In short

A galvanometer is a sensitive current detector but a useless ammeter or voltmeter as it stands — it reads only microamperes full-scale and can handle only a fraction of a volt across its own coil resistance. You convert it into a practical instrument with one extra resistor.

Ammeter = galvanometer + shunt (a small resistor S in parallel with the galvanometer). To measure a current up to I amperes on a galvanometer of full-scale current I_g and coil resistance G,

\boxed{\;S \;=\; \frac{I_g\,G}{I - I_g}\;}

The shunt carries most of the current (I - I_g); the galvanometer still sees only its own safe full-scale I_g and its needle reads full-scale when the total through the circuit equals I.

Voltmeter = galvanometer + multiplier (a large resistor R in series with the galvanometer). To measure a voltage up to V volts,

\boxed{\;R \;=\; \frac{V}{I_g} - G\;}

The multiplier drops the bulk of the voltage; the galvanometer itself carries only its safe I_g at full-scale deflection.

Ideal vs real. An ideal ammeter has zero resistance, so inserting it in series changes nothing in the circuit. An ideal voltmeter has infinite resistance, so connecting it across a component draws zero current and again changes nothing. Real instruments have small but non-zero ammeter resistance and large but finite voltmeter resistance, so they always perturb the circuit slightly — the loading effect.

Range extension. Need to measure bigger currents? Reduce the shunt. Bigger voltages? Increase the multiplier. The same galvanometer can serve as a 0\text{–}100\ \mu\text{A} ammeter, a 0\text{–}5\ \text{A} ammeter, and a 0\text{–}300\ \text{V} voltmeter — just by swapping one external resistor. This is exactly how the multi-range Mastech analogue multimeters in every Indian hardware shop work.

Take a CBSE class-12 physics practical kit off the shelf. Inside: a lead-acid cell, a rheostat, a resistance box, two lengths of copper wire, and — side by side in their wooden cradles — an ammeter reading 0–3 A and a voltmeter reading 0–15 V. The ammeter goes in series with the resistor you are studying (the current you want to measure must flow through it). The voltmeter goes in parallel across the resistor (the voltage you want to measure must appear across its terminals). Connect one on the wrong side and the reading is meaningless — or worse, you blow a fuse.

Open the back cover of either instrument (gently, with a jeweller's screwdriver, in a moment when no teacher is watching) and you find the same thing inside both of them: a moving coil galvanometer. The ammeter has one extra piece of thick wire soldered across the galvanometer terminals — a shunt — and that is all. The voltmeter has one extra stick of carbon-film resistor in series with the galvanometer coil — a multiplier — and that is all. The difference between an ammeter and a voltmeter is not the meter mechanism. It is one resistor.

This chapter shows you exactly which resistor, why that value, how to derive the shunt and multiplier formulas from Kirchhoff's laws, why a "good" ammeter has low resistance and a "good" voltmeter has high resistance, and how to quantify the disturbance a real instrument causes when you insert it into a circuit. The same sequence of ideas gets you the auto-ranging Mastech or Fluke digital multimeter that the electrician in your colony uses — only in those, the switch-selectable resistors are multiplexed by a rotary dial.

What a galvanometer can and cannot do

Recall the galvanometer equation from the previous chapter: a current I through the coil produces a proportional deflection \phi = (NAB/k)\,I. Every galvanometer is specified by two numbers:

A current larger than I_g will peg the needle against the mechanical stop and may burn out the hairspring. A voltage bigger than I_g G across its terminals (by Ohm's law) will drive a current bigger than I_g and do the same. For the typical numbers above,

V_g \;=\; I_g\,G \;=\; 10^{-4}\ \text{A} \times 40\ \Omega \;=\; 4\times 10^{-3}\ \text{V} \;=\; 4\ \text{mV}.

So this galvanometer, on its own, is an ammeter reading 0\text{–}100\ \mu\text{A} full-scale and a voltmeter reading 0\text{–}4\ \text{mV} full-scale. Useful for detecting whether a Wheatstone bridge is balanced. Useless for reading the 230 V mains or the 0.8 A draw of a small DC motor.

The idea is simple. Divert most of the current around the galvanometer so only I_g goes through the coil even when I amperes flow through the line (that is the shunt). Or drop most of the voltage across a big series resistor so only V_g = I_g G appears across the coil even when V volts sit across the instrument's terminals (that is the multiplier).

