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Figure of Merit Formula

Electricity and magnetism formulasClass 12PracticalUpdated 1 Nov 2026

k=Iθk = \frac{I}{\theta}
SI unit
A/division
Dimensions
[M⁰ L⁰ T⁰ I¹]
where
  • k figure of merit of the galvanometer A/division
  • I current through the coil A
  • θ deflection of the pointer divisions

The figure of merit formula for a moving coil galvanometer is k=Iθk = \frac{I}{\theta}: divide the current through the coil by the deflection, counted in scale divisions, that the current produces. The result is the current needed for a one division swing, so a small k means a more sensitive meter.

The idea appears in Class 12 magnetic effects of current and in a standard board practical, where you find k by the half deflection method. It is simply the reciprocal of current sensitivity. Once you know k and the number of divisions on the scale, you know the full scale current, which is the starting point for turning the galvanometer into an ammeter or a voltmeter. This page belongs to the electricity formulas section.

Quick facts

  • Formula

    k=Iθk = \frac{I}{\theta}

  • From the coil

    k=cNABk = \frac{c}{NAB}, c is the torsion constant

  • SI unit

    ampere per division (A/div), or A/rad when θ is in radians

  • Dimensional formula

    [M⁰ L⁰ T⁰ I¹], since a division has no dimension

  • Link to sensitivity

    k=1current sensitivityk = \frac{1}{\text{current sensitivity}}

  • Practical form

    k=E(R+G) θk = \frac{E}{(R + G)\,\theta}

  • Typical size

    a few μA to tens of μA per division for a lab galvanometer

Symbols in the figure of merit formula

Terms in k = I / θ
SymbolMeaningUnit
kkFigure of merit, current per divisionA/div
IISteady current through the galvanometer coilA
θ\thetaDeflection read on the scaledivisions (or rad)
ccTorsion constant of the spring or suspensionN m/rad
N,A,BN, A, BTurns, coil area, radial magnetic fieldno unit, m², T

Inside the meter the magnetic torque on the coil, NIABNIAB, is balanced by the restoring torque of the spring, cθc\theta. Setting them equal gives I=cNAB θI = \frac{c}{NAB}\,\theta, so the deflection grows in direct proportion to the current. That straight line relation is why the scale can be evenly marked. The torque comes from the magnetic force on each side of the coil, and NIANIA is the coil's magnetic moment.

Deriving k from the torque balance

  1. Torque on the coil in a radial field: τ=NIAB\tau = NIAB.
  2. Restoring torque of the spring at deflection θ: τ=cθ\tau = c\theta.
  3. At equilibrium NIAB=cθNIAB = c\theta, so Iθ=cNAB\frac{I}{\theta} = \frac{c}{NAB}.
  4. The left side is the figure of merit, so k=cNABk = \frac{c}{NAB}.

The result shows how to build a sensitive instrument: many turns, a large coil, a strong magnet and a weak spring all lower k. Current sensitivity is θI=NABc\frac{\theta}{I} = \frac{NAB}{c}, the exact inverse. Voltage sensitivity divides the current sensitivity by the coil resistance G, giving NABcG\frac{NAB}{cG}. Adding turns raises the current sensitivity but also raises G, so voltage sensitivity need not improve.

Measuring it in the lab: half deflection method

In the board practical you join a cell of emf E, a high resistance box R and the galvanometer in series. Take out a resistance R and note the deflection θ. The current is set almost entirely by R and G, so Ohm's law gives I=ER+GI = \frac{E}{R + G} and

k=E(R+G) θk = \frac{E}{(R + G)\,\theta}

The galvanometer resistance G is found first by the half deflection method. A shunt S is placed across the meter and adjusted until the deflection falls to θ/2. Then G=RSR−SG = \frac{RS}{R - S}. When R is thousands of ohms and S only tens of ohms, G is close to S. Take readings for several values of R and average k, because a single reading carries the full error of the scale.

Choose R so the deflection is large, ideally over half the scale. Small deflections give large percentage reading errors.

