(3.3)

Self excited DC machines :

The process of generating a magnetic field by means of an electric current is called excitation. Modern generators with field coils are self-excited, where some of the power output from the rotor is used to power the field coils.

In these type of machines field coils are internally connected with the armature.

Due to residual magnetism some flux is always present in the poles. When the armature is rotated some emf is induced producing small amount of current.

This small current flows through the field coil producing armature emf, which cause further increase of current through the field. This increased field current raises armature emf and this phenomenon continues until the excitation reaches to the rated value.

According to the position of the field coils the Self-excited DC generator may be classified as

A. Series wound generators

B. Shunt wound generators

C. Compound wound generators

(3.4)

Purpose of a DC motor starter :

The primary issue is that DC motors require a large amount of starting current which must be limited to prevent the internal circuit of the armature winding of the DC motor. Hence we make use of DC motor starter is used to prevent the motor.

(4.1)

Step 1:

Primary winding voltage is 320 V.

Secondary winding voltage is 332 V.

Net area of the conductor is 50 sq-cm.

Maximum flux density is 1 tesla.

Calculate Magnetic flux.

Area of the conductor is .

.

is the Magnetic flux.

is the magnetic flux density.

Wb.

Step 2:

Calculate number of turns of each winding.

EMF induced by the primary winding is .

Hence Number of primary turns .

Transformer ratio : .

Number of primary turns .

Solution :

Number of primary turns .

Number of primary turns .

Step 1:

(4.2)

Maximum voltage of a sinusoidal wave is 400 V.

Frequency of a sinusoidal wave is 50 Hz.

4.2.1)

RMS value of the voltage :

RMS voltage of a sinusoidal wave is V.

Step 2:

(4.2.2)

Find the periodic time.

Periodic time is 20 ms.

Solution :

(4.2.1) RMS voltage of a sinusoidal wave is V.

(4.2.2) Periodic time is 20 ms.

Step 1:

(4.3)

The two branches current are A, A and A.

If the polar form of equation is then complex form of .

Complex form of equation is

Complex form of equation is

Complex form of equation is

For parallel connection : .

Hence the resultant current is .

in polar form can be written as .

Hence

Step 2 :

Phasor diagram :

Plot the phasor diagram .

Solution :

Step 1:

(4.4.1)

The two impedance of the circuit are  and .

Voltage applied is 240 V.

The two impedance are in series .

Find the current flowing in the circuit.

Current flowing in the circuit is .

Current flowing in the circuit .

Step 2:

(4.4.2)

Find Power Factor.

Current flowing in the circuit .

in polar form can be written as .

Power Factor of the circuit is .

.

Power Factor is 0.866 lag.

Solution :

Current flowing in the circuit .

Power Factor is 0.866 lag.