Ammeter versus voltmeter — circuit topologyTwo circuits. Left: an ammeter, shown as a galvanometer G in parallel with a small shunt resistor S, wired in series with the load resistor R in a battery circuit. Right: a voltmeter, shown as a galvanometer G in series with a large multiplier resistor R_m, wired in parallel across the load resistor R in a battery circuit. Ammeter — in series E A (ammeter) G S (shunt) R Voltmeter — in parallel E R V (voltmeter) R (mult.) G
Left: an ammeter is a galvanometer $G$ with a small shunt $S$ in parallel; the combination is inserted in series with the load $R$ so that every electron passing through the circuit passes through it. Right: a voltmeter is a galvanometer $G$ with a large multiplier $R$ in series; the combination is placed in parallel across the component whose voltage you want to read.

The ammeter — shunt in parallel

Here is the ammeter circuit, drawn as a two-node network. Current I enters the left terminal, splits between the galvanometer (resistance G, carrying current I_g) and the shunt (resistance S, carrying current I_s), and recombines at the right terminal.

Step 1. Current conservation at the junction

At the left node, current in equals current out:

I \;=\; I_g + I_s. \tag{1}

Why: this is Kirchhoff's current law. The line current I has nowhere else to go; it must split between the two parallel paths.

Step 2. Equal voltage across parallel branches

The galvanometer and the shunt sit between the same two nodes, so they have the same voltage drop:

V_G \;=\; I_g\,G \;=\; I_s\,S \;=\; V_S. \tag{2}

Why: potential difference is single-valued between two points. Whatever voltage sits between the two ammeter terminals, both the galvanometer and the shunt see it.

Step 3. Fix the galvanometer current at full-scale

You want the pointer to read full-scale when the line current is the chosen range I. That means I_g should equal the galvanometer's maximum safe current at that moment. From (2),

I_s \;=\; \frac{I_g\,G}{S}. \tag{3}

Step 4. Substitute into current conservation

Use (3) in (1):

I \;=\; I_g + \frac{I_g\,G}{S} \;=\; I_g\left(1 + \frac{G}{S}\right) \;=\; I_g\,\frac{S + G}{S}.

Rearrange to solve for S:

I\,S \;=\; I_g\,(S + G)
I\,S - I_g\,S \;=\; I_g\,G
S(I - I_g) \;=\; I_g\,G
\boxed{\;S \;=\; \frac{I_g\,G}{I - I_g}.\;} \tag{4}

Why: equation (4) is the master shunt formula. The denominator I - I_g is the current that the shunt must carry — everything the galvanometer refuses to handle. The shunt must be small enough that the full line current minus the tiny I_g still produces only the voltage I_g G across the parallel combination. Because I \gg I_g in any practical situation, S \ll G: a tiny fraction of an ohm for ammeters reading amperes, a few milliohms for ammeters reading hundreds of amperes.

Step 5. The effective resistance of the ammeter

From outside, the ammeter looks like a single two-terminal resistor whose resistance is G in parallel with S:

R_A \;=\; \frac{G\,S}{G + S}. \tag{5}

Substituting (4) into (5) and simplifying:

R_A \;=\; \frac{G\cdot \dfrac{I_g G}{I - I_g}}{G + \dfrac{I_g G}{I - I_g}} \;=\; \frac{G\cdot I_g G}{G(I - I_g) + I_g G} \;=\; \frac{I_g\,G^2}{G\,I} \;=\; \frac{I_g}{I}\,G.
\boxed{\;R_A \;=\; \frac{I_g}{I}\,G.\;} \tag{6}

Why: equation (6) says the ammeter's total resistance is the galvanometer's resistance G scaled down by the factor I_g/I — the ratio of the old full-scale current to the new one. A 100\ \muA galvanometer with G = 40\ \Omega converted to a 1 A ammeter gets a resistance of (10^{-4}/1) \times 40 = 4\ \text{m}\Omega. Converted to a 10 A ammeter it drops to 0.4\ \text{m}\Omega. The higher the current range, the lower the ammeter resistance must be — which is why big ammeters use thick-bar shunts the size of a plumbing clamp.

Why an ideal ammeter has zero resistance

An ammeter sits in series with the component whose current you want to measure. It carries the same current as the rest of the line. The voltage it drops is wasted — it does not appear across the component being studied, so that component sees less voltage than the source intended, and therefore draws less current than it would in the ammeter's absence. The reading is too low by exactly the factor you introduce.

