Electric Circuits II - Philadelphia University...The Resistor The angles ΞΈ and π are equal, so...
Transcript of Electric Circuits II - Philadelphia University...The Resistor The angles ΞΈ and π are equal, so...
Electric Circuits II Phasor Diagram
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Dr. Firas Obeidat
Dr. Firas Obeidat β Philadelphia University
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Phasor diagram for the Passive Circuit Elements
Let
In polar form
But VmβΞΈ and Imβ π merely represent the
general voltage and current phasors V and I. Thus
π(π) = π°ππππ(ππ + π)
The Resistor
The angles ΞΈ and π are equal, so that the current
and voltage are always in phase.
The Inductor
Let π(π) = π°πππ π ππ + π = π°πππ(ππ+π)
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Phasor diagram for the Passive Circuit Elements
We obtain the desired phasor relationship
Note that the angle of the factor jΟL is exactly
+90 and that I must therefore lag V by 90Β° in
an inductor.
The Capacitor
Let π(π) = π½ππππ(ππ + π) = π½πππ(ππ+π)
π π = πͺπ π(π)
π π π°ππ
ππ½ = πππͺπ½ππππ
π° = πππΆπ½ π½ =π
πππΆ I
Note that the angle of the factor 1/jΟC is
exactly -90 and that I must therefore lead V
by 90Β° in an Capacitor.
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Phasor diagram for series RL circuit
Example: for the circuit shown in figure (a), draw the phasor
circuit , impedance diagram and voltages phasor diagram.
V=100β0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL=3Ξ©+j4Ξ© =5β53.13o.
Impedance diagram is shown in figure (c).
πΌ =π
ππ
=100β0π
5β53.13o= 20ββ53.13o
VR=IZR=(20β-53.13o A)(3β0Ξ©)=60β-53.13o V.
VL=IZL=(20β-53.13o A)(4β90Ξ©)=80β36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=60β-53.13o =36-j48V.
VL=80β36.87o =64+j48V.
V=VR+VL=36-j48+64+j48=100+j0V=100β0 V.
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Phasor diagram for series RC circuit
Example: for the circuit shown in figure (a), draw the phasor
circuit , impedance diagram and voltages phasor diagram.
I=5β0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZC=6Ξ©-j8Ξ© =10β-53.13o.
Impedance diagram is shown in figure (c).
π = πΌππ= (5β53.13o)(10ββ53.13o)=50β0 V
VR=IZR=(5β53.13o)(6β0o)=30β53.13o0 V
VC=IZC=(5β53.13o A)(8β-90Ξ©)=40β-36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=30β53.13o=18+j24 V
VC=40β-36.87o =32-j24V.
V=VR+VC=18+j24+32-j24=50+j0=50β0 V.
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Phasor diagram for series RLC circuit
Example: for the circuit shown in figure (a), draw the
phasor circuit , impedance diagram and voltages phasor
diagram.
V=50β0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL+ZC=3Ξ©+7Ξ©-j3Ξ© =3+j4= 5β53.13o.
Impedance diagram is shown in figure (c).
VR=IZR=(10ββ53.13o)(3β0o)=30ββ53.13o0 V
VC=IZC=(10β-53.13o A)(3β-90Ξ©)=30β-143.13o V.
Phasor diagram is shown in figure (d).
In rectangular form
V=VR+VL+VC=18-j24+56+j42-24-j18
V=50+j0=50β0 V.
πΌ =π
ππ
=50β0π
5β53.13o= 10ββ53.13o
VL=IZL=(10β-53.13o A)(7β90Ξ©)=70β36.87o V.
VR=30ββ53.13o0 V=18-j24 V
VC=30β-143.13o V=-24-j18.
VL=70β36.87o V=56+j42 V.
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Phasor diagram for parallel RL circuit
Example: for the circuit shown in figure (a), draw the
phasor circuit , impedance diagram and currents
phasor diagram.
V=20β53.13, so the phasor circuit is shown in figure
(b).
YT=YR+YL=1/3.33+1/j2.5=0.3-j0.4 =0.5β-53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
I=IR+IL=3.6+j4.8+6.4-j4.8=10+j0=10 β0.
ππ =1
ππ
=1
0.5ββ53.13=2 β53.13
πΌ =π
ππ
=20β53.13π
2β53.13o= 10β0o
πΌπ =π
ππ
=20β53.13π
3.33β0o = 6β53.13o
πΌπΏ =π
ππΏ
=20β53.13π
2.5β90o = 8ββ36.87o
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Phasor diagram for parallel RC circuit
Example: for the circuit shown in figure (a), draw the
phasor circuit , impedance diagram and currents
phasor diagram.
I=10β0, so the phasor circuit is shown in figure (b).
YT=YR+YC=1/1.67+1/-j2.5=0.6+j0.8 =1β53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
ππ =1
ππ
=1
1β53.13=1β-53.13
π = πΌππ = (10β0π)(1β-53.13)= 10ββ53.13o
πΌπ =π
ππ
=10ββ53.13π
1.67β0o = 6ββ53.13o
πΌπΆ =π
ππΆ
=10ββ53.13π
1.25ββ90o= 8β36.87o
Dr. Firas Obeidat β Philadelphia University
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Phasor diagram for parallel RLC circuit
Example: for the circuit shown in figure (a),
draw the phasor circuit , impedance diagram
and currents phasor diagram.
V=100β53.13, so the phasor circuit is shown
in figure (b).
YT=YR+YL+YC=1/3.33+1/j1.43+1/-j3.33
=0.3+j0.4 =0.5β-53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
ππ =1
ππ
=1
0.5ββ53.13=2β53.13o=1.2+j1.6
πΌπ =π
ππ
=100β53.13π
3.33β0o = 30β53.13o
πΌπΆ =π
ππΆ
=100β53.13π
3.33ββ90o= 30β143.13o
πΌ =π
ππ
=100β53.13π
2β53.13o = 50β0o
πΌπΏ =π
ππΏ
=100β53.13π
1.43β90o = 70ββ36.87o
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Series-parallel AC circuit
Example: for the circuit, calculate ZT, Is, VR, VC,
IL and IC
ππ =ππ
ππ=πππ. ππβ β π. πππ¨
πβ πππ¨= ππ. ππβ β ππ. πππ¨
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Series-parallel AC circuit
Example: for the circuit, calculate Is and Vab
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Series-parallel AC circuit
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Series-parallel AC circuit
Example: Determine the current I and
the voltage V.
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Series-parallel AC circuit
Example: calculate I, I1, I2, I3 and ZT.
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Series-parallel AC circuit
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Series-parallel AC circuit
Example: calculate ZT, I, I1, andI2.
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Series-parallel AC circuit
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