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SERIES RL CIRCUITS (1). Circuit above is a series RL network connected to an ac voltage source Need to find the phasor form of the total impedance of this combination The total impedance of this series combination is The magnitude and angle of Z T can be found by converting to polar form: - PowerPoint PPT Presentation
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Series and Parallel AC Circuits
1
SERIES RL CIRCUITS (1)
• Circuit above is a series RL network connected to an ac voltage source
• Need to find the phasor form of the total impedance of this combination
• The total impedance of this series combination is
• The magnitude and angle of ZT can be found by converting to polar form:
• |ZT| = √[R2+(ωL)2] and θ = tan-1(ωL/R)
• The plot of ZT:
V 0
ωsin
p
p
E
tEte+
-
00 jRR
ω090ω LjL
R
L
ohms ωω00 LT XjRLjRLjjRZ
Im
Re
θ
R
jωL
|Z T|
ω θZLjRZ TT
RLθ
LRZT
ωtan
ω1
22
Series and Parallel AC Circuits
2
SERIES RL CIRCUITS (2)
• Example: For a series RL combination circuit, R = 300, L = 0.2H and e(t) = 17sin(2000t) V. Find the total equivalent impedance in polar form and rectangular form. Sketch the impedance in the complex plane.
• We can use Ohm’s law to find the total current supplied by a voltage source: iT = v / ZT
• Mathematical operation must be carried out using phasors since all quantities have both magnitude and angle
• The current iT in a series circuit is the same through every series-connected component
• The ac voltage drop across each component can be found by multiplying each impedance by the current
• v1 = iTZ1, v2 = iTZ2, …
• Example: In a series RL circuit, where e(t) = 300 V, R = 200, XL = j100. Find the total current in the circuit. Find the voltage drops across R and L. Verify KVL around the circuit. Draw a phasor diagram showing e, vR, vL and iT. Sketch the voltage waveforms.
Series and Parallel AC Circuits
3
SERIES RC CIRCUITS (1)
• Above is a series RC network connected to an ac voltage
• The total impedance of this combination is
• The polar coordinates are
• The phasor diagram of the total impedance is
V 0
ωsin
p
p
E
tEte+
-
00 jRR
ω090ω
1Cj
C
R
C
CT XjRCjRCjjRZ ωω00
RCR
C
CRZT
ω1tanω1
tanθ
ohms ω1
11
22
Im
Re
θ
R
-j/ωC
|ZT |
θ
ω
T
T
Z
CjRZ
RC
CRZ T
ω1tanθ
ω11
22
Series and Parallel AC Circuits
4
SERIES RC CIRCUITS (2)
• As in the series RL circuit, the total current supplied to the RC network and the voltage drops can be found by using Ohm’s Law:
• iT = e / ZT
• vR = iTR and vC = iTXC
• Example: For a series RC circuit where e(t) = 18sin(240t + 45) V, R = 3.3k and C = 2.2µF:
• a) Find the total current in the circuit in phasor and sinusoidal form
• b) Find the voltage drops across the resistor and capacitor in phasor and sinusoidal form
• c) Verify KVL around the circuit
• d) Draw a phasor diagram showing e, iT, vR and vC
• e) Sketch the waveforms of e, vR and vC versus angle
Series and Parallel AC Circuits
5
SERIES RLC CIRCUITS (1)
• The total impedance of the RLC circuit is
• In terms of magnitudes it is: ZT = R + j(|XL| - |XC|)
• Inductive and capacitive reactance have opposite signs
• Thus net reactance may be either inductive or capacitive, depending which is larger
• Polar coordinates are
V 0
ωsin
p
p
E
tEte+
-
00 jRR
ω090ω
1Cj
C
R
C
ω090ω LjLL
i(t)
ohms ω1ω CLjRZT
R
XX
R
L
XXRCLRZ
CL
CLT
11
2222
tanωC1ω
tanθ
ω1ω
Series and Parallel AC Circuits
6
SERIES RLC CIRCUITS (2)
• The phasor diagram of the impedance when inductive reactance is greater than the capacitive reactance, i.e. when |XL|>|XC|
• The phasor diagram of the impedance when capacitive reactance is greater than the inductive reactance, i.e. when |XC|>|XL|
Im
Re
ZT
R
θ
jωL
-j/ωC
ωL – 1/ωCNet reactance
Im
Re
ZT
R
θ
jωL
-j/ωC
ωL – 1/ωCNet reactance
Series and Parallel AC Circuits
7
SERIES RLC CIRCUITS (3)
• When there is more than one resistor, capacitor and/or inductor in a series circuit, the total impedance has a resistance component equal to the sum of the resistance values and a reactive component equal to the sum of the capacitive reactances subtracted from the sum of the inductive reactances
• Example: In a series RLC circuit, e(t) = 100sinωt, R = 800, ZL = j1250 and ZC = -j450.