Quantitatively: a circuit with source EMF E, internal resistance r, and load R originally draws I_0 = E/(R + r). Insert an ammeter of resistance R_A, and the current drops to

I \;=\; \frac{E}{R + r + R_A} \;=\; I_0 \cdot \frac{R + r}{R + r + R_A}.

The fractional error in the current reading is

\frac{I_0 - I}{I_0} \;=\; \frac{R_A}{R + r + R_A}.

For the reading to be accurate to 1%, the ammeter resistance must be less than 1% of (R + r). An ideal ammeter would have R_A = 0 — a perfect short that measures without disturbing. Real ammeters never reach zero, but by making R_A \ll R you get as close as you need.

Explore the shunt

The interactive figure below lets you slide the desired full-scale range I from the galvanometer's own 100\ \muA up to 5 A, and watch three quantities respond in real time: the required shunt resistance S, the ammeter's overall resistance R_A, and the fractional loading error on a 10\ \Omega load driven from a 1.5 V cell.

Interactive: how shunt and ammeter resistance depend on chosen range Two curves. One shows log10 of shunt resistance S versus chosen full-scale range I. The other shows log10 of ammeter resistance R_A versus I. A draggable dot on the horizontal axis selects the current range, and readouts report S, R_A, and the fractional error on a 10-ohm load. full-scale range I (A) log₁₀(ohms) −3 −1 0 2 4 0 1 2 3 5 log₁₀ S log₁₀ Rₐ drag the red dot along the axis
Drag the red dot to change the desired full-scale range. As $I$ grows, the required shunt $S$ (red) shrinks roughly as $1/I$, and the ammeter's overall resistance $R_A$ (dark) shrinks at exactly the same rate. The readout shows the loading error on a $10\ \Omega$ load — tiny even for a modest $I = 1$ A, because $R_A \approx 4\ \text{m}\Omega$ is negligible next to $10\ \Omega$. The galvanometer here is $I_g = 100\ \mu$A, $G = 40\ \Omega$.

The voltmeter — multiplier in series

Now repeat the analysis for a voltmeter. A voltmeter must accept a voltage up to V across its two terminals, and at that maximum it should drive exactly I_g through the galvanometer so the pointer reads full-scale. The trick: put a large resistor R in series with the galvanometer, inside the same two-terminal package.

Step 1. One series loop — the current is the galvanometer current

Current flows from the "+" terminal, through the multiplier R, through the galvanometer coil G, out of the "−" terminal. Series: the same current everywhere. At full-scale,

I \;=\; I_g \quad\text{(through both } R \text{ and } G \text{).} \tag{7}

Step 2. Kirchhoff's voltage law around the instrument

The total voltage across the instrument equals the sum of voltage drops across R and G:

V \;=\; I_g\,R + I_g\,G \;=\; I_g\,(R + G). \tag{8}

Why: this is Kirchhoff's voltage law. Walking from the "+" terminal to the "−" terminal inside the box, the potential drops by I_g R across the multiplier and then by I_g G across the coil. The total drop must equal the externally applied voltage V.

Step 3. Solve for R

R + G \;=\; \frac{V}{I_g}
\boxed{\;R \;=\; \frac{V}{I_g} - G.\;} \tag{9}

Why: equation (9) is the master multiplier formula. The quantity V/I_g is the total resistance you need; you already have G from the galvanometer, so the extra resistor you must add is V/I_g - G. For any practical voltmeter, V/I_g \gg G, so R \approx V/I_g — a large number in kilo-ohms or mega-ohms.

Step 4. The voltmeter resistance

From the outside, the voltmeter is a single series combination:

R_V \;=\; R + G \;=\; \frac{V}{I_g}. \tag{10}

Why: two resistors in series add. Notice that the voltmeter's total resistance is just the range over the full-scale current. For a 100\ \muA galvanometer turned into a 10 V voltmeter, R_V = 10/10^{-4} = 10^5\ \Omega = 100\ \text{k}\Omega. For a 300 V voltmeter, R_V = 3\ \text{M}\Omega. The industry quotes this as "\Omega per volt": a 100\ \muA movement gives 10{,}000\ \Omega/\text{V}.

