Chapter 4 AC Circuit Network theorems Prof. Dr. Fahmy ElKhouly
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Chapter 4 AC Circuit Network theorems
Prof. Dr. Fahmy ElKhouly
1 Mesh-current analysis
Mesh and nodal analysis
1212111 EIZZI
0323222121 IZIZZI
2333232 EIZIZ
2
1
3
2
1
3332
232212
1211
0
0
0
E
E
I
I
I
ZZ
ZZZ
ZZ
EIZ
2
0
0
0
332
2312
111
Det
ZE
ZZ
EZ
Det
3
0
0
232
2212
11211
Det
EZ
ZZ
EZZ
Det
ZDet
DetI
11
ZDet
DetI
22 ZDet
DetI
33
10
0
33322
2322
121
Det
ZZE
ZZ
ZE
Det
ZDet
ZZ
ZZZ
ZZ
Det
3332
232212
1211
0
0
Z
XL
R
Φ
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
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