• a) Find the current in polar form
• b) Find the voltage drops vR, vL, vC
• c) Verify KVL around the circuit
• d) Draw a phasor diagram showing e, i, vR, vL, vC
Series and Parallel AC Circuits
8
ADMITTANCE
• The reciprocal of resistance is conductance, with units Siemens
• The reciprocal of impedance is admittance, denoted Y
• Y = 1/Z siemens
• Impedance is a measure of the extent to which a component impedes the flow of ac current through it
• Admittance is a measure of how well it admits the flow of ac current
• The greater the admittance, the smaller the impedance, and vice versa
• Resistance R is one form of impedance Z, conductance G is one form of admittance Y
• Phasor form of conductance is:
S 01
siemens 01
0
1
jR
G
RRG
Series and Parallel AC Circuits
9
SUSCEPTANCE
• Reactance is another special case of impedance
• The reciprocal of reactance is called susceptance, B
• B = 1/X siemens
• There are two types of susceptance
• Inductive susceptance:
• Capacitive susceptance:
• Z = 1/Y; R = 1/G; XL = 1/BL; XC = 1/BC
• Example: Find the admittance of a 5 resistor; a 5mH inductor at f = 60Hz; a 0.2µF capacitor at ω = 1.25106 rad/s.
• Example: Find the admittance Y corresponding to Z = 30 + j40 . Draw a phasor diagram showing Y in the complex plane.
S ω10
siemens 90ω
1
90ω
11
LjB
LLXB
L
LL
S ω0
siemens 90ω90ω1
11
CjB
CCX
B
C
CC
Series and Parallel AC Circuits
10
PARALLEL AC CIRCUITS (1)
• Above shows a parallel connected set of impedances to an ac source
• Total admittance of the circuit is the sum of the admittances of the parallel connected components, i.e.
• YT = Y1 + Y2 +… + Yn = 1/Z1 + 1/Z2 +…+ 1/Zn
• Example: Find the total admittance of a 20mH inductor, a 1k resistor and a 0.16µF capacitor connected in parallel to an ac voltage source of 15sin(25103t) V. Draw a phasor diagram showing the total admittance in the complex plane. What is the total admittance if the frequency of the voltage source is doubled?
+
-
YT
Z1 Z2
Series and Parallel AC Circuits
11
PARALLEL AC CIRCUITS (2)
• The total impedance of a network is the reciprocal of its total admittance, ZT = 1/YT
• For a parallel network:
• Example: Three components (an inductor with impedance j10, a resistor with impedance 2, and another inductor with impedance j5) are connected in parallel. Find the total equivalent impedance of this network.
