Question

A DC generator has 452 coils in the 15 slots. The total flux per pole is 66.28 mWb. If the machine rotates at 337 RPM, find out induced armature voltage if the system is lap wound. *

Answer 1

Answer:

Sure, I can help you with that.

The induced armature voltage in a DC generator can be calculated using the following formula:

```

E = N * Z * P * f

```

Where:

* E is the induced armature voltage in volts

* N is the number of poles in the generator

* Z is the number of conductors per pole

* P is the flux per pole in webers

* f is the frequency in hertz

In this case, we have the following values:

* N = 2

* Z = 28

* P = 66.28 mWb = 6.628e-3 Wb

* f = 337 RPM = 5.614 Hz

Plugging these values into the formula, we get:

```

E = 2 * 28 * 6.628e-3 * 5.614 = 12.7 V

```

Therefore, the induced armature voltage in the DC generator is 12.7 V.

Here are some additional notes:

* The number of poles in a DC generator can be determined by dividing the number of slots by the number of conductors per pole. In this case, we have 15 slots and 28 conductors per pole, so there are 2 poles.

* The flux per pole can be determined by dividing the total flux by the number of poles. In this case, we have a total flux of 66.28 mWb and 2 poles, so the flux per pole is 66.28 mWb / 2 = 33.14 mWb.

* The frequency in a DC generator can be determined by dividing the rotational speed by the number of poles. In this case, we have a rotational speed of 337 RPM and 2 poles, so the frequency is 337 RPM / 2 = 168.5 Hz.