EE 201 ELECTRIC CIRCUITS
LECTURE 29
The material covered in this lecture will be as follows:
⇒ Series and Parallel Impedance
⇒ Series and Parallel Admittance
⇒ Resistance, Reactance, Conductance, Susceptance
⇒ Phase Relations in Sinusoidal Circuits
⇒ More Examples of Sinusoidal Circuits
At the end of this lecture you should be able to:
⇒ Combine Impedances and Admittances in series and in parallel
⇒ Calculate the resistance, reactance, conductance and susceptance
⇒ Identify resistive, capacitive and inductive impedances
⇒ Determine the phase relationship between the circuit variables
⇒ Recognize that current and voltage in a resistor are always in phase
⇒ Recognize that current always leads voltage in the capacitor by 90o
⇒ Recognize that current always lags voltage in the inductor 90o
⇒ Solve more involved sinusoidal circuit problems
Series and Parallel Impedance:
For the series impedances Z1 , Z 2 , … Z n
Z1
⇒
Z eq = Z1 + Z 2 + .... + Z n
Z2
Z eq
Figure 1
Zn
For the parallel impedances Z1 , Z 2 , … Z n
Z eq
⇒
Z1
1
1
1
1
= +
+ .... +
Z eq Z1 Z 2
Zn
Z2
Zn
Figure 2
Series and Parallel Admittance:
For the series admittances
Y1 , Y2 , … Yn
⇒
Y1
1
1 1
1
= + + .... +
Yeq Y1 Y2
Yn
Y2
Yn
Y eq
Figure 3
For the parallel admittances Y1 , Y2 , … Yn
Y eq
Y1
⇒
Yeq = Y1 + Y2 + .... + Yn
Y2
Yn
Figure 4
Example 1:
Calculate the impedance to the right of “a-b” at the angular frequency ω = 1000 [ rad / s ]
Figure 5
Solution:
Calculate the impedance of each element:
4Ω
⇒
4Ω
3Ω ⇒ 3Ω
0.25mF
⇒
1
−j
−j
=
=
= − j 4Ω
jωC ωC 1000 × (0.25 ×10−3 )
0.5mF
⇒
−j
−j
=
= − j 2Ω
ωC 1000 × (0.5 ×10−3 )
3mH
⇒
jω L = j × 103 (3 × 10−3 ) = j 3Ω
a
Z eq
b
-j4
j3
4
3
-j2
ω = 103 [rad / s ]
Figure 6
− j 4Ω & j 3Ω (in series) ⇒
Z1 = − j 4 + j 3 = − jΩ
3Ω − j 2Ω ⇒
Z2 =
3 × (− j 2)
−6 j
6 −90o
=
=
= 1.664 −90o + 33.690o = 1.664 −56.310o Ω
o
3 + (− j 2) 3 − j 2 3.606 −33.690
Z1 = − j Ω
a
Z 2 = 1.66∠− 56.31o Ω
Z eq
b
4
Figure 7
Z1 = − jΩ & Z 2 = 1.664 −56.310o Ω (in series) ⇒
Z 3 = Z1 + Z 2 = (− j ) + (1.664 −56.310o )
Z 3 = (− j ) + (0.923 − j1.385) = 0.923 − j 2.385Ω
Z 3 = 0.923 − j 2.385Ω
Z eq
Figure 8
4Ω Z 3
⇒
Z eq =
4 × Z 3 4 × (0.923 − j 2.385) 4 × (2.557 −68.843o ) 10.228 −68.843o
=
=
=
4 + Z 3 4 + (0.923 − j 2.385)
4.923 − j 2.385
5.470 −25.848o
∴ Z eq = 1.870 −68.843o + 25.848o = 1.870 −42.995o = 1.368 − j1.275 Ω
Z eq
1.368 − j1.275Ω
Figure 9
Resistance, Reactance, Conductance, Susceptance:
In general, the impedance Z is complex, it can be written as:
Z = R + jX
Where:
R = resistance (the real pat of Z )
X = reactance (imaginary part of Z )
Z = R + jX
Figure 10
Also, the admittance Y is complex in general:
Y = G + jB
Where:
G = conductance (the real pat of Y )
B = susceptance (imaginary part of Y )
Y = G + jB
Figure 11
Resistive, Capacitive, Inductive Impedance:
The impedance Z = R + jX has the following special cases:
1) X = 0
⇒ (e.g. Z = 6 Ω )
2) X < 0
⇒ (e.g. Z = 3 − j 5 Ω ) (capacitive impedance)
3) X > 0
⇒ (e.g. Z = 4 + j10 Ω ) (inductive impedance)
Z = 6Ω
(resistive impedance)
Z = 3-j5 Ω
Figure 12
Example 2:
Assuming a source frequency of 60 Hz , calculate:
a) the impedance
b) the resistance & the reactance
Z = 4 + j10 Ω
c) is the impedance resistive, capacitive or inductive?
