SERIES RL CIRCUITS (1)
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Transcript of SERIES RL CIRCUITS (1)
Series and Parallel AC Circuits
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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 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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