Chapter 4 AC Circuit Network theorems - Delta Univ · 2016. 3. 8. · AC Circuit Network theorems...
Transcript of Chapter 4 AC Circuit Network theorems - Delta Univ · 2016. 3. 8. · AC Circuit Network theorems...
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Chapter 4 AC Circuit Network theorems
Prof. Dr. Fahmy ElKhouly
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1 Mesh-current analysis
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2 Nodal analysis
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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
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(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
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(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.
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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
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(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
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(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
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(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
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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.
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This value may be checked since total active power dissipated is given
by:
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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
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(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
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(iv) The Thévenin equivalent circuit
is shown in Figure 21 connected to
the (48+ j 144)W load
Figure 21
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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,
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(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
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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
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(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
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(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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