Why an ideal voltmeter has infinite resistance

A voltmeter sits in parallel with the component whose voltage you want to read. It adds a parallel path for current to flow, so some fraction of the current that was flowing through the component now leaks through the voltmeter. Less current through the component means less voltage across it — so the voltmeter reads low.

Take a simple divider: two resistors R_1 and R_2 in series across an ideal EMF E, with no voltmeter present. The voltage across R_2 is

V_0 \;=\; E\,\frac{R_2}{R_1 + R_2}.

Connect a voltmeter of resistance R_V across R_2. The parallel combination R_2 \| R_V = R_2 R_V/(R_2 + R_V) replaces R_2, and the reading becomes

V \;=\; E\,\frac{R_2 R_V/(R_2 + R_V)}{R_1 + R_2 R_V/(R_2 + R_V)} \;=\; E\,\frac{R_2 R_V}{R_1(R_2 + R_V) + R_2 R_V}.

For R_V \to \infty this returns the true voltage V_0. For finite R_V, the reading is too low. The fractional error depends on the ratio R_V/R_2: if R_V is 100 times R_2, the error is about 1%.

An ideal voltmeter draws no current and therefore does not disturb the circuit. Real voltmeters never reach infinite R_V, but by making R_V \gg R_2 (the component you are measuring across) you get as close as you need. A good-quality CBSE-lab voltmeter has R_V = 10{,}000\ \Omega/\text{V}; a modern Mastech digital multimeter has R_V \approx 10\ \text{M}\Omega on all voltage ranges, which is essentially ideal for any circuit in a school lab.

Range extension — one galvanometer, many instruments

The same galvanometer (fixed I_g, fixed G) can serve as many different ammeters and voltmeters. The trick is to switch in different external resistors.

Multi-range ammeter. Use several shunts in a rotary switch. Position 1: a 4 mΩ shunt → 1 A range. Position 2: a 0.4 mΩ shunt → 10 A range. Position 3: 40 mΩ shunt → 100 mA range. Each position uses formula (4) with a different I.

Multi-range voltmeter. Use several multipliers in a rotary switch. Position 1: a 100 kΩ multiplier → 10 V range. Position 2: 1 MΩ → 100 V range. Position 3: 3 MΩ → 300 V range. Each position uses formula (9) with a different V.

Multimeter. Combine both switches and the same galvanometer serves as ammeter, voltmeter, and (with a battery added) ohmmeter — all in one handheld unit. Open the back of the analogue Mastech MS7010 in your physics lab and you will see this exact architecture: a central galvanometer, a rotary switch, a PCB with a rack of carbon-film multipliers and a row of metal-strip shunts.

Worked examples

Example 1: Converting a galvanometer to a 3 A ammeter

A moving-coil galvanometer has full-scale current I_g = 150\ \mu\text{A} and coil resistance G = 50\ \Omega. Find the shunt resistance needed to make it read 0\text{–}3 A full-scale. Also compute the ammeter's overall resistance and the loading error when this ammeter is placed in series with a 2\ \Omega load driven by a 6 V cell of negligible internal resistance.

Galvanometer with shunt for 3 A ammeterA galvanometer G with resistance 50 ohms and 150 microamperes full-scale, in parallel with a shunt resistor S of unknown value. Line current I of 3 amperes splits into 150 microamperes through the galvanometer and the rest through the shunt. I = 3 A G 50 Ω, 150 µA Iₘ = 150 µA S = ? Iₛ I = 3 A
The line current $I = 3$ A splits at the left junction. Only $I_g = 150\ \mu$A of it goes through the galvanometer; the remaining $I - I_g \approx 3$ A goes through the shunt $S$. Both branches must drop the same voltage, which fixes $S$.

Step 1. Write the known values.

I_g = 150\ \mu\text{A} = 1.5\times 10^{-4} A. G = 50\ \Omega. I = 3 A.

Why: list every quantity in SI units before substituting. Mixing up microamperes and amperes is the single most common error in these problems.

Step 2. Apply the shunt formula (4).

S \;=\; \frac{I_g\,G}{I - I_g} \;=\; \frac{(1.5\times 10^{-4})(50)}{3 - 1.5\times 10^{-4}}.

Step 3. Simplify. Since I_g \ll I, I - I_g \approx I = 3 A to four significant figures.

S \;\approx\; \frac{7.5\times 10^{-3}}{3} \;=\; 2.5\times 10^{-3}\ \Omega \;=\; 2.5\ \text{m}\Omega.

