Electrical Circuits I Lecture 8

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Transcription:

Electrical Circuits I Lecture 8 <Dr hmed El-Shenawy>

Thevenin and Norton theorems Thevenin theorem tells us that we can replace the entire network, exclusive of the load resistor, by an equivalent circuit that contains only an independent voltage source in series with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged. Norton theorem tells us that we can replace the entire network, exclusive of the load resistor, by an equivalent circuit that contains only an independent current source in parallel with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged. i o i o i o v o - v o - v o - (a) complex network including a load resistor R L. (b) Thevenin equivalent network (c) Norton equivalent network 2

i o v o - How to determine v TH (t) and R TH for a particular circuit. It is helpful to note that if we connect no load and therefore i o (t) = 0, then we can determine v TH (t) from v TH ( t) v ( t) v ( t) opencircui where v oc (t) is called the open circuit voltage If we short circuit the two terminals to force v o (t) = 0, then we get v R TH ( t) TH i shortcircu If v TH (t) 0, then i shortcircuit (t) 0 and we find it t ( t) i sc oc ( t) R TH v i oc sc ( t ) ( t ) 3

Case 1 example 1 - Find the Thevenin equivalent circuit at terminal pair a and b for the circuit shown. This specific problem can be solved by using different approaches. We solve the problem by using source transformation technique. Thus we have R TH = 12 and v TH = -8V 4

Network 1 B We now deactivate all sources of Network 1. To deactivate a voltage source, we remove the source and replace it with a short circuit. To deactivate a current source, we remove the source.

Find V X by first finding V TH and R TH to the left of -B. 12 4 30 V _ 6 2 V X _ B First remove everything to the right of -B.

12 4 30 V _ 6 B VB (30)(6) 6 12 10V Notice that there is no current flowing in the 4 resistor (-B) is open. Thus there can be no voltage across the resistor.

We now deactivate the sources to the left of -B and find the resistance seen looking in these terminals. 12 4 6 R TH B We see, R TH = 12 6 4 = 8

fter having found the Thevenin circuit, we connect this to the load in order to find V X. R TH 8 V TH _ 10 V 2 V X _ B VX ( 10)( 2) 28 2V

For the circuit below, find V B by first finding the Thevenin circuit to the left of terminals -B. 1.5 5 10 20 V _ 20 17 B We first find V TH with the 17 resistor removed. Next we find R TH by looking into terminals -B with the sources deactivated.

1.5 5 10 20 V _ 20 B V V V (1.5)(10) OS B TH V TH 31V 20(20) (20 5)

5 10 20 B 5(20) RTH 10 14 (5 20)

R TH 14 V TH _ 31 V 17 V B _ B We can easily find that, VB 17V

Norton theorem i o Based on source transformation we have learned, we can determine i N (t) and R N (c) v o - i R N N R TH v i oc sc ( t ) ( t ) vth ( t) ishortcircuit ( t) isc ( t) R TH

Norton theorem Find the Norton equivalent circuit to the left of terminals -B for the network shown below. Connect the Norton equivalent circuit to the load and find the current in the 50 resistor. 10 20 40 _ 50 V 60 50 B

Norton theorem 10 20 40 _ 50 V 60 I SS It can be shown by standard circuit analysis that I SS 10.7

Norton theorem It can also be shown that by deactivating the sources, We find the resistance looking into terminals -B is R 55 N R N and R TH will always be the same value for a given circuit. The Norton equivalent circuit tied to the load is shown below. 10.7 55 50

Norton theorem

Norton theorem