Lab II Lecture Notes: Single Phase Power, Lab Reports of Electrical and Electronics Engineering

Detailed lecture notes for a university lab session on single phase power, including calculations for finding voltage and current phasors, complex power, power angle, power factor, active and reactive power, and instantaneous power.

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Pre 2010

Uploaded on 08/19/2009

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D. Niebur Lab II Lecture on Single Phase Power 8/9/2004
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D. Niebur Lab II Lecture on Single Phase Power 8/9/

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Lab II-Examples

D. Niebur

Monday, August 9, 2004

Example 1.)

The following single phase circuit is given:

Z

Load

ZLine

I

v(t)

The time dependent ac voltage is

vt   339. 4 cos t  60º V

The line impedance is

Z Line ^5 j

and the load impedance is

Z Load  8. 66

Find

1. the voltage phasor V ,
2. the current phasor I ,
3. the time-dependent current signal i  t ,
4. the complex power S injected by the source,
5. the power angle of the complex power S
6. the power factor of the source power.

Is the power factor leading or lagging?

7. Compute the active and reactive power.
8. What power is consumed by the reactive line?
9. What power is consumed by the resistive load?
10. What is the instanteneous power?
11. What is the average instanteneous power?

Solution:

1. the voltage phasor V :

V max  339. 4 V

| V | ^ V^ eff ^ V^ rms ^ V max /^2 ^ 339. 4/^2 ^ 240. 0

V  240 e j /3^ V

2. the current phasor I :
I  V

Z total

We thus have to find the total impedance of the circuit as the sum of 2 serial impendances Z line and Z Load and convert the complex impedance from cartesian to polar coordinates:

Z totalZ lineZ Load  8. 66  5 j  

| Z | ^ 8. 66^2 ^25  ^ 10. 000

angleZ   arctan 5

  1. 66

 0. 523 6 rad  0. 5236  180/  30. 0º  /

I  V

Z total

A  24 ∡30º A
3. the time-dependent current signal i  t 

I max  (^) | I | 2 

it   I max cos tangleZ 

 24 2 cos t  30º

 33. 94 cos t  30º A

4. the complex power S injected by the source:
S  VI ∗^  240  24 ∡60º − 30º VA  5760 ∡30º VA
5. the power angle of the complex power S :
6. the power factor of the circuit.

pt   vtit   240 2 cos t  60º  24 2 cos t  30º

 5760 2 2

cos t^ ^ 60º^ −^ t^ −^ 30º ^ cos t^ ^ 60º^ ^ t^ ^ 60º

 5760 cos30º  cos 2 t  120º

 5760 cos 1 6

 cos 4 60 t  2 3

 4988. 3  5760. 0 cos753. 98 t  2. 094 4

11. What is the average instanteneous power?

We have two ways to compute it:

a. P  T^1 

0

T ptdt

with T  (^22)   1201 

P  1

0

120 cos 1 6

 cos 4 60 t  2 3

 dt

 48 ^1
 4988. 3 W
b. P  Re  S   4988. 3 W