Worked examples

WORKED EXAMPLES

Case (a): Balanced star connected loads

Example 1 A balanced star connected load of (8+ j6) phase is connected to a 3 phase, 230V, 50c/s supply. Find the line current, power factor, power, reactive volt amperes and total volt amperes.

Solution:

 

Example 2 A balanced star connected load of (3 - j4)2 impedance is connected to 400 V three phase supply. What is the real power consumed?

Solution: Note: For Star system, unless specified, the given voltage is taken as line voltage.

Case (b): Balanced Delta Connected Load: wel 'mdo ve

Example 3 A synimetrical 3 phase, 400 V system supplies a balanced mesh connected load. The current in each branch circuit is 20A and the phase angle is 40° lag. Find (a) the line current (b) the total power.

Solution: Data: Type of connection = delta

 

Example 4 Three impedances each of 10 resistance and 52 inductive reactance are connected in delta to a 400 V, 36 supply. Determine the current in each phase and in each line. Calculate also the total power drawn from the supply and the p.f of the load.

Solution:

 

Case (c): 3-phase unbalanced delta load fed by 3 phase balanced supply.

(i) In this case, for a given phase sequence, the phase voltages are expressed in polar form.

(ii) Load impedance in each phase is expressed in polar form.

(iii) The phase current is obtained by ohm's law. The line currents are calculated by applying Kirchoff's current law.

Example 5 Determine the line currents for the unbalanced delta connected load of figure given. Assume the phase sequence to be RYB. E = 200 volts.

Solution :

E_RY = 200 ∠ 0^o

I_R, I_Y and I_B are the line currents as shown in the circuit diagram.

Example 6 A three phase delta connected loud has Z_ab= (100+ j0) olms. Z_bc = (-j100) and Z_ca = (70.7+j70.7) ohms and is connected to a balanced 3 phase 400V supply. Determine the line currents I_a, I_b and I Assume the phase sequence as abc.

Solution:

For a phase sequence abc

E_ab = 400 ∠ 0°V

E_bc = 400 ∠ -120°V, and

E_ca 400 ∠ 120°V

The phase currents lab, Ibc, Ica are calculated as below.

Now, the line currents Ia, Ib, Ic are computed by applying KCL.

 [Note: From equations (i), (ii), (iii), we observe that Ia+Ib+ Ic=0. That is the phasor sum of the three line currents is zero.]

 

Example 7 The two line currents taken by an unbalanced delta connected load are Ia = 10-120° amps and I = 5/150° amps. What is the current I_C?

Solution:

I_a + I_b+ I_c = 0

I_c = - I_a – I_b

= -10 - 120°- 5150°

= 10 ∠ 60° - 5 ∠ 150°

= (5 + j8.66) - (-4.33 + j 2.5)

= 9.33 + j 6.16

= 11.18 ∠ + 33.4° Amps

 

Example 8 A delta load with Z_AB = 10 ∠ 30° Ω, Z_BC=25 20° and Z_CA = 20 ∠ -30° is connected to a 3 phase 500 volt. ABC system. Find the line currents and total power.

Solution: Let the phase sequence be ABC..

Let E_AB = 500 ∠ 0° volts, E_BC = 500 ∠ -120° and E_CA = 500 ∠ 120°.

To calculate the phase currents I_AB I_BC and I_CA:

The line currents IA, IB and IC are calculated as below by applying KCL.

Total power = Power in Z_AB + Power in Z_BC + Power in Z_CA

= (500 × 50 cos 30°) + (500 × 20) cos 0° + 500 × 25 cos 30°

= 42476 W = 42.476 KW

[To check: Total power = sum of powers dissipated in the resistance parts of the impedances

= 50² × 8.66 + 20² (25) + 25² (17.32) = 42476 W]

Case (d): Unbalanced-Star connected load fed by 3 phase balanced supply.

 (i) 4 wire star connected unbalanced load: (When load star point is connected to source neutral)

In a 3 phase, 3 wire star connected load circuits, the star point of load and generator are tied together through neutral wire (conductor) of zero impedance. So, both star points are at the same potential. Voltage across each impedance is same and equal to phase voltage of the source, irrespective of the fact that the circuit is balanced or unbalanced. The three phase currents or line currents are determined by ohm's law as below.

I_R = E_RN / I_RN, I_Y = E_YN / Z_YN and I_B = E_BN / Z_BN

By applying KCL at the star point, the current through the O neutral wire is given by

I_N + I_R + I_Y + I_B = 0

⇒ I_N = (IR + I_Y + I_B)

If this load is balanced I_N = 0, hence the neutral wire is omitted.

[Note: If the neutral wire is disconnected in a 3 phase 4 wire unbalanced star load, the potential difference across high resistance is increased and that across low resistance is decreased.]

(ii) Unbalanced star connected load: (without neutral):

When a star connected load is unbalanced, having no neutral wire, then its star point is isolated from the star point of generator. The potential of the load star point is different from that of generator star point. The potential of the former is subjected to variations according to the imbalance of the load. This isolated load star point or neutral point is called a floating neutral point, because its potential is always changing and is not fixed.

Such unbalanced star connected loads having isolated neutral points can be analysed by the following four method:

(a) Star - delta conversion

(b) Application of Kirchoff's laws

(c) By Millmans theorem

(d) Loop analysis or Nodal analysis

 

Example 9 A 3-phase, 4 wire, 208 V, ABC system supplies a star connected load in which Z_A = 10 ∠ 10°Ω Z_B = 15 ∠ -30° Ω Find the line currents, the neutral currents and the load power.

Solution: Let us assume that the phase sequence be positive i.e., ABC, let the angle of voltage EAN be taken at angle of 90°.

[Note: We can take EAN as at angle 0 also.]

As the supply system is 4-wire, it will have neutral point. The star point of the load system is shown. It is assumed to be connected to neutral point of the supply. Hence, the load phase voltage is same as supply phase voltage. So, we can calculate the currents IA, IB, IC and IN as shown below: The given voltage 208 volts is the line voltage.

This current is in the direction shown.

Calculation of powers in the load:

Powers are calculated using the formula, power = P = I²_R

Power in load Z_A = | IA |2 RA

 

Example 10 An unbalanced star connected load shown in the fig. 4.17 is supplied from a 3 phase to the 440 volts symmetrical system. Determine the line currents and the power input to the circuit.

Solution: Let us solve this problem by mesh current method. Let I1 and I2 be the loop currents as shown. Assume the phase sequence to be RYB. Taking ERN as reference, we can express the various voltages as below:

 

Example 11 A star connected load with neutral isolated, has impedances of Z_A = 15 = Ω ; Z_B = 10 ∠ 30° Ω ; and Zc =15 ∠ 0° Ω. The supply voltage is 230V, ABC system. Taking E_BC as reference, determine the line currents and the voltage across the load impedances.

Solution: Given that EBC is the reference. So, the various supply voltages are shown in the polar form as below.

E_BC = 230 ∠ °V

E_CA = 230 ∠ -120°V

and E_AB = 230 ∠ 120° V

As the neutral is isolated, the load phase voltage is not equal to supply phase voltage. Hence, this type of problem can be solved by many ways like loop current method, Star Delta conversion method, Kirchoff's laws methods and application of Milliman's theorem. Let us apply here Star-Delta conversion method.

Let Z_AB, Z_BC, Z_CA be the impedances of equivalent Delta.

Now, refer the following fig. 4.19,

Now, the line currents are found as below by applying KCL :

These line currents will be the currents flowing in the given circuit also. i.e., star connection given.

To determine the voltages across the load impedances:

 [Note: Actually this method is a laborious one. We can prefer loop current method or Milliman's theorem method. The student is advised to solve the problem by other methods also.]

 

Example 12 A wye load with Z_A = 3 + j0, Z_B = 2 + j3 and Z_C = 2 - j1 Ω is connected to a 3 phase 4 wire, 100 volts, CBA system. Find the currents in all the four lines.

Solution:

As the star point of load is connected to neutral point of supply, potential between supply neutral point and load star point is zero. [Assuming that the impedance of neutral wire is zero.]