Why: the approximation I - I_g \approx I introduces a fractional error of I_g/I = 5\times 10^{-5}, which is far below the accuracy of any physical resistor. Keeping the subtraction exact gives S = 2.50013\ldots\ \text{m}\Omega, a difference of micro-ohms that no laboratory ohmmeter can measure anyway.

Step 4. Compute the ammeter's overall resistance via (6).

R_A \;=\; \frac{I_g}{I}\,G \;=\; \frac{1.5\times 10^{-4}}{3}\times 50 \;=\; 2.5\times 10^{-3}\ \Omega \;=\; 2.5\ \text{m}\Omega.

Why: R_A equals S to three significant figures, because S \ll G forces the parallel combination to be essentially S itself. The galvanometer's 50\ \Omega is dwarfed by the 2.5\ \text{m}\Omega shunt in parallel.

Step 5. Compute the loading error.

Without the ammeter, the circuit draws I_0 = 6/2 = 3 A. With the ammeter inserted,

I \;=\; \frac{6}{2 + 0.0025} \;=\; \frac{6}{2.0025} \;\approx\; 2.9963\ \text{A}.

Fractional error: (I_0 - I)/I_0 = 0.0025/2.0025 \approx 0.125\%.

Result: Shunt S = 2.5\ \text{m}\Omega. Ammeter resistance R_A = 2.5\ \text{m}\Omega. Loading error 0.125%.

What this shows: A quarter-percent error is invisible to a classroom ammeter whose scale has divisions of 0.1 A (3% of full-scale). In other words, the instrument is more than accurate enough for the load it measures. The shunt is genuinely a piece of thick, short wire — 2.5 mΩ is the resistance of a few centimetres of 2.5 mm² copper wire (the kind used in CBSE-lab connecting leads). That is why when you look inside the back of a lab ammeter, the shunt is often just a visible loop of bare copper bar soldered between the two terminals.

Example 2: Converting the same galvanometer to a 150 V voltmeter

Using the same galvanometer (I_g = 150\ \muA, G = 50\ \Omega), design a voltmeter reading 0\text{–}150 V full-scale. Find the multiplier resistance, the voltmeter's total resistance, and the loading error when measuring the voltage across a 100\ \text{k}\Omega resistor in a voltage divider where the other resistor is also 100\ \text{k}\Omega and the source is 12 V.

Galvanometer with multiplier for 150 V voltmeterA series combination of a multiplier resistor R in series with a galvanometer G of 50 ohms resistance, with voltage V of 150 volts applied across the ends. + R (multiplier) G 50 Ω, 150 µA − Iₘ = 150 µA V = 150 V (across both terminals at full-scale)
At full-scale, 150 V is applied across the two terminals. The multiplier $R$ and galvanometer $G$ in series must together carry exactly $I_g = 150\ \mu$A. That fixes $R + G = V/I_g$.

Step 1. Write the known values.

I_g = 1.5\times 10^{-4} A. G = 50\ \Omega. V = 150 V.

Step 2. Apply the multiplier formula (9).

R \;=\; \frac{V}{I_g} - G \;=\; \frac{150}{1.5\times 10^{-4}} - 50 \;=\; 10^6 - 50 \;\approx\; 1{,}000{,}000\ \Omega \;=\; 1.00\ \text{M}\Omega.

Why: the galvanometer's 50\ \Omega is six orders of magnitude smaller than the 10^6\ \Omega you need, so to three significant figures R = V/I_g = 1.00\ \text{M}\Omega. In practice, you would solder a 1\ \text{M}\Omega \pm 1\% metal-film resistor in series with the galvanometer terminals.

Step 3. Total voltmeter resistance.

R_V \;=\; R + G \;=\; 10^6 + 50 \;\approx\; 10^6\ \Omega \;=\; 1.00\ \text{M}\Omega.

Why: R_V is the full series resistance, which equals V/I_g exactly. For this instrument the "ohms per volt" rating is R_V/V = 10^6/150 \approx 6.67\ \text{k}\Omega/\text{V}.

Step 4. Loading error on the voltage divider.

Without the voltmeter: V_0 = 12 \times 100\text{k}/(100\text{k} + 100\text{k}) = 6.00 V.