d) the admittance
e) the conductance & the susceptance
Figure 13
Solution:
ω = 2π f = 2π × 60 = 120π = 377 [ rad / s ]
a) For the resistors ⇒ impedance = resistance
2mF ⇒
5mH
⇒
−j
−j
=
= − j1.326Ω
ωC 377 × (2 × 10−3 )
jω L = j × 377(5 ×10−3 ) = j1.885Ω
Figure 14
Y1 =
1
= 0.333 S
3
Y2 =
1
= j 0.754 S
− j1.326
Y3 =
1
= − j 0.531 S
j1.885
Y3 = − j 0.531S
Y2 = j 0.754 S
Y1 = 0.333 S
Figure 15
Yx = Y1 + Y2 + Y3 = 0.333 + j 0.754 − j 0.531 = 0.333 + j 0.223 = 0.4 33.809o S
Yx = 0.4∠ 33.81o S
Figure 16
1
1
1 0o
Zx = =
=
= 2.5 −33.809o = 2.077 − j1.391 Ω
o
o
Yx 0.4 33.809
0.4 33.809
Z x = 2.077 − j1.391Ω
Figure 17
Z eq = 2 + Z x = 2 + (2.077 − j1.391) = 4.077 − j1.391 Ω
b) Z eq = Req + jX eq = 4.077 − j1.391
⇒
Req = 4.077 Ω & X eq = −1.391 Ω
c) the impedance is capacitive, because X eq = −1.391 < 0 (negative reactance)
4.077 − j1.391 Ω
Z eq
Figure 18
d) Yeq =
1
1
1
=
=
= 0.232 18.839o = 0.220 + j 0.075 S
o
Z eq 4.077 − j1.391 4.308 −18.839
e) Yeq = Geq + jBeq = 0.220 + j 0.075
⇒
Geq = 0.220 S
& Beq = 0.075 S
0.220 + j 0.075 S
Y eq
Figure 19
For sinusoidal circuits (in the phasor representation), it is easy to conclude:
1) The three basic circuit laws are applicable:
V =I Z
(equivalent to Ohm’s law)
KCL
(Kirchhoff’s Current Law)
KVL
(Kirchhoff’s Voltage Law)
2) Almost all the techniques covered in the D.C. part of this course are applicable. This includes:
VDR (Voltage Divider Rule)
CDR (Current Divider Rule)
NA (Nodal Analysis)
MA (Mesh Analysis)
SP (Superposition)
ST (Source Transformation)
TEC (Thevenin Equivalent Circuit)
NEC (Norton Equivalent Circuit)
To gain further experience with sinusoidal circuits, we need to solve more examples
Example 3:
Calculate:
a) i (t )
b) vL (t )
Figure 20
Solution:
a) Transform to phasor representation:
3H ⇒
j 90Ω
1mF ⇒
− j 33.333Ω
v1 (t ) = 5sin(30t ) = 5 cos(30t − 90o ) ⇒ V1 = 5 −90o
v1 (t ) = 4 cos 30t ⇒ V2 = 4 0o
i (t ) ⇒
I
vL (t ) ⇒ VL
KVL & Ohm’s Law ⇒
−V1 + ( j 90) × I + 60 × I + (− j 33.333) × I + V2 = 0
−5 −90o + ( j 90) × I + 60 × I + (− j 33.333) × I + 4 0o = 0
( j 90 + 60 − j 33.333) × I + 4 0o − 5 −90o = 0
(60 + j 56.667) × I + 4 − (− j 5) = 0
(60 + j 56.667) × I + 4 + j 5 = 0
(60 + j 56.667) × I = −(4 + j 5)
(4 + j 5)
6.403 51.340o
I =−
=−
= −0.0776 7.976o
o
60 + j 56.667
82.530 43.364
The minus sign can be removed by adding ±180o (add −180o , because the resulting angle has the
lowest magnitude)
∴ I = −0.0776 7.976o = 0.0776 7.976o − 180o = 0.0776 −172.024o [ A]
I = 0.0776 −172.024o [ A] ⇒
i (t ) = 0.0776 cos(30t − 172.024o ) [ A]
b) VL = ( j 90) × I = ( j 90) × (0.0776 −172.024o ) = (90 90o ) × (0.0776 −172.024o )
∴ VL = 6.984 −82.024o [V ]
⇒
vL (t ) = 6.984 cos(30t − 82.024o ) [V ]
VL
I
V1 = 5∠ − 90o
V 2 = 4∠0o
Figure 21
