circuit analisis i

8/13/2019 Circuit Analisis I

1/4

Circuit Analysis I with MATLAB Applications 7-9

Orchard Publications

Thevenin’s and Norton’s Theorems

Figure 7.8. Circuit for Example 7.4

Solution:

With the resistor disconnected, the circuit reduces to that shown in Figure 7.9.

Figure 7.9. Circuit for Example 7.4 with the resistor disconnected

By application of the voltage division expression,

(7.16)

and

(7.17)

Then, from (7.16) and (7.17),

or

(7.18)

Next, we find the Thevenin equivalent impedance by shorting the voltage source.

The circuit then reduces to that shown in Figure 7.10.

170 0 V

j100– 85

50

100

j200

I X

100

170 0 V

j100– 85

50 j200

V 1 V 2

100

V 1 j200

85 j200+-----------------------170 0

200 90

217.31 67 ------------------------------170 0 156.46 23 144 j61.13+= = = =

V 250

50 j100–-----------------------170 0

50

111.8 63.4– --------------------------------------- 170 0 76 63.4 34 j68+= = = =

V TH

V OC

V 12

V 1 V

2– 144 j61.13 34 j68+ –+= = = =

V TH 110 j 6.87 – 110.21 3.6 – = =

Z TH 170 0 V


2/4

Chapter 7 Phasor Circuit Analysis

7-10 Circuit Analysis I with MATLAB Applications


Figure 7.10. Circuit for Example 7.4 with the voltage source shorted

We observe that the parallel combinations and are in series as shown in Figure

7.11.

Figure 7.11. Network for the computation of for Example 7.4

From Figure 7.11,

and with MATLAB,

Zth=85*200j/(85+200j) + 50*(100j)/(50100j)

Zth =

1.1200e+002 + 1.0598e+001i

or

The Thevenin equivalent circuit is shown in Figure 7.12.

j100– 85

50 j200

X Y

j100– 85

50 j200

X Y

j100– 85

50 j200

X Y

j100– 85

50 j200

X Y

j100– 85

50 j200

X Y Z

TH

85

50

j200

X

Y

j100–

200 || 85 50 || j100

85 j200

j100 50

Y

X

Z TH

Z TH

Z TH 85 j200

85 j200+-----------------------

50 j100–

50 j100–-------------------------------+=

Z TH 112.0 j10.6 + 112.5 5.4 = =


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Circuit Analysis I with MATLAB Applications 7-11


Thevenin’s and Norton’s Theorems

Figure 7.12. Thevenin equivalent circuit for Example 7.4

With the resistor connected at X-Y, the circuit becomes as shown in Figure 7.13.

Figure 7.13. Simplified circuit for computation of in Example 7.4

We find using MATLAB:

Vth=1106.87j; Zth=112+10.6j; Ix=Vth/(Zth+100);

fprintf(' \n'); disp('Ix = '); disp(Ix); fprintf(' \n');

Ix = 0.5160 - 0.0582i

that is,

(7.19)

The same answer is found in Example C.18 of Appendix where we applied nodal analysis to find

.

Norton’s equivalent is obtained from Thevenin’s circuit by exchanging and its series with

in parallel with as shown in Figure 7.14. Thus,

and

V TH 110.213.6

112 j10.6 X

Y

Z TH

110 j6.87

100

V TH 110 j6.87 100

j10.6

X

Y

112 I X

I X

I X

I X

V TH

Z TH 100 +--------------------------------- 0.516 j0.058– 0.519 6.4 A–= = =

C

I X

V TH Z TH

I N Z N

I N

V TH

Z TH ----------

110.21 3.6 –

112.5 5.4---------------------------------- 0.98 9 A–= = =

Z N Z TH 112.5 5.4 = =


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Chapter 7 Phasor Circuit Analysis

7-12 Circuit Analysis I with MATLAB Applications


Figure 7.14. Norton equivalent circuit for Example 7.4

7.5 Phasor Analysis in Amplifier Circuits

Other circuits such as those who contain op amps and op amp equivalent circuits can be analyzed

using any of the above methods.

Example 7.5

Compute for the circuit below where .

Figure 7.15. Circuit for Example 7.5

Solution:

As a first step, we perform the , to transformation. Thus,

and

Also,

and the phasor circuit is shown in Figure 5.16.

I N

Z N

i X t vin t 2 30000t V cos=

+

+

0.2 mH

8

2

10

50

4

i X t

vC t vin t

5vC t 10 3 F

t domain– domain–

X L j L j0.2 103–

30 103

j6 = = =

jX C – j–1

C -------- j–

1

30 103

10

3------ 10

6 –

-------------------------------------------------- j10–= = =

V IN 2 0=

circuit analisis i

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