PMMC instruments, shunts, and multipliers

A permanent-magnet moving-coil (PMMC) movement is the tiny, delicate mechanism hiding inside almost every analogue DC ammeter and voltmeter you have ever seen — a light coil of wire, suspended in a magnet's field, that swings a pointer in exact proportion to the current through it. Left alone, that coil can safely carry only a few milliamperes before it burns out, which seems to make it useless for measuring the tens or hundreds of amperes flowing in a real circuit, or the hundreds of volts across a real supply. The fix is two humble resistors. A shunt, wired in parallel with the coil, diverts almost all of a large current around the coil so only a safe trickle passes through it — turning the movement into an ammeter of any range you like. A multiplier, wired in series, forces the coil to see only a safe fraction of a large voltage — turning it into a voltmeter. Both designs fall straight out of Ohm's law and Kirchhoff's laws applied to the coil-plus-resistor combination, and this article works through the torque mechanism that makes the coil deflect proportionally to current, then derives and applies both design formulas.

The core insight

A PMMC coil deflects linearly with DC current because $T_d = NBAI$ balances a spring torque $T_c = K\theta$ ; a low parallel shunt $R_{sh} = I_mR_m/(I-I_m)$ extends its current range, and a high series multiplier $R_{se} = V/I_m - R_m$ extends its voltage range.

The core idea

How the coil deflects. A PMMC instrument has a light rectangular coil of $N$ turns and area $A$ , wound on an aluminium former and suspended in the radial field of a permanent magnet; soft-iron pole pieces shape that field so the flux density $B$ is uniform wherever the coil sits. When a current $I$ flows through the coil, each side experiences a force $F = BIL$ , and the pair of forces on opposite sides of the coil combine into a deflecting torque proportional to the current:

T_d = NBAI = GI

where: $N$ is the number of turns, $B$ is the (uniform, radial) flux density in tesla, $A$ is the coil area in square metres, $I$ is the coil current in amperes, and $G = NBA$ bundles the fixed properties of the instrument into a single constant.

Why the scale is uniform. The deflecting torque is opposed by a controlling torque from hairsprings (or a taut-band suspension), which grows in direct proportion to the deflection angle $\theta$ : $T_c = K\theta$ , with $K$ the spring constant. The pointer settles wherever the two torques balance, $T_d = T_c$ , giving

\theta = \frac{G}{K}\,I

Deflection is therefore directly proportional to current, which is exactly why PMMC scales are evenly spaced — unlike, say, a hot-wire meter's cramped, non-uniform scale.

Why PMMC reads DC only. Because $T_d = GI$ changes sign whenever $I$ reverses, an alternating current produces a torque that alternates at the supply frequency. The coil's own inertia cannot follow such rapid reversals, so the average torque over a full cycle is zero and the pointer simply sits near zero — a PMMC movement gives a genuine, steady deflection only for direct current.

Figure

A PMMC coil shown between the poles of a permanent magnet with the shunt resistor drawn in parallel across the coil and the multiplier resistor drawn in series with the coil

Derivation ADesigning the shunt and multiplier resistors

The goal is to turn a fragile PMMC coil that can only tolerate a full-scale current $I_m$ into an ammeter reading a much larger current $I$ , and separately into a voltmeter reading a much larger voltage $V$ , by choosing exactly one extra resistor in each case, using nothing beyond Ohm's law and Kirchhoff's laws applied to the coil-plus-resistor combination.

Step 1. Set up the shunt circuit. The coil, of resistance $R_m$ , is placed in parallel with a shunt resistor $R_{sh}$ , and the combination carries the full line current $I$ . By KCL, the current splits: $I_m$ flows through the coil and the remainder, $I - I_m$ , flows through the shunt.

Step 2. Apply Ohm's law to both branches. Because the coil and the shunt are in parallel, they share the same voltage across their terminals: $V_{\text{parallel}} = I_m R_m$ (across the coil) and also $V_{\text{parallel}} = (I - I_m)R_{sh}$ (across the shunt). Setting these equal gives $I_m R_m = (I - I_m)R_{sh}$ .

Step 3. Solve for the shunt resistance. Rearranging, $R_{sh} = \dfrac{I_m R_m}{I - I_m}$ . Since $I \gg I_m$ for any useful ammeter range, the denominator is close to $I$ , and $R_{sh}$ comes out very small compared with $R_m$ — the shunt is a low-resistance, heavy-gauge element precisely because it must carry nearly all of the line current.

Step 4. Now set up the multiplier circuit. Here the coil and a series resistor $R_{se}$ share the same current, $I_m$ , because they are in series across the full voltage $V$ to be measured. By KVL, the applied voltage splits across the two: $V = I_m R_m + I_m R_{se} = I_m(R_m + R_{se})$ .

