• Supply Frequency (f)
• Number of Poles (P)

The relationship between these factors and the synchronous speed (Ns) is given by the formula:

$$N_{s}=\frac{120f}{P}$$

Where Ns is the synchronous speed in revolutions per minute (RPM), f is the supply frequency in hertz, and P is the number of poles.

For example,

• A 2-pole motor (P=2) running on a 50 Hz supply (f=50) will have a synchronous speed of 3000 RPM.
• A 4-pole motor (P=4) on the same supply will have a synchronous speed of 1500 RPM.

In practical applications, especially for induction motors, the actual operating speed is slightly less than the synchronous speed due to a phenomenon known as slip, which is necessary for torque production.

For synchronous motors, the rotor locks in step with the stator's rotating magnetic field, and thus operates precisely at the synchronous speed.

(b) Given:

• Number of poles, P = 12
• Speed, N = 600 rev/min
• Line voltage, VL = 440 V
• Resistance per phase, R = 35 Ω
• Inductive reactance per phase, XL = 25 Ω
• The alternator is delta connected and supplies a balanced star-connected load.

(i) Frequency of Supply

The frequency of an alternator is given by:

$$f = \frac{P \times N}{120}$$

Substituting the given values:

$$f = \frac{12 \times 600}{120} = 60\ Hz$$

Answer: Frequency = 60 Hz

(ii) Current in Each Coil

Since the load is star connected, the phase voltage is:

$$V_{ph} = \frac{V_L}{\sqrt{3}}$$

$$V_{ph} = \frac{440}{1.732} \approx 254.03\ V$$

The impedance of each coil is:

$$Z = \sqrt{R^2 + X_L^2}$$

$$Z = \sqrt{35^2 + 25^2}$$

$$Z = \sqrt{1850} \approx 43.01\ \Omega$$

The current through each coil is:

$$I_{coil} = \frac{V_{ph}}{Z}$$

$$I_{coil} = \frac{254.03}{43.01} \approx 5.91\ A$$

Answer: Current in each coil = 5.91 A

(iii) Current in Each Phase of the Alternator

For a star-connected load:

$$I_L = I_{coil} = 5.91\ A$$

Since the alternator is delta connected, the phase current is:

$$I_{phase} = \frac{I_L}{\sqrt{3}}$$

$$I_{phase} = \frac{5.91}{1.732} \approx 3.41\ A$$

Answer: Current in each phase of the alternator = 3.41 A

(iv) Total Power Supplied to the Load

First, calculate the power factor:

$$cos\phi = \frac{R}{Z}$$

$$cos\phi = \frac{35}{43.01} \approx 0.814$$

Total three-phase power is given by:

$$P = \sqrt{3} \times V_L \times I_L \times cos\phi$$

$$P = 1.732 \times 440 \times 5.91 \times 0.814$$

$$P \approx 3662.4\ W$$

$$P \approx 3.66\ kW$$

Answer: Total power supplied = 3662.4 W (approximately 3.66 kW)