4-ac Circuit Analysis

ELECTRIC CIRCUITS THEORY 1 These lecture slides have been compiled by Mohammed LECTURE 4 contd... SalahUdDin Ayubi. AC Circuit Analysis 17 August 2005

Engineer M S Ayubi

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8) AC Circuit Analysis Sinusoidal voltages and currents 8.1) Description of Sinusoidal Signal 3 parameters Magnitude Phase Frequency

Vm θ ω = 2π f

v (t ) =Vm ⋅ cos( ωt +θ) Ac circuit analysis is conducted at one frequency at a time. Only two variables left, magnitude and phase. If more than one frequency source is present. Use the principle of superposition 17 August 2005 Engineer M S Ayubi

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8.2 Advantages of Sinusoidal Signals • Sinusoidal functions yield further sinusoidal functions when integrated or differentiated

v(t ) = Vm ⋅ cos(ωt + θ ) Vm v ( t ) dt = ⋅ sin( ω t + θ ) ∫ ω

dv (t ) = − Vm ⋅ ω ⋅ sin( ωt + θ ) dt

17 August 2005

Engineer M S Ayubi

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Signal v(t) and integral (represented by iv(t)) and differential (represented by dv(t)) 10V rm s ac signal at 0.5 H z

44.429 60 48

Voltage in volts

36 v( t n)

24

dv( t n)

12

iv( t n)

0 12 24 36

− 44.429

48 60

0 0

17 August 2005

0.5

1

1.5

2

tn in seconds Engineer MTime S Ayubi

2.5

3

3.5

4 4 4

8.3 Representation of Amplitude and Phase of a Signal on the Complex Plane imaginary + j

θ = 90 or π /2

Vm ωt

θ = 180 or π - real

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θ =0 + real

- j θ = -90 or -π /2

Engineer M S Ayubi

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v = Vm cos( 2π ⋅ f ⋅ t ) = 10 cos( 2π ⋅ 0.5 ⋅ t ) In this particular case

+ j θ = 90 or π /2

10V rms ac signal at 0.5 Hz

14.142 15 12 9

Vm ωt

θ = 180 or π

θ =0 + real

- real

Voltage in volts

6 v imag ( t n)

3 0 3 6 9 12

− 14.142

- j θ = -90 or -π /2 angular frequency times time in radians

0

10V rm s ac signal at 0.5 Hz

0

15

0

5

10

15

0 ( ω ⋅ t) n 12.566 angular frequency times time in radians

The projection of the rotating phasor on the j (imaginary) axis is

1 .5

vimag = Vm ⋅ sin(ωt )

3 4 .5 − ω⋅t n

6 7 .5

The projection of the rotating

9 1 0 .5

phasor on the real axis is

12 − 1 2 .5 6 61 3 .5 15

17 August

vreal = Vm ⋅ cos(ωt ) 15

− 1 4 .1 4 2 2005

10

5

0

5

v real( t ) n v o ltag e in v o lts

10

15

1 4 .1 4M 2 Engineer

S Ayubi

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8.4 Relationship between rms and peak and i and j • Electrical engineers generally use a slightly different nomenclature to mathematicians for complex notation – Firstly we use j, rather than i, for the square root of –1 (i gets confused with current) – Secondly, we normally use rms values rather than peak values to describe the amplitude. For sine waves the peak is always 1.414 times the rms. (rms values rather than peak are used to describe voltages and currents) 17 August 2005

Engineer M S Ayubi

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8.5 Representation of d/dt by jω • We know that a signal represented by the differentiation of a sinusoidal voltage or current is the equivalent of the original signal – Multiplied by ω – Advanced in phase by 90 degrees or π /2

v (t ) =Vm ⋅cos( ωt +θ) dv(t ) π = − Vm ⋅ ω ⋅ sin(ωt + θ ) = Vm ⋅ ω ⋅ cos(ωt + θ + ) dt 2 • Now multiplying by j gives a phase shift of +90 degrees • Thus the operation of d/dt on a sinusoidal signal is like multiplying by j ω (inEngineer the Mcomplex plane) 17 August 2005 S Ayubi 8

8.6 Impedance of resistors, inductors and capacitors

• Impedance is a term used to describe the relationship between sinusoidal voltage and current for – a single passive component – a group of passive components

(passive components are R, L and C. They are passive because they are linear and have no gain) 17 August 2005

Engineer M S Ayubi

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• Impedance is given the symbol Z

i v

Z

v=iR

di v =L = jωL ⋅ i dt

v=iZ dv i =C = jωC ⋅ v dt thus 1 j v= ⋅i = − ⋅i jωC ωC

Note v and i are complex quantities here. (Some books use bold type to distinguish between complex and absolute values of sinusoidal 17 August 2005 and currents or between Engineer M S Ayubi and dc values) voltages complex

10

•

Thus

ZR = R Z L = jωL ZC

j =− ωC

We have two components to the impedance. One has no j term and is referred to as real The other is all j terms and is referred to as imaginary The same is true for voltage and current. All complex values can be expressed as a magnitude and phase angle Or as a complex quantity. Complex voltages and currents called phasors 17 August 2005 Engineerare M S Ayubi

