Phasor Examples PDF

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Teacher: Jones
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EET 300 || Network Analysis || Chapter 8 (B) Lesson Notes || Phasor Domain Example 12/2/2011

PART A)

Find the Thevenin equivalent circuit of the circuit shown using Nodal analysis.

v (t )  5 sin(1000t )

100  F

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10 mH

SOLUTION 1st, convert to the phasor domain: The rotational 1   j 10 Zc  frequency,  , is equal to 1000 radians/sec. So, we j  1000  (100 f ) can convert the capacitor and inductor into the Z L  j  1000  (10mH )  j10 phasor domain: 5

v (j  )  5 0

+ -

10

 j 10

+ vth(t) -

j 10

Next, if we were to short out the source and find the total impedance, ZT , from the viewpoint of the load (also known as Z Thevenins), we would get:

Z th   R1 || ZC R1 || Z C

 R1 || ZC   R2

  R2  || ZL 5   j 10  

5  j 10  4.4721  26.565

 4.4721  26.565  10  14.14   8.13   14  j 2

Z th   R1 || ZC

  R2  || ZL

Zth  8.7715 2.125



 14 .14  8.13  j 10  14  j 2  j 10

EET 300 || Network Analysis || Chapter 8 (B) Lesson Notes || Phasor Domain Example

2 OF 4

12/2/2011

Next, we need to find VTh. Since there is a single node, it might be smart to use KCL.

VA  5v

5 0

 j 10

VA VA  5  j 10 10  j 10 V V VA 0  A  1 A   j 10 10  j 10 5 0 



1  1 1 1   V  j 10 10  j10  A 5 1   254.95m 11.31  VA 1  3.922   11.31  254 .95m  11.31 3.922   11.3 1  j 10    2.7735   33.69 10  j 10

VA  Vth

Zth  5.385  j 6.923

2.774 33.69

+ -

8.771 52.125

j 10

EET 300 || Network Analysis || Chapter 8 (B) Lesson Notes || Phasor Domain Example

3 OF 4

12/2/2011

PART B) Find the Thevenin equivalent circuit of the previous circuit by first finding the Norton equivalent circuit thru the use of Mesh Analysis. SOLUTION When you find the Norton equivalent, Z Thev doesn’t change so we won’t recalculate it.

5 5 0

+ -

I1  j 10

Isc

10 I2

j 10

For I norton , the output is shorted out. Note that when that is done in this circuit, the inductor is shorted out, so we will just leave it out of the future calculations.

5 It should also be noted that I norton is equal to I sc which is equal to I2.

5 0

+ -

I1  j 10

0  5  5I1    j 10  I1  I2  5  5I1    j 10 I1   j 10 I2 5   5  j 10 I1  j 10I 2 0   j 10  I2  I1   10 I2

0    j 10  I2    j 10  I1  10 I2 0  j 10I1   10  j 10 I2

Isc

10 I2

EET 300 || Network Analysis || Chapter 8 (B) Lesson Notes || Phasor Domain Example 12/2/2011

4 OF 4

5  5  j 10  I1  j 10 I2 0  j 10 I 1   10  j 10  I 2 5   5  j10      j 10  0    5  j 10   j 10  5 

 I    1  10  j 10  I2 

j10

j 10

 10  j 10 j 10  10  j 10    j 10   j 10 

  50  j 150  ( 100)   50  j 150  158.11  71.565  2 

5 

j 10 5

j 10

0

 0  5  j 10 

 2   j 50  50  90 2 50   90   158.11  71.56 5 I2  316.23m   18.435

I 2  I sc  I N 

Now that the Norton equivalent has been found, a source conversion is used to convert to the Thevenin equivalent.

Vth  Isc * Zth

 316.23m   18.435  8.77152.125 

Vth  2.773V  33.69 Zth 

2.774 33.69

+ -

5.385  j 6.923 8.771 52.125...