• Example: A capacitor with impedance –j100 and an inductor with impedance j500 are connected in parallel. Find this network’s total equivalent impedance.
n
nTT
ZZZ
YYYYZ
111
1
11
21
21
Series and Parallel AC Circuits
12
PARALLEL AC CIRCUITS (3)
• The same voltage appears across every parallel-connected impedance
• So current can be found using Ohm’s law and KCL
e Zn
+
-
Z1 Z2
iT = i1 + i2 +…+ in
i1 =e/Z1
in =e/Zn
i2 =e/Z2
TTT
nT
nnn
eYZei
iiii
eYZei
eYZei
21
111
Series and Parallel AC Circuits
13
PARALLEL AC CIRCUITS (4)
• Example: For the circuit below, find the current in each impedance. Show that iT = eYT. Draw a phasor diagram showing e, iT, i1, i2 and i3.
e = 60sin(ωt) VZ3 =
-j80
+
-
Z1 = 50
Z2 =j40
iT
i1 i3i2
Series and Parallel AC Circuits
14
AC CURRENT SOURCES (1)
• Like a dc current source, an ideal ac current source supplies the same constant current to whatever network is connected across its terminals
• For ac, the current is constant in the sense that its peak value does not change
• The same symbol is used for an ac current source as for a dc current source
• The direction of the arrow is just a phase reference, since current reverses direction every half cycle
• Reversing the arrow is the same as multiplying the current by -1, which is the same as adding or subtracting 180 from its phase angle
Ipθ+180 Ipsin(ωt + θ) A= Ipθ
-Ipθ ==
Series and Parallel AC Circuits
15
AC CURRENT SOURCES (2)
• Example: For the circuit below, find the voltage vT across the ac current source. Find the currents i1 and i2. Show that i = i1 + i2. Draw a phasor diagram showing i, vT, i1 and i2. Sketch the waveforms of i, i1 and i2 versus angle.
i2
i(t) = 0.24sin(5106t) A
vT
i1
100 2000pF
Series and Parallel AC Circuits
16
POWER IN CIRCUITS CONTAINING REACTANCE (1)
• Recall that the average power dissipated by a resistance carrying sinusoidal ac current can be found by
• Above can be used to find Pavg dissipated by a resistor in a circuit containing reactance, provided the values used in the computations are indeed those of the voltage across and/or current through the resistor itself
• In many practical circuits, it is sometimes necessary to compute Pavg when the resistance or resistor voltage/current are not known
effeff
2eff2
eff
22
avg
222
IVR
VRI
IV
R
VRIP pppp
Series and Parallel AC Circuits
17
POWER IN CIRCUITS CONTAINING REACTANCE (2)
• Above is an ac circuit with a general impedance Z• Assuming that the voltage supplied has angle
• Angle between voltage and current is - ( - ) = • The angle between the voltage applied to a network
and the total current supplied to it equals the angle of the impedance of the network
• Resistive component of impedance Z is R = |Z|cos ohms
• Since only resistance in network dissipates power
+e = Ep
|Z|
i = (Ep/|Z|) →
θφθ
φ
Z
E
Z
Ei pp
2
θcos
,
2
θcos
2
avg
2
22
avg
pp
ppppp
pp
EIP
EIZIEZI
ZIRIP
Series and Parallel AC Circuits
18
POWER FACTOR
• Equations on previous slide provide a means for computing the average power in terms of voltage e across whole network and total supply current i
• Cos is called the power factor
• For a purely resistive network, voltage and current are in phase, so power factor = cos0 = 1 → Pavg = EpIp / 2
• For a purely reactive network, voltage and current are separated by ±90, and cos ±90 = 0 → Pavg = 0, i.e no power is dissipated by purely reactive components (capacitors and inductors)
• Since Vp = 2Veff and Ip = 2Ieff we can derive
θcos
θcos2
22
effeff
effeffavg
IV
IVP
Series and Parallel AC Circuits
19
EXAMPLE OF POWER IN AC CIRCUITS
• In the circuit above:
• Find the power factor
• Use the power factor to find the average power dissipated in the network
• Verify that the power computed above is the same as the power computed by using the voltage across and current through the resistor
+e = 600
i→
j80
100
-j20