Phase Relationship:
Given the two sinusoidal functions:
f1 (t ) = A1 cos(ωt + θ1 ) &
f1 (t ) = A2 cos(ωt + θ 2 )
where A1 & A2 are assumed to be positive quantities, then:
The positive quantity θ1 − θ 2 = the phase difference between f1 (t ) &
f 2 (t )
1) if θ1 > θ 2
f1 (t ) leads f 2 (t ) by θ1 − θ 2
⇒
[or equivalently, we can say that f 2 (t ) lags f1 (t ) by θ1 − θ 2 ]
2) if θ1 = θ 2
⇒
f1 (t ) & f 2 (t ) are in phase [the phase difference is zero]
3) if θ1 − θ 2 = 180o
⇒
f1 (t ) & f 2 (t ) are 180o out of phase [the phase difference is 180o ]
Example 4:
a) Calculate all the unknown quantities
b) Describe the phase relationship between is (t ) & iR (t )
c) Repeat (b) for iR (t ) & vR (t )
d) Repeat (b) for i (t ) & vL (t )
e) Repeat (b) for i (t ) & vc (t )
f) Repeat (b) for vc (t ) & vL (t )
Figure 22
Solution:
a) 0.2H ⇒ j 2Ω
0.02F ⇒ − j 5Ω
is (t ) = 5 cos(10t − 40o ) [ A] ⇒
I s = 5 −40o [ A]
iR (t ) ⇒ I R
&
i (t ) ⇒ I
vL (t ) ⇒ VL
&
vc (t ) ⇒ Vc
Z = ( j 2) + (− j 5) = − j 3Ω [because j 2Ω & − j 5Ω are in series]
4Ω Z
IR =
⇒ use CDR (Current Divider Role)
Z
− j3
3 −90o
× Is =
× (5 −40o ) =
= 0.6 −53.130o [ A]
o
4+Z
4 + (− j 3)
5 −36.870
I R = 0.6 −53.130o [ A]
⇒
iR (t ) = 0.6 cos(10t − 53.130o ) [ A]
VR = 4 I R = 4 × (0.6 −53.130o ) = 2.4 −53.130o ) [V ]
∴ vR (t ) = 2.4 cos(10t − 53.130o ) [V ]
CDR
⇒
4
4
4 0o
o
I =
× Is =
× (5 −40 ) =
= 0.8 36.870o [ A]
o
4+Z
4 + (− j 3)
5 −36.870
I = 0.8 36.870o [ A]
⇒
i (t ) = 0.8cos(10t + 36.870o ) [ A]
VL = ( j 2) × I = ( j 2) × (0.8 36.870o ) = (2 90o ) × (0.8 36.870o )
∴ VL = 1.6 126.870o ) [V ]
⇒
vL (t ) = 1.6 cos(10t + 126.870o ) [V ]
Vc = ( − j 5) × I = (− j 5) × (0.8 36.870o ) = (5 −90o ) × (0.8 36.870o )
∴ Vc = 4.0 −53.130o ) [V ]
⇒
vc (t ) = 4.0 cos(10t − 53.130o ) [V ]
b) is (t ) = 5 cos(10t − 40o ) [ A] & iR (t ) = 0.6 cos(10t − 53.130o ) [ A]
∴ is (t ) leads iR (t ) by −40 − (−53.130) = 13.130o
[we can also say that: iR (t ) lags is (t ) by 13.130o ]
c) iR (t ) = 0.6 cos(10t − 53.130o ) [ A] & vR (t ) = 2.4 cos(10t − 53.130o ) [V ]
∴
iR (t ) & vR (t ) are in phase [in the resistor iR (t ) & vR (t ) are always in phase, why?]
d) i (t ) = 0.8cos(10t + 36.870o ) [ A] & vL (t ) = 1.6 cos(10t + 126.870o ) [V ]
∴
i (t ) lags vL (t ) by 126.870 − 36.870 = 90o
[in the inductor, the current always lags the voltage by 90o . Why?]
e) i (t ) = 0.8cos(10t + 36.870o ) [ A] & vc (t ) = 4.0 cos(10t − 53.130o ) [V ]
∴
i (t ) leads vc (t ) by 36.870 − (−53.870) = 90o
[in the capacitor, the current always leads the voltage by 90o . Why?]
f) vL (t ) = 1.6 cos(10t + 126.870o ) [V ] & vc (t ) = 4.0 cos(10t − 53.130o ) [V ]
The phase difference = 126.870 − (−53.870) = 180o
∴
vL (t ) & vc (t ) are 180o out of phase
I
IR
I S = 5∠ − 90o
VL
VR
VC
Figure 23
Z