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### Transcript of Electric Circuits II - Philadelphia ... The Resistor The angles ®¸ and...

• Electric Circuits II Phasor Diagram

1

Dr. Firas Obeidat

• Dr. Firas Obeidat – Philadelphia University

2

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 = 𝑰𝒎𝒆 𝒋(𝝎𝒕+𝝓)

• Dr. Firas Obeidat – Philadelphia University

3

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) = 𝑽𝒎𝒆 𝒋(𝝎𝒕+𝝓)

𝒊 𝒕 = 𝑪 𝒅𝒗(𝒕)

𝒅𝒕 𝑰𝒎𝒆

𝒋𝜽 = 𝒋𝝎𝑪𝑽𝒎𝒆 𝒋𝝓

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

4

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∟53.13 o.

Impedance diagram is shown in figure (c).

𝐼 = 𝑉

𝑍 𝑇

= 100∟0𝑜

5∟53.13o = 20∟−53.13o

VR=IZR=(20∟-53.13 o A)(3∟0Ω)=60∟-53.13o V.

VL=IZL=(20∟-53.13 o A)(4∟90Ω)=80∟36.87o V.

Phasor diagram is shown in figure (d).

In rectangular form

VR=60∟-53.13 o =36-j48V.

VL=80∟36.87 o =64+j48V.

V=VR+VL=36-j48+64+j48=100+j0V=100∟0 V.

• Dr. Firas Obeidat – Philadelphia University

5

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∟-53.13 o.

Impedance diagram is shown in figure (c).

𝑉 = 𝐼𝑍𝑇= (5∟53.13 o)(10∟−53.13o)=50∟0 V

VR=IZR=(5∟53.13 o)(6∟0o)=30∟53.13o0 V

VC=IZC=(5∟53.13 o A)(8∟-90Ω)=40∟-36.87o V.

Phasor diagram is shown in figure (d).

In rectangular form

VR=30∟53.13 o=18+j24 V

VC=40∟-36.87 o =32-j24V.

V=VR+VC=18+j24+32-j24=50+j0=50∟0 V.

• Dr. Firas Obeidat – Philadelphia University

6

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∟53.13 o.

Impedance diagram is shown in figure (c).

VR=IZR=(10∟−53.13 o)(3∟0o)=30∟−53.13o0 V

VC=IZC=(10∟-53.13 o 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

𝑍 𝑇

= 50∟0𝑜

5∟53.13o = 10∟−53.13o

VL=IZL=(10∟-53.13 o A)(7∟90Ω)=70∟36.87o V.

VR=30∟−53.13 o0 V=18-j24 V

VC=30∟-143.13 o V=-24-j18.

VL=70∟36.87 o V=56+j42 V.

• Dr. Firas Obeidat – Philadelphia University

7

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

𝑌 𝑇

= 1

0.5∟−53.13 =2 ∟53.13

𝐼 = 𝑉

𝑍 𝑇

= 20∟53.13𝑜

2∟53.13o = 10∟0o

𝐼𝑅 = 𝑉

𝑍 𝑅

= 20∟53.13𝑜

3.33∟0o = 6∟53.13o

𝐼𝐿 = 𝑉

𝑍 𝐿

= 20∟53.13𝑜

2.5∟90o = 8∟−36.87o

• Dr. Firas Obeidat – Philadelphia University

8

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

𝑌 𝑇

= 1

1∟53.13=1∟-53.13

𝑉 = 𝐼𝑍𝑇 = (10∟0 𝑜)(1∟-53.13)= 10∟−53.13o

𝐼𝑅 = 𝑉

𝑍 𝑅

= 10∟−53.13𝑜

1.67∟0o = 6∟−53.13 o

𝐼𝐶 = 𝑉

𝑍 𝐶

= 10∟−53.13𝑜

1.25∟−90o= 8∟36.87 o

• Dr. Firas Obeidat – Philadelphia University

9

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

𝑌 𝑇

= 1

0.5∟−53.13=2∟53.13 o=1.2+j1.6

𝐼𝑅 = 𝑉

𝑍 𝑅

= 100∟53.13𝑜

3.33∟0o = 30∟53.13 o

𝐼𝐶 = 𝑉

𝑍 𝐶

= 100∟53.13𝑜

3.33∟−90o= 30∟143.13 o

𝐼 = 𝑉

𝑍 𝑇

= 100∟53.13𝑜

2∟53.13o = 50∟0o

𝐼𝐿 = 𝑉

𝑍 𝐿

= 100∟53.13𝑜

1.43∟90o = 70∟−36.87 o

• Dr. Firas Obeidat – Philadelphia University

10

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

11

Series-parallel AC circuit

Example: for the circuit, calculate Is and Vab

• Dr. Firas Obeidat – Philadelphia University

12

Series-parallel AC circuit

• Dr. Firas Obeidat – Philadelphia University

13

Series-parallel AC circuit

Example: Determine the current I and

the voltage V.

• Dr. Firas Obeidat – Philadelphia University

14

Series-parallel AC circuit

Example: calculate I, I1, I2, I3 and ZT.

• Dr. Firas Obeidat – Philadelphia University

15

Series-parallel AC circuit

• Dr. Firas Obeidat – Philadelphia University

16

Series-parallel AC circuit

Example: calculate ZT, I, I1, andI2.

• Dr. Firas Obeidat – Philadelphia University

17

Series-parallel AC circuit

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