, same scale, two different numbers, because RMS and peak are not the same thing.

Competitive stretch

A multimeter's DC voltmeter section uses a PMMC movement with full-scale coil current $I_m = 50\ \mu\text{A}$ and coil resistance $R_m = 2\ \text{k}\Omega$ . Find the multiplier resistance for a 0–300 V range and the meter's sensitivity in ohms per volt.

Step 1. Apply the multiplier relation carried over from the PMMC/shunt topic, $R_{se} = V/I_m - R_m$ : $R_{se} = \dfrac{300}{50\times10^{-6}} - 2000 = 6{,}000{,}000 - 2000 = 5{,}998{,}000\ \Omega \approx 6\ \text{M}\Omega$ .

Step 2. Compute sensitivity as $1/I_m = 1/(50\times10^{-6}) = 20{,}000\ \Omega/\text{V}$ , the "20 k $\Omega$ /V" figure printed on analogue multimeters. On the 300 V range this gives a total input resistance of $20{,}000\times300 = 6\ \text{M}\Omega$ , matching $R_m + R_{se}$ found above — a higher $\Omega/\text{V}$ rating means the voltmeter loads the circuit under test less.

Common traps

Assuming an MI instrument has a linear scale like a PMMC movement — the $T_d \propto I^2$ law forces a compressed, non-uniform scale, cramped near zero.
Reading a sinusoid's peak value as the meter's displayed value — MI (and rectifier-based) AC scales are calibrated in RMS, so a 10 A peak sine reads 7.07 A, not 10 A.
Treating the ohmmeter range as working like the voltmeter or ammeter ranges — it uses its own internal battery rather than the circuit under test, and typically has a non-linear scale valid only within its calibrated span.
Believing MI instruments need a known current direction — since torque depends on $I^2$ , polarity is irrelevant, unlike a PMMC movement, whose pointer jams against the stop on reversed current.

Ask the tutor: want to see why the moving-iron scale looks the way it does on a real dial, or work through another multimeter range-design problem?

Why does a moving-iron instrument work on both AC and DC while a PMMC instrument does not?

A PMMC movement's torque is proportional to current directly ( $T_d \propto I$ ), so it reverses direction whenever the current reverses — fine for steady DC, but on AC the pointer cannot follow the rapid reversals and jams against a stop or reads nothing useful. A moving-iron instrument's torque instead depends on the square of the current, $T_d \propto I^2$ , which stays positive whichever way the current flows. That single difference in the force law is why MI instruments read correctly on AC and DC alike, while PMMC instruments need a rectifier bolted on for AC use.

Why is the moving-iron scale cramped at the low end instead of being evenly spaced like a PMMC scale?

The pointer settles where the spring's restoring torque, growing linearly with angle ( $T_c \propto \theta$ ), balances the deflecting torque, growing with the square of the current ( $T_d \propto I^2$ ). Near zero current, small increases in current produce only tiny increases in $I^2$ , so the pointer barely moves and the low end looks compressed; at higher currents the same-sized increase produces a much larger jump in $I^2$ , so the scale opens out. This crowding near zero is a direct fingerprint of the $T_d \propto I^2$ law.

If I turn a multimeter to its ohmmeter setting and accidentally connect it across a live voltage source, what happens conceptually?

The ohmmeter range assumes no external source across the test leads — its own internal battery is meant to be the only source of current through the unknown resistance, and the scale is calibrated on that assumption. Connecting it across a live circuit adds an unexpected external current, so the deflection no longer corresponds to any real resistance and the reading is meaningless. It can also push far more current through the delicate movement than it is rated for, which is why "never use the ohms range on a live circuit" is standard practice.

Multimeters and moving-iron instruments

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads