EMF Equation of Synchronous Generator or Alternator

Voltage Generation in Alternator

The rotor of the alternator is run at its proper speed by the prime mover. The prime mover is a device which supplies mechanical energy input to the alternator. For slow and medium speed alternators, the water turbines are used as prime mover while the steam and gas turbines are used for the large and high speed alternators.

When the poles of the rotor move under the armature conductors on the stator, the magnetic field cuts the armature conductors. Therefore, EMF is induced in these conductors by the electromagnetic induction. This induced EMF is of alternating nature because field poles of alternate polarity successively pass by the armature conductor. This induced EMF in the synchronous generator or alternator is given by the equation that is given below.

EMF Equation of Alternator


  • 𝑃 = Number of poles

  • φ = Flux per pole in webers

  • 𝑁 = Rotor speed in RPM

  • 𝑓 = Frequency of the induced EMF in Hz

  • 𝑍 = Number of conductors in series per phase

  • 𝑇 = Number of coils or turns per phas𝑒

As the flux per pole is φ, hence, in one revolution, each stator conductor cut a flux of,

$$\mathrm{𝑑_{πœ‘} = 𝑃_{πœ‘}}$$

And the time taken to complete one revolution is,

$$\mathrm{𝑑𝑑 =\frac{60}{𝑁}}$$

Therefore, according electromagnetic induction, the average EMF induced in one stator conductor is given by,

$$\mathrm{EMF\:per\:conductor =\frac{𝑑φ}{𝑑𝑑}=\frac{𝑃φ𝑁}{60}}$$

Since there are Z conductors in series per phase, thus,

$$\mathrm{EMF\:per\:phase = EMF\:per\:conductor \times Z =(\frac{𝑃φ𝑁}{60})\times 𝑍}$$

$$\mathrm{∡\:Speed\:of\:the\:rotor,\:𝑁 =\frac{120𝑓}{𝑃}}$$

$$\mathrm{∴\:EMF\:per\:phase =\frac{𝑃φ𝑍}{60}\times\frac{120𝑓}{𝑃}= 2\:𝑓\:πœ‘\:𝑍\:… (1)}$$

Equation (1) gives the value of average induced EMF per conductor per phase. Since, one turn has two conductors, i.e., 𝑍 = 2𝑇 and hence the expression for the average induced EMF per phase can be written as,

$$\mathrm{EMF\:per\:phase = 4\:𝑓\:φ\:𝑇\:… (2)}$$

For any voltage wave, the form factor is given by,

$$\mathrm{F. F. =\frac{RMS\:Value}{Average\: Value}}$$

$$\mathrm{∡\:RMS\:value\:of\:EMF/phase = (Average\:Value/Phase) \times Form\:Factor}$$

$$\mathrm{∴\:RMS\:value\:of\:EMF/phase = 4\:𝑓\:φ\:𝑇\: \times(\frac{RMS\:Value}{Average\: Value})}$$

For a sinusoidal voltage wave, form factor = 1.11.

Thus, the RMS value of the induced EMF per phase can be written as,

$$\mathrm{𝐸_{𝑅𝑀𝑆}/π‘ƒβ„Žπ‘Žπ‘ π‘’ = 1.11 \times 4\:𝑓\:φ\:𝑇 = 4.44\:𝑓\:φ\:𝑇 …(3)}$$

Equation (3) gives the RMS value of the induced EMF with the following assumptions −

  • All the armature conductors are concentrated in one stator slot.

  • Coils got have full pitch.

Again, taking the coil span factor and distribution factor into account, the actual induced EMF per phase is given by,

$$\mathrm{𝐸_{π‘β„Ž} = 4.44\:π‘˜_{𝑐}\:π‘˜_{𝑑}\:𝑓_{φ}\:𝑇 … (4)}$$

$$\mathrm{(∡\:𝑍 = 2𝑇)}$$

$$\mathrm{∴\:𝐸_{π‘β„Ž} = 2.22\:π‘˜_{𝑐}\:π‘˜_{𝑑}\:𝑓\:{φ}\:𝑍 … (5)}$$

Equations (3) and (4) are called as the EMF equation of an alternator.

Sometimes, the coil span factor and the distribution factor of a winding are combined into a single factor, known as winding factor (kw) which is given by,

$$\mathrm{π‘˜_{𝑀} = π‘˜_{𝑐}\:π‘˜_{𝑑}}$$

Hence, the EMF equation can also be written as,

$$\mathrm{𝐸_{π‘β„Ž} = 4.44\:π‘˜_{𝑀}\:𝑓_{φ}\:𝑇 … (6)}$$

For a star-connected alternator, the line voltage is the $\sqrt{3}$ times of the phase voltage, thus,

$$\mathrm{𝐸_{𝐿} =\sqrt{3}𝐸_{π‘β„Ž} … (7)}$$

Numerical Example

A 3-phase, 50 Hz, star connected alternator has 200 conductors per phase and the flux per pole is 0.0654 Wb. Calculate the following −

  • EMF induced per phase, and

  • EMF between line terminals.

Assume the winding to be full pitched and the distribution factor to be 0.86.


  • EMF induced per phase −

$$\mathrm{𝐸_{π‘β„Ž} = 2.22\:π‘˜_{𝑐}\:π‘˜_{𝑑}\:𝑓\:{φ}\:𝑍}$$

$$\mathrm{\Rightarrow\:𝐸_{π‘β„Ž} = 2.22 \times 1 \times 0.86 \times 50 \times 0.0654 \times 200 = 1248.6\:V}$$

  • EMF between line terminals −

$$\mathrm{𝐸_{𝐿} =\sqrt{3}𝐸_{π‘β„Ž} =\sqrt{3}\times 1248.6 = 2162.6\:V}$$