Electric Circuits II - Philadelphia University...The Resistor The angles ฮธ and ๐ are equal, so...
Transcript of Electric Circuits II - Philadelphia University...The Resistor The angles ฮธ and ๐ are equal, so...
Electric Circuits II Phasor Diagram
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Dr. Firas Obeidat
Dr. Firas Obeidat โ Philadelphia University
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Phasor diagram for the Passive Circuit Elements
Let
In polar form
But Vmโฮธ and Imโ ๐ merely represent the
general voltage and current phasors V and I. Thus
๐(๐) = ๐ฐ๐๐๐๐(๐๐ + ๐)
The Resistor
The angles ฮธ and ๐ are equal, so that the current
and voltage are always in phase.
The Inductor
Let ๐(๐) = ๐ฐ๐๐๐ ๐ ๐๐ + ๐ = ๐ฐ๐๐๐(๐๐+๐)
Dr. Firas Obeidat โ Philadelphia University
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Phasor diagram for the Passive Circuit Elements
We obtain the desired phasor relationship
Note that the angle of the factor jฯL is exactly
+90 and that I must therefore lag V by 90ยฐ in
an inductor.
The Capacitor
Let ๐(๐) = ๐ฝ๐๐๐๐(๐๐ + ๐) = ๐ฝ๐๐๐(๐๐+๐)
๐ ๐ = ๐ช๐ ๐(๐)
๐ ๐ ๐ฐ๐๐
๐๐ฝ = ๐๐๐ช๐ฝ๐๐๐๐
๐ฐ = ๐๐๐ถ๐ฝ ๐ฝ =๐
๐๐๐ถ I
Note that the angle of the factor 1/jฯC is
exactly -90 and that I must therefore lead V
by 90ยฐ in an Capacitor.
Dr. Firas Obeidat โ Philadelphia University
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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=100โ0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL=3ฮฉ+j4ฮฉ =5โ53.13o.
Impedance diagram is shown in figure (c).
๐ผ =๐
๐๐
=100โ0๐
5โ53.13o= 20โโ53.13o
VR=IZR=(20โ-53.13o A)(3โ0ฮฉ)=60โ-53.13o V.
VL=IZL=(20โ-53.13o A)(4โ90ฮฉ)=80โ36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=60โ-53.13o =36-j48V.
VL=80โ36.87o =64+j48V.
V=VR+VL=36-j48+64+j48=100+j0V=100โ0 V.
Dr. Firas Obeidat โ Philadelphia University
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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=5โ0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZC=6ฮฉ-j8ฮฉ =10โ-53.13o.
Impedance diagram is shown in figure (c).
๐ = ๐ผ๐๐= (5โ53.13o)(10โโ53.13o)=50โ0 V
VR=IZR=(5โ53.13o)(6โ0o)=30โ53.13o0 V
VC=IZC=(5โ53.13o A)(8โ-90ฮฉ)=40โ-36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=30โ53.13o=18+j24 V
VC=40โ-36.87o =32-j24V.
V=VR+VC=18+j24+32-j24=50+j0=50โ0 V.
Dr. Firas Obeidat โ Philadelphia University
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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=50โ0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL+ZC=3ฮฉ+7ฮฉ-j3ฮฉ =3+j4= 5โ53.13o.
Impedance diagram is shown in figure (c).
VR=IZR=(10โโ53.13o)(3โ0o)=30โโ53.13o0 V
VC=IZC=(10โ-53.13o A)(3โ-90ฮฉ)=30โ-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=50โ0 V.
๐ผ =๐
๐๐
=50โ0๐
5โ53.13o= 10โโ53.13o
VL=IZL=(10โ-53.13o A)(7โ90ฮฉ)=70โ36.87o V.
VR=30โโ53.13o0 V=18-j24 V
VC=30โ-143.13o V=-24-j18.
VL=70โ36.87o V=56+j42 V.
Dr. Firas Obeidat โ Philadelphia University
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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=20โ53.13, so the phasor circuit is shown in figure
(b).
YT=YR+YL=1/3.33+1/j2.5=0.3-j0.4 =0.5โ-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 โ0.
๐๐ =1
๐๐
=1
0.5โโ53.13=2 โ53.13
๐ผ =๐
๐๐
=20โ53.13๐
2โ53.13o= 10โ0o
๐ผ๐ =๐
๐๐
=20โ53.13๐
3.33โ0o = 6โ53.13o
๐ผ๐ฟ =๐
๐๐ฟ
=20โ53.13๐
2.5โ90o = 8โโ36.87o
Dr. Firas Obeidat โ Philadelphia University
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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=10โ0, so the phasor circuit is shown in figure (b).
YT=YR+YC=1/1.67+1/-j2.5=0.6+j0.8 =1โ53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
๐๐ =1
๐๐
=1
1โ53.13=1โ-53.13
๐ = ๐ผ๐๐ = (10โ0๐)(1โ-53.13)= 10โโ53.13o
๐ผ๐ =๐
๐๐
=10โโ53.13๐
1.67โ0o = 6โโ53.13o
๐ผ๐ถ =๐
๐๐ถ
=10โโ53.13๐
1.25โโ90o= 8โ36.87o
Dr. Firas Obeidat โ 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=100โ53.13, 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 =0.5โ-53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
๐๐ =1
๐๐
=1
0.5โโ53.13=2โ53.13o=1.2+j1.6
๐ผ๐ =๐
๐๐
=100โ53.13๐
3.33โ0o = 30โ53.13o
๐ผ๐ถ =๐
๐๐ถ
=100โ53.13๐
3.33โโ90o= 30โ143.13o
๐ผ =๐
๐๐
=100โ53.13๐
2โ53.13o = 50โ0o
๐ผ๐ฟ =๐
๐๐ฟ
=100โ53.13๐
1.43โ90o = 70โโ36.87o
Dr. Firas Obeidat โ Philadelphia University
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Series-parallel AC circuit
Example: for the circuit, calculate ZT, Is, VR, VC,
IL and IC
๐๐ =๐๐
๐๐=๐๐๐. ๐๐โ โ ๐. ๐๐๐จ
๐โ ๐๐๐จ= ๐๐. ๐๐โ โ ๐๐. ๐๐๐จ
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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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