Worked examples

Example 1: reading k from the scale

A galvanometer has 30 divisions on each side of zero. A current of 60 μA gives a full scale swing. Find the figure of merit.

Given: I = 60 × 10⁻⁶ A, θ = 30 divisions.

k=60×10−630=2.0×10−6 A/divk = \frac{60 \times 10^{-6}}{30} = 2.0 \times 10^{-6}\ \text{A/div}

Each division stands for 2.0 μA.

Example 2: practical readings

A 2.0 V cell, a 4970 Ω resistance and a galvanometer of resistance 30 Ω are in series. The deflection is 20 divisions. Find k.

Current: I=2.04970+30=2.05000=4.0×10−4 AI = \frac{2.0}{4970 + 30} = \frac{2.0}{5000} = 4.0 \times 10^{-4}\ \text{A}

k=4.0×10−420=2.0×10−5 A/divk = \frac{4.0 \times 10^{-4}}{20} = 2.0 \times 10^{-5}\ \text{A/div}

Example 3: from the coil design

A coil has 50 turns, each 2.0 cm by 2.0 cm, in a radial field of 0.20 T. The torsion constant is 2.0 × 10⁻⁸ N m/rad. Find k in A/rad.

Convert the area: A = 0.020 × 0.020 = 4.0 × 10⁻⁴ m². Then NAB=50×4.0×10−4×0.20=4.0×10−3NAB = 50 \times 4.0 \times 10^{-4} \times 0.20 = 4.0 \times 10^{-3} in SI units.

k=2.0×10−84.0×10−3=5.0×10−6 A/radk = \frac{2.0 \times 10^{-8}}{4.0 \times 10^{-3}} = 5.0 \times 10^{-6}\ \text{A/rad}

Using k to convert a galvanometer

The full scale current is Ig=k×nI_g = k \times n, where n is the number of divisions on one side of the scale. With Example 2, a 30 division scale gives Ig=2.0×10−5×30=6.0×10−4I_g = 2.0 \times 10^{-5} \times 30 = 6.0 \times 10^{-4} A. That number feeds two standard conversions.

Turning a galvanometer into a meter
New meterAddValue
Ammeter of range ILow shunt S in parallelS=IgGI−IgS = \frac{I_g G}{I - I_g}
Voltmeter of range VHigh resistance R in seriesR=VIg−GR = \frac{V}{I_g} - G

Both rest on the formula of Ohm's law; the resistance and current pages revise those basics. The calculator above handles the core relation k = I / θ in SI units.

Mistakes students make

  • Confusing k with sensitivity. A sensitive galvanometer has a high current sensitivity and a low figure of merit. They move in opposite directions.
  • Leaving current in μA or mA. Convert to amperes before you divide, then state A/div.
  • Ignoring G in the practical formula. Use R + G, not R alone, unless R is so large that G is negligible and the question says so.
  • Counting both sides of the scale. The full scale current uses the divisions from zero to one end only.
  • Mixing up the two k values. The torsion constant c and the figure of merit k are different quantities, so give them different letters in your working.

Frequently asked questions

What is the figure of merit of a galvanometer?

It is the current that produces a deflection of one scale division, k = I / θ. A smaller figure of merit means the galvanometer responds to smaller currents.

What is the SI unit of figure of merit?

It is ampere per division. If the deflection is measured in radians, the unit becomes ampere per radian. A division is a pure count, so the dimensional formula is that of current, [M⁰ L⁰ T⁰ I¹].

How is figure of merit related to current sensitivity?

They are reciprocals. Current sensitivity is θ / I = NAB / c, and figure of merit is I / θ = c / NAB.

What is the formula used in the figure of merit practical?

With a cell of emf E, a series resistance R and a galvanometer of resistance G, the figure of merit is k = E / ((R + G)θ). G itself comes from the half deflection method, G = RS / (R − S).

How can the figure of merit of a galvanometer be reduced?

Use more turns, a larger coil area, a stronger radial field or a suspension with a smaller torsion constant. Each of these makes k = c / NAB smaller.