Step 1:

A sinusoidal alternating current supply has a maximum value of 396.04 V and a periodic time of 50 milliseconds.

The maximum/peak voltage is .

Time period of sinusoidal function is ms.

(2.1.1)

Find RMS value of the voltage.

The RMS voltage is .

Where is peak voltage.

(2.1.2)

Find the average value of the voltage.

The average voltage is .

Where is peak voltage.

(2.1.3)

Find the frequency.

Frequency is resiprocal of time period.

Where is time period.

(2.1.4)

Find the instantaneous value after two milliseconds.

The instantaneous voltage V_i value after a time of 2ms is given as:

Where is angular frequency,

             t  is time.

The instantaneous voltage V_i value after a time of 2ms

 A 48 kVA 4 800/48 V, 50 Hz single-phase transformer has 3000 turns on the primary winding

 (2.2)

Transformer rating 48 kVA.

Frequnecy is .

Number of turns in primary winding is .

Primary voltage is .

Secondary voltage is .

(2.2.1)

Find the turn ration.

Turn ratio is given as .

The turns ratio is 1:100.

(2.2.2)

Find the number of secondary winding turns.

The number turns can be find using the formula .

Substitute , and .

The number of turns in secondary winding is 30.

(2.2.3)

Find the secondary full-load current.

The formula for full-load current is .

Secondary voltage is .

Transformer rating 48 kVA.

Secondary full-load current is

The secondary full-load current is 577.35 A.

(2.2.4)

 Find the maximum value of the core flux.

Formula for the maximum core flux is .

Substitute , and .

The maximum value of the core flux is 7.202 m Wb.

(2.3.1)

The two branches current are and .

Write the equation in complex form.

Polar form to complex form :

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

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 :

.

(1.1)

Step 1:

For a four pole generator :

The number of conductors Z is 315 conductors.

Armature resistance is 0.6 .

Brush resistance is 0.4 .

The total armature resistance R_a is 0.6 + 0.4 = 1 .

Flux per pole is 0.1 Wb.

Supplies load of 50 kW at 1000 v.

Armature current is .

The EMF of a DC generator : ,

is the emf of the DC Generator.

.

where Z is the number of conductors.

N is the speed of the DC generator.

is the magnetic flux.

P is the number of magnetic poles.

a is the number of parallel circuits.

Step 2:

Calculate the emf of the DC Generator.

Emf of the DC Generator is 950 v.

EMF of a DC generator : .

For a four pole generator : P = 4

For wave wound winding : a = 2

Speed of the DC generator is 905 rpm.

Soltuion :

Speed of the DC generator is 905 rpm.

  A short-shunt compound generator supplies a load of 100 A. It has a shunt-field resistance of 20 ohms, an armature resistance of 0.3 ohm and a field resistance of 0.2 ohm. Determine the armature EMF if the terminal voltage is 180 V.    

The short-shunt compound generator supplies a load current of 100 A.

Shunt field resistance is 20 .

Armature resistance is 0.3 .

Field resistance is 0.2 .

Terminal voltage is 180 v.

Construct a circuit daigram of short-shunt compound generator with specifications.

EMF generated by the DC generator is , where .

Step 2:

Calculate .

Armature current .

EMF generated by the DC generator is .

.

EMF generated by the DC generator is .

Solution:

EMF generated by the DC generator is .