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8.7 Phasors, Reference Phasor and CIVIL • A phasor is a term given to a voltage or current which has a real and complex part. Represented as one of the following forms

v = v real + j vimag

θ = 90 or π /2

v = v m ⋅ e jθ

v = v m ⋅ cos(θ ) + j v m ⋅ sin (θ ) v = v m ∠θ where

θ = 180 or π 2

v m = v real + vimag

2

-real

 v real   v real  −1  v real       θ = arccos = a cos = cos     vm   vm   vm  Note Using the cos function requires the sign of θ to be the sign of the imaginary part In circuits with several voltages and currents We need to define θ we need to take 17 August 2005 M S Ayubi a current or voltage as a reference (i.e.Engineer let θ =0 for that variable)

vm

θ

θ =0 + real

- jθ = -90 or -π /2 12

• Taking the same voltage applied across a parallel combination of a resistor, a capacitor and an inductor

iR ZR

v •

iL ZL

iC ZC

v i= Z

ZR = R Z L = jωL j ZC = − ωC

Plotting the voltage and three current phasors on the complex plane produces a phasor diagram. Voltage is common to all so we take that as the reference phasor in this case

+ j θ = 90 or π /2 v

θ = 180 or π - real 17 August 2005

Engineer M S Ayubi

θ =0 + real

- jθ = -90 or -π /2

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• Taking the same voltage applied across a parallel combination of a resistor, a capacitor and an inductor

iR ZR

v •

iL ZL

iC ZC

v i= Z

ZR = R Z L = jωL j ZC = − ωC

Plotting the voltage and three current phasors on the complex plane produces a phasor diagram. Voltage is common to all so we take that as the reference phasor in this case

+ j θ = 90 or π /2

θ = 180 or π

iR

- real 17 August 2005

Engineer M S Ayubi

v

θ =0 + real

- jθ = -90 or -π /2

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• Taking the same voltage applied across a parallel combination of a resistor, a capacitor and an inductor

iR ZR

v •

iL ZL

iC ZC

v i= Z

ZR = R Z L = jωL j ZC = − ωC

Plotting the voltage and three current phasors on the complex plane produces a phasor diagram. Voltage is common to all so we take that as the reference phasor in this case

+ j θ = 90 or π /2 iC

θ = 180 or π

iR

- real 17 August 2005

Engineer M S Ayubi

v

θ =0 + real

- jθ = -90 or -π /2

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• Taking the same voltage applied across a parallel combination of a resistor, a capacitor and an inductor

iR ZR

v •

iL ZL

iC ZC

v i= Z

ZR = R Z L = jωL j ZC = − ωC

Plotting the voltage and three current phasors on the complex plane produces a phasor diagram. Voltage is common to all so we take that as the reference phasor in this case

+ j θ = 90 or π /2 iC iR

θ = 180 or π - real

v

θ =0 + real

iL 17 August 2005

Engineer M S Ayubi

- jθ = -90 or -π /2

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• We can see that the phase angle between current and voltage is – 0 for the resistor, – + 90 degrees for the capacitor – and –90 degrees for the inductor

• A memory aid is the acronym CIVIL • CIV, for a capacitor (C), current (I) leads voltage (V) • VIL, voltage (V) leads current (I) for an inductor (L) 17 August 2005

Engineer M S Ayubi

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8.8 Phasor Addition and Complex Arithmetic to Find the Combined Current for the Previous Circuit i

v

iR ZR

iL ZL

iC ZC j

• Find i i = iR + iC + iL    1 1 1 1  1 1   = v ⋅  + i = v ⋅  + + + 1 Z Z Z R jωL   C L   R jωC   1 1   i = v ⋅  + j ⋅  ωC −  R ω L   

v - real

By complex algebra and arithmetic 17 August 2005

By phasor addition

Engineer M S Ayubi

+ real iL -j

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8.8 Phasor Addition and Complex Arithmetic to Find the Combined Current for the Previous Circuit i

v

iR ZR

iL ZL

iC ZC j

• Find i i = iR + iC + iL    1 1 1 1  1 1   = v ⋅  + i = v ⋅  + + + 1 Z Z Z R jωL   C L   R jωC   1 1   i = v ⋅  + j ⋅  ωC −  R ω L   

v - real

By complex algebra and arithmetic 17 August 2005

By phasor addition

Engineer M S Ayubi

+ real iL iR -j

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8.8 Phasor Addition and Complex Arithmetic to Find the Combined Current for the Previous Circuit i

v

iR ZR

iL ZL

iC ZC j

• Find i i = iR + iC + iL    1 1 1 1  1 1   = v ⋅  + i = v ⋅  + + + 1 Z Z Z R jωL   C L   R jωC   1 1   i = v ⋅  + j ⋅  ωC −  R ω L   

v - real

By complex algebra and arithmetic 17 August 2005

By phasor addition

Engineer M S Ayubi

iC + real iL iR -j

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8.8 Phasor Addition and Complex Arithmetic to Find the Combined Current for the Previous Circuit i

v

iR ZR

iL ZL

iC ZC j

• Find i

By phasor addition

i = iR + iC + iL    1 1 1 1  1 1   = v ⋅  + i = v ⋅  + + + 1 Z Z Z R jωL   C L   R jωC   1 1   i = v ⋅  + j ⋅  ωