ECE320, Spring 2006,Exam 1 25Thjanuary, 2006

ECE320, Spring 2006,Exam 1 25thJanuary, 2006

Duration 1 hour

Name SOLUTIONS

Instructions

• Read all the problems first.
• Attempt to solve first the problems you can do-don’t spend too much time on one problem.
• Read carefully the statement of the problems.
• Use the previous page’s back if more space is needed.
• Write neatly and show all the steps, no marks otherwise.

Problem 1. For the circuit below, calculate the load using Thevenin’s theorem.

Solution:

The voltages at nodes A and B can be found by voltage division.

Open Circuit Voltage ()

VAD

VBD

= 0.

Thevenin Impedance ()

Since , no voltage drop appears across terminals A and B; therefore, no current flows through .

.

Problem 2. For a balanced 3-phase circuit below,the source voltage is 208 V. Calculate (1) the real (active) and reactive power of each load, (2) the total real and reactive power of both loads, and (3) the real and reactive power provided by the source. Explain why the source reactive power differs from the total load reactive power. What about the source real power and the total load real power (should they be equal)?

Solution:

Converting delta connected load to star and then considering only a single phase, the circuit will be as shown in figure.

(1)

Total Impedance

We haveA

Using Current DivisionA

A

Load Voltages

V

V

Transmission Line Voltage Drop

A

Load 1

Real PowerkW

Reactive PowerkVAR

Load 2

Real PowerkW

Reactive PowerkVAR

(2)Total Real and Reactive Power of Both Loads:

kW

(3)The Real and Reactive Power Provided by the Source:

kW

kVAR

Source Reactive Power differs from the total load reactive power (which is zero in this case) because it is the power associated with the transmission line.

Transmission Line Reactive Power = 3 (11.94) (59.7) sin (84.2+5.72) = 2.138 kVAR

Or = 3 (59.7 A)2 (0.2 ) = 2.138 kVAR

Since, Transmission Line Real Power = 0

Therefore, Source Real Power = Total Load Real Power.

Problem 3

In a 3-phase system below (Fig. a), the source is 230V and the load (Load_1) is drawing 50kW at 0.866pflagging.

a)A second load (Load_2) is added and its real power is equal to 30kW. What should the pf of the second load be in order for the total pf of the both loads together to be unity?

b)Draw a power triangle diagram to represent the above-obtained real, reactive, and apparent power of the source, Load_1, and Load_2.

Solution:

(a)kW

kVA

kVAR

kW[given]

kW

kVA

Therefore,, since

(b)Power Triangles

Load 1Load 2