Step 5. Solve for the multiplier resistance. Rearranging, $R_{se} = \dfrac{V}{I_m} - R_m$ . Because $I_m$ is tiny (milliamperes) while $V$ can be tens or hundreds of volts, $R_{se}$ comes out very large compared with $R_m$ — the multiplier is a high-resistance element so that it, not the coil, drops nearly all of the measured voltage.

Worked example

A PMMC galvanometer has coil resistance $R_m = 5\ \Omega$ and gives full-scale deflection for $I_m = 1\ \text{mA}$ . Design (a) a shunt to convert it into a 0 to 1 A ammeter, and (b) a multiplier to convert the same coil into a 0 to 150 V voltmeter.

Step 1. Shunt for the 1 A range. The full-scale ammeter current is $I = 1\ \text{A} = 1000\ \text{mA}$ , so the shunt must carry $I - I_m = 999\ \text{mA}$ while the coil carries $1\ \text{mA}$ . Using $R_{sh} = I_mR_m/(I - I_m)$ :

R_{sh} = \frac{(1\times10^{-3})(5)}{1 - 1\times10^{-3}} = \frac{5\times10^{-3}}{0.999} \approx 5.005\times10^{-3}\ \Omega

So a shunt of about $5.0\ \text{m}\Omega$ in parallel with the coil diverts $999\ \text{mA}$ around it while $1\ \text{mA}$ flows through the coil, giving full-scale deflection at a total line current of 1 A. Notice how small $R_{sh}$ is next to $R_m$ , confirming that shunts must be low-resistance.

Step 2. Multiplier for the 150 V range. Using $R_{se} = V/I_m - R_m$ :

R_{se} = \frac{150}{1\times10^{-3}} - 5 = 150{,}000 - 5 = 149{,}995\ \Omega \approx 150\ \text{k}\Omega

The total voltmeter resistance is $R_m + R_{se} = 150\ \text{k}\Omega$ . Check: at full scale, $I_m = V/(R_m + R_{se}) = 150/150{,}000 = 1\ \text{mA}$ , matching the coil's rated current — the design is consistent.

Step 3. Read off the sensitivity. The voltmeter's "ohms-per-volt" figure is $150{,}000/150 = 1000\ \Omega/\text{V}$ , which is simply $1/I_m$ . A smaller $I_m$ always gives a more sensitive (higher-resistance) voltmeter, which disturbs the circuit being measured less. These two results, $R_{sh} \approx 5.0\ \text{m}\Omega$ for the ammeter and $R_{se} \approx 150\ \text{k}\Omega$ for the voltmeter, are the final designed values for this galvanometer.

Common traps

Writing the shunt denominator as just $I$ instead of $I - I_m$ ; the shunt only carries the difference current, since the coil itself still takes $I_m$ .
Assuming a PMMC ammeter or voltmeter can read AC directly; without an added rectifier stage (out of scope here), the average torque over a cycle is zero and the pointer will not deflect meaningfully.
Swapping shunt and multiplier placement: shunt is always parallel (low resistance, for current), multiplier is always series (high resistance, for voltage). Reversing them can destroy the range and possibly the coil.
Forgetting to subtract $R_m$ when computing $R_{se} = V/I_m - R_m$ — the coil is already part of the series path, so its resistance must come out of the total, not be added on top.

Ask the tutor: Want to try designing a shunt or multiplier for a galvanometer with your own $R_m$ and $I_m$ values, or see what happens if you accidentally swap the shunt and multiplier connections?

Why does the shunt resistance come out so much smaller than the coil resistance?

Because the shunt must carry almost the entire line current while the coil carries only its tiny rated current $I_m$ . Since the two are in parallel and share the same voltage, a much larger current through the shunt requires a proportionally much smaller resistance, from $R_{sh} = I_mR_m/(I-I_m)$ . The bigger the ratio of the ammeter's full-scale current to $I_m$ , the smaller $R_{sh}$ must be relative to $R_m$ .

Why is a more sensitive voltmeter one with a smaller Im?

Voltmeter sensitivity is quoted as ohms-per-volt, which works out to exactly $1/I_m$ . A smaller $I_m$ means the coil needs a larger multiplier resistance to still draw only $I_m$ at full scale, so the total voltmeter resistance is higher. A higher-resistance voltmeter draws less current from the circuit under test, disturbing it less and giving a reading closer to the true, unloaded voltage.

Can the same shunt formula be used if the galvanometer's Im is given in microamperes instead of milliamperes?

Yes, the formula $R_{sh} = I_mR_m/(I-I_m)$ is unit-independent as long as $I_m$ and $I$ are expressed in the same units before you subtract or divide. The usual slip is mixing microamperes and amperes in the same calculation; convert both currents to one consistent unit first, then apply the formula exactly as shown in the worked example.

PMMC instruments, shunts, and multipliers

The core idea

Worked example

Common traps

Before this

Where this leads