With the voltmeter across the lower 100\ \text{k}\Omega: the parallel combination is

R_2\|R_V \;=\; \frac{100\text{k} \times 1000\text{k}}{100\text{k} + 1000\text{k}} \;=\; \frac{10^8}{1100} \;\approx\; 90.91\ \text{k}\Omega.

The new reading:

V \;=\; 12 \times \frac{90.91}{100 + 90.91} \;=\; 12 \times \frac{90.91}{190.91} \;\approx\; 5.714\ \text{V}.

Fractional error: (6.00 - 5.714)/6.00 \approx 4.76\%.

Why: when the voltmeter's resistance (1\ \text{M}\Omega) is only 10 times the resistance it measures across (100\ \text{k}\Omega), loading is significant. This is why measuring voltages across high-resistance components requires a high-impedance meter. A 10 MΩ digital multimeter would give an error of 0.5%, a factor of 10 better.

Result: Multiplier R = 1.00\ \text{M}\Omega. Voltmeter resistance R_V = 1.00\ \text{M}\Omega. Loading error on a 100 kΩ divider: ≈ 4.76%.

What this shows: The voltmeter is never more accurate than the ratio of its resistance to the component's resistance. For low-resistance circuits (a battery driving a few ohms) any voltmeter is effectively ideal; for high-resistance circuits (biasing an op-amp, measuring across a microphone capsule) you need a voltmeter that dwarfs the circuit. This single idea is why every practical modern multimeter is a FET-input design with \geq 10\ \text{M}\Omega input resistance.

Example 3: Designing a dual-range instrument

The same galvanometer (I_g = 100\ \muA, G = 40\ \Omega) is to be wired so that one pair of external terminals gives a 0\text{–}500\ \text{mA} ammeter, and a second pair of terminals gives a 0\text{–}20\ \text{V} voltmeter. Calculate both external resistors and the ohms-per-volt rating of the voltmeter range.

Dual-range meter: ammeter and voltmeter terminalsSchematic with a central galvanometer G of 40 ohms. One pair of terminals labelled A includes a shunt S in parallel with G for the 500 mA range. A second pair of terminals labelled V includes a multiplier R in series with G for the 20 V range. G 40 Ω, 100 µA A₊ S R (mult.) V₊ Ammeter range 0–500 mA Voltmeter range 0–20 V
One galvanometer, two pairs of terminals. The A-terminal pair shorts in the shunt $S$ for the 500 mA current range; the V-terminal pair puts the multiplier $R$ in series for the 20 V voltage range. Real multimeters are exactly this idea with a rotary switch.

Step 1. Ammeter shunt for I = 500 mA = 0.5 A.

S \;=\; \frac{I_g\,G}{I - I_g} \;=\; \frac{(10^{-4})(40)}{0.5 - 10^{-4}} \;\approx\; \frac{4\times 10^{-3}}{0.5} \;=\; 8\times 10^{-3}\ \Omega \;=\; 8\ \text{m}\Omega.

Why: once you have the master formulas, every ammeter range is a two-line calculation — plug in I_g, G, and the desired I.

Step 2. Voltmeter multiplier for V = 20 V.

R \;=\; \frac{V}{I_g} - G \;=\; \frac{20}{10^{-4}} - 40 \;=\; 200{,}000 - 40 \;\approx\; 2.00\times 10^{5}\ \Omega \;=\; 200\ \text{k}\Omega.

Step 3. Ohms-per-volt rating.

\frac{R_V}{V} \;=\; \frac{R + G}{V} \;=\; \frac{1}{I_g} \;=\; \frac{1}{10^{-4}} \;=\; 10{,}000\ \Omega/\text{V}.

Why: the ratio \Omega/\text{V} is a property of the galvanometer alone, not of the chosen range. A 100\ \muA movement always gives 10{,}000\ \Omega/\text{V}, whether you set it up for 1 V full-scale (10 kΩ multiplier) or 1000 V full-scale (10 MΩ multiplier). This is what printed labels like "20{,}000 Ω/V DC" on the face of a Mastech analogue meter actually mean.

Result: Shunt S = 8\ \text{m}\Omega (for the 500 mA ammeter range); Multiplier R = 200\ \text{k}\Omega (for the 20 V voltmeter range); Ohms-per-volt rating = 10{,}000 Ω/V.

