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The frequency of the generated voltage by the alternator depends upon the number of field poles and the speed at which the field poles are rotated. One complete cycle of the voltage being generated in an armature coil when a pair of field poles i.e. one north pole and one south pole passes over the coil.

Let,

𝑃 = Number of rotor field poles

𝑁 = Speed of rotor or field poles in RPM

𝑓 = Frequency of the generated voltage in Hz

In one revolution of the rotor, an armature coil is cut by (P/2) north poles and (P/2) south poles. Since one cycle of the voltage is generated in the armature coil when a pair of field poles passes over the coil. Thus, the number of cycles generated in one revolution of the rotor will be equal to the number of pairs of poles i.e.

$$\mathrm{Number\:of\:cycles/Revolution = Number \:of\:pole\:pairs =\frac{𝑃}{2}}$$

$$\mathrm{∵\:Number\:of\:revolutions/second =\frac{𝑁}{60}}$$

$$\mathrm{∴\:Number\:of\:cycles/second =\frac{𝑃}{2} \times \frac{𝑁}{60}}$$

The frequency is defined as the number of cycles per seconds. Therefore,

$$\mathrm{Frequency,\:𝑓 =\frac{Number\:of \:cycles}{second}=\frac{𝑃𝑁}{120}… (1)}$$

Equation (1) gives the relationship between the number of poles, speed and the frequency of an alternator.

The synchronous speed is defined as the speed at which the rotating magnetic field of an electric machine rotates. It is denoted by N_{s} and is given by (from Eq. (1))

$$\mathrm{Synchronous\:Speed,\:𝑁_{𝑆} =\frac{120𝑓}{𝑃}… (2)}$$

For a given alternator, the number of field poles on the rotor are fixed and hence the rotor must be run at synchronous speed to produce an output of desired frequency. Thus, for an alternator, the rotor speed (N) bears a constant relationship with the field poles and the frequency of the generated voltage in the armature winding. The speed given by Eqn. (2) is known as **synchronous speed**. An electrical machine that runs at synchronous speed is known as **synchronous machine**.

What is the frequency of the generated voltage by an alternator having 6 poles and rotating at 1200 RPM?

**Solution**

$$\mathrm{Frequency,\:𝑓 =\frac{𝑁𝑃}{120}=\frac{1200 × 6}{120}= 60\:Hz}$$

Calculate the highest speed at which a 50 Hz alternator can be operated.

**Solution **

Since the alternator operates at synchronous speed which is given by,

$$\mathrm{𝑁_{𝑆} =\frac{120𝑓}{𝑃}}$$

The speed of the alternator will be maximum, when the number of pole (P) is minimum. The minimum possible number of poles is 2. Hence, the maximum speed for a 50 Hz alternator will be,

$$\mathrm{𝑁_{𝑆} =\frac{120𝑓}{𝑃}=\frac{120 \times 50}{2}= 3000\:RPM}$$

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