Electric Circuits II Phasor Diagram
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Dr. Firas Obeidat
Dr. Firas Obeidat – Philadelphia University
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Phasor diagram for the Passive Circuit Elements
Let
In polar form
But Vm∟θ and Im∟ 𝝓 merely represent the
general voltage and current phasors V and I. Thus
𝒊(𝒕) = 𝑰𝒎𝒄𝒐𝒔(𝝎𝒕 + 𝝓)
The Resistor
The angles θ and 𝝓 are equal, so that the current
and voltage are always in phase.
The Inductor
Let 𝒊(𝒕) = 𝑰𝒎𝒄𝒐 𝒔 𝝎𝒕 + 𝝓 = 𝑰𝒎𝒆𝒋(𝝎𝒕+𝝓)
Dr. Firas Obeidat – 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 jωL is exactly
+90 and that I must therefore lag V by 90° in
an inductor.
The Capacitor
Let 𝒗(𝒕) = 𝑽𝒎𝒄𝒐𝒔(𝝎𝒕 + 𝝓) = 𝑽𝒎𝒆𝒋(𝝎𝒕+𝝓)
𝒊 𝒕 = 𝑪𝒅𝒗(𝒕)
𝒅𝒕 𝑰𝒎𝒆
𝒋𝜽 = 𝒋𝝎𝑪𝑽𝒎𝒆𝒋𝝓
𝑰 = 𝑗𝜔𝐶𝑽 𝑽 =𝟏
𝑗𝜔𝐶 I
Note that the angle of the factor 1/jωC is
exactly -90 and that I must therefore lead V
by 90° in an Capacitor.
Dr. Firas Obeidat – 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=100∟0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL=3Ω+j4Ω =5∟53.13o.
Impedance diagram is shown in figure (c).
𝐼 =𝑉
𝑍𝑇
=100∟0𝑜
5∟53.13o= 20∟−53.13o
VR=IZR=(20∟-53.13o A)(3∟0Ω)=60∟-53.13o V.
VL=IZL=(20∟-53.13o A)(4∟90Ω)=80∟36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=60∟-53.13o =36-j48V.
VL=80∟36.87o =64+j48V.
V=VR+VL=36-j48+64+j48=100+j0V=100∟0 V.
Dr. Firas Obeidat – 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=5∟0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZC=6Ω-j8Ω =10∟-53.13o.
Impedance diagram is shown in figure (c).
𝑉 = 𝐼𝑍𝑇= (5∟53.13o)(10∟−53.13o)=50∟0 V
VR=IZR=(5∟53.13o)(6∟0o)=30∟53.13o0 V
VC=IZC=(5∟53.13o A)(8∟-90Ω)=40∟-36.87o V.
Phasor diagram is shown in figure (d).
In rectangular form
VR=30∟53.13o=18+j24 V
VC=40∟-36.87o =32-j24V.
V=VR+VC=18+j24+32-j24=50+j0=50∟0 V.
Dr. Firas Obeidat – 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=50∟0, so the phasor circuit is shown in figure (b).
ZT=ZR+ZL+ZC=3Ω+7Ω-j3Ω =3+j4= 5∟53.13o.
Impedance diagram is shown in figure (c).
VR=IZR=(10∟−53.13o)(3∟0o)=30∟−53.13o0 V
VC=IZC=(10∟-53.13o A)(3∟-90Ω)=30∟-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=50∟0 V.
𝐼 =𝑉
𝑍𝑇
=50∟0𝑜
5∟53.13o= 10∟−53.13o
VL=IZL=(10∟-53.13o A)(7∟90Ω)=70∟36.87o V.
VR=30∟−53.13o0 V=18-j24 V
VC=30∟-143.13o V=-24-j18.
VL=70∟36.87o V=56+j42 V.
Dr. Firas Obeidat – 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=20∟53.13, so the phasor circuit is shown in figure
(b).
YT=YR+YL=1/3.33+1/j2.5=0.3-j0.4 =0.5∟-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 ∟0.
𝑍𝑇 =1
𝑌𝑇
=1
0.5∟−53.13=2 ∟53.13
𝐼 =𝑉
𝑍𝑇
=20∟53.13𝑜
2∟53.13o= 10∟0o
𝐼𝑅 =𝑉
𝑍𝑅
=20∟53.13𝑜
3.33∟0o = 6∟53.13o
𝐼𝐿 =𝑉
𝑍𝐿
=20∟53.13𝑜
2.5∟90o = 8∟−36.87o
Dr. Firas Obeidat – 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=10∟0, so the phasor circuit is shown in figure (b).
YT=YR+YC=1/1.67+1/-j2.5=0.6+j0.8 =1∟53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
𝑍𝑇 =1
𝑌𝑇
=1
1∟53.13=1∟-53.13
𝑉 = 𝐼𝑍𝑇 = (10∟0𝑜)(1∟-53.13)= 10∟−53.13o
𝐼𝑅 =𝑉
𝑍𝑅
=10∟−53.13𝑜
1.67∟0o = 6∟−53.13o
𝐼𝐶 =𝑉
𝑍𝐶
=10∟−53.13𝑜
1.25∟−90o= 8∟36.87o
Dr. Firas Obeidat – 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=100∟53.13, 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 =0.5∟-53.13
Impedance diagram is shown in figure (c).
Currents Phasor diagram is shown in figure
(d).
𝑍𝑇 =1
𝑌𝑇
=1
0.5∟−53.13=2∟53.13o=1.2+j1.6
𝐼𝑅 =𝑉
𝑍𝑅
=100∟53.13𝑜
3.33∟0o = 30∟53.13o
𝐼𝐶 =𝑉
𝑍𝐶
=100∟53.13𝑜
3.33∟−90o= 30∟143.13o
𝐼 =𝑉
𝑍𝑇
=100∟53.13𝑜
2∟53.13o = 50∟0o
𝐼𝐿 =𝑉
𝑍𝐿
=100∟53.13𝑜
1.43∟90o = 70∟−36.87o
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Example: for the circuit, calculate ZT, Is, VR, VC,
IL and IC
𝐈𝐋 =𝐕𝐋
𝐙𝐋=𝟏𝟏𝟖. 𝟒𝟒∠ − 𝟗. 𝟒𝟔𝐨
𝟑∠𝟗𝟎𝐨= 𝟑𝟗. 𝟒𝟖∠ − 𝟗𝟗. 𝟒𝟔𝐨
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Example: for the circuit, calculate Is and Vab
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Example: Determine the current I and
the voltage V.
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Example: calculate I, I1, I2, I3 and ZT.
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
Example: calculate ZT, I, I1, andI2.
Dr. Firas Obeidat – Philadelphia University
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Series-parallel AC circuit
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