What this shows: The same galvanometer serves two radically different instruments — a low-resistance ammeter (8 mΩ) and a high-resistance voltmeter (200 kΩ) — purely through the choice of external resistor. Looking at a multi-range analogue multimeter, you are looking at the same movement wired through a rotary switch to a bank of these shunts and multipliers. The instrument's precision is set by the precision of the resistors, not the galvanometer.

Common confusions

If you came here to understand ammeters and voltmeters, use the two formulas, and wire up a lab circuit, you have what you need. What follows is for JEE Advanced readers who want the errors quantified, the Ayrton shunt derived, and the correct placement of the two instruments chosen.

Quantifying ammeter and voltmeter errors together

Consider a simple Ohm's-law experiment: a battery of EMF E drives current I through a resistor R, you want to determine R from measured V and I. Two circuit topologies are possible.

Topology A — ammeter "outside", voltmeter "inside". The ammeter reads the current through R plus the current through the voltmeter (because the voltmeter is in parallel with R, and the ammeter is outside that parallel combination). The voltmeter reads the true V across R. Your computed value R_\text{meas} = V_\text{read}/I_\text{read} is too small because I_\text{read} > I_\text{true}.

Topology B — voltmeter "outside", ammeter "inside". The voltmeter reads V across R plus the voltage drop across the ammeter (because the ammeter is in series with R, inside the voltmeter's loop). The ammeter reads the true current through R. Your computed R_\text{meas} is too large because V_\text{read} > V_\text{true}.

Fractional errors (derivation by the same Kirchhoff's law manipulations as above):

  • Topology A: \delta R/R \approx -R/R_V
  • Topology B: \delta R/R \approx R_A/R

For small R (smaller than \sqrt{R_A R_V}), topology A is better (the voltmeter-loading error is tiny because R \ll R_V). For large R (larger than \sqrt{R_A R_V}), topology B is better (the ammeter's series drop is tiny because R_A \ll R). The cross-over is at R = \sqrt{R_A R_V}, often called the geometric mean rule. For a lab with R_A = 0.1\ \Omega and R_V = 10^4\ \Omega, the cross-over is \sqrt{10^3} = 31.6\ \Omega. Below that, use topology A; above it, topology B. This is the reason the layout matters in a practical's observation column.

The Ayrton shunt — a multi-range ammeter without switching contact errors

Switching between shunts with a rotary switch has a problem: the switch contact itself has a small, variable resistance (fractions of a milliohm), which adds in series with the shunt and changes the calibration. For cheap school ammeters this is acceptable; for precision lab instruments it is not.

The Ayrton shunt solves this by using a single chain of resistors with multiple tap points. Let R_\text{total} be the total resistance of the chain (in parallel with the galvanometer always), and let the taps divide it into segments R_1, R_2, R_3 with R_1 + R_2 + R_3 = R_\text{total}. Depending on which tap you connect to the external terminal, a different portion of the chain acts as the "shunt proper" and the rest acts in series with the galvanometer.

Working out the arithmetic: for range n, the effective shunt is R_1 + R_2 + \ldots + R_n and the effective series-with-galvanometer segment is R_{n+1} + \ldots + R_\text{total}. The full-scale current for each range works out to

I_n \;=\; I_g\,\frac{G + R_\text{total}}{R_1 + R_2 + \ldots + R_n}.

Pick any desired I_1, I_2, I_3 set of ranges and solve the linear system for the segment values R_1, R_2, R_3. The contact resistance of the rotary switch adds in series with the galvanometer, where it is negligible compared to G — not in series with the shunt, where it would corrupt the reading.

Why the galvanometer deflection is truly linear in I, even with the shunt

You might worry that the shunt changes the linear \phi \propto I relation derived in the galvanometer chapter. It does not. At any line current I, the fraction I_g/I = S/(S+G) flows through the galvanometer, and this fraction is a constant — it does not depend on I. So the galvanometer deflection \phi = (NAB/k)\,I_g = (NAB/k)\cdot(S/(S+G))\cdot I is linear in I, with a slope reduced by exactly the factor S/(S+G). The scale stays uniform; only the "constant" multiplying it changes. The same argument applies to the voltmeter (I_g = V/(R+G) is linear in V).

This linearity is the reason your analogue ammeter has equal-spaced divisions from 0 to full-scale, and it is a direct consequence of the radial magnetic field in the galvanometer — another reminder that the dead-beat, uniform-scale moving coil galvanometer is the silent hero behind every lab current and voltage measurement in India.

Where this leads next