Chapter 4 AC Circuit Network theorems - Delta Univdeltauniv.edu.eg/new/engineering/wp-content/uploads/chap...If the network of Figure 18 is superimposed on the network of Figure 17,

Chapter 4 AC Circuit Network theorems

Prof. Dr. Fahmy ElKhouly

1 Mesh-current analysis

Mesh and nodal analysis

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0323222121 IZIZZI

2333232 EIZIZ

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I

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2

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3

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11

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22 ZDet

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Poller form

Complex form

2 Nodal analysis

Superposition analysis of AC Circuit

Problem 3.

Use the superposition theorem to obtain the current flowing in the (4+ j 3) impedance of Figure16.

Figure16

(i) The network is redrawn with V2 removed

(ii) Current I1 and I2 are shown in Figure17. (4+ j 3) in

parallel with −j 10 gives an equivalent impedance of

Total impedance of Figure 17 is Figure17

(iii) The original network is redrawn with V1 removed, as shown in Figure

18.

(iv) Currents I3 and I4 are shown in Figure 18. 4W in parallel with (4+ j 3)W gives an equivalent impedance of

Total impedance of Figure 18 is

I3

Figure 18.

If the network of Figure 18 is superimposed on the network of Figure

17, it can be seen that the current in the (4 + j 3)W impedance is given

by I2 − I4.

Problem 4.

For the a.c. network shown in Figure 21 determine, using the superposition

theorem, (a) the current in each branch, (b) the magnitude of the voltage

across the (6+ j 8)W impedance, and (c) the total active power delivered to

the network.

Figure 21

(a) (i) The original network is redrawn with E2 removed, as shown in

Figure 22.

(ii) Currents I1, I2 and I3 are labelled as shown in Figure 22. (6+ j 8)W in

parallel with (2− j 5) W gives an equivalent impedance of

Figure 22 Figure 23

(iii) The original network is redrawn with E1 removed, as shown in Figure 24

(iv) Currents I4, I5 and I6 are shown labelled in Figure 24, (3+ j 4)W in

parallel with (6+ j 8)W gives an equivalent impedance of

Figure 24 Figure 25

(v) If Figure 24 is superimposed on Figure22, the resultant currents

are as shown in Figure 26.

(vi) Resultant current flowing from

(5+ j 0)V source is given by

Resultant current flowing from (2+ j 4)V source is given by

Figure 26

Resultant current flowing through the (6+ j 8)W impedance is given by

(b) Voltage across (6+ j 8)W impedance is given by

(c) Total active power P delivered to the network is given by

P = E1(I1 + I6)cosφ1 + E2(I3 + I4)cosφ2

where φ1 is the phase angle between E1 and (I1 + I6) and φ2 is the phase

angle between E2 and (I3 + I4), i.e.

This value may be checked since total active power dissipated is given

by:

Thévenin’s and Norton’s analysis of

AC Circuit

Problem 3. Use Thévenin’s theorem to determine the power

dissipated in the 48 W resistor of the network shown in Figure 19.

Figure 19

The power dissipated by a current I flowing through a resistor R is given by I 2R, hence initially the current flowing in the 48W resistor is required.

(i) The (48+ j 144)W impedance is initially removed from the

network as shown in Figure 20

Figure 20

(ii) From Figure 20

Figure 20

(iii) When the 50∠0◦V source shown in Figure 20 is removed, the

impedance, z, is given by

(iv) The Thévenin equivalent circuit

is shown in Figure 21 connected to

the (48+ j 144)W load

Figure 21

Norton’s theorem:

Problem 9. Use Norton’s theorem to determine the value of current I in

the circuit shown in Figure 56.

Figure 56

(i) The branch containing the 2.8 W resistor is

short circuited, as shown in Figure 57

Figure 57

(ii)The network reduces to that shown in

Figure 58, where ISC=5/2=2.5A Figure 58,

(iii) If the 5V source is removed from the network the input impedance,

z, ‘looking-in’ at a break made in AB gives z=(2×3)/(2+3)=1.2

Figure 59

Figure 60

(iv) The Norton equivalent network is shown in Figure 60, where

current I is given by

Problem 11. Use Norton’s theorem to determine the magnitude of

the p.d. across the 1 resistance of the network shown in Figure 64.

Figure 64

(i) The branch containing the 1 W resistance is initially short-

circuited, as shown in Figure 65

Figure 65

(ii) 4 W in parallel with −j 2 W in parallel with

0 giving the equivalent circuit of Figure 66.

Hence

ISC=10/4=2.5A.

Figure 66

(iii) The 10V source is removed from the network of Figure 64, as

shown in Figure 67, and the impedance z, ‘looking in’ at a break made

in AB is given by

Figure 67

(iv) The Norton equivalent network is shown in Figure 68, from

which current I is given by

Figure 68

Hence the magnitude of the p.d. across the 1W resistor is given

by

IR = (1.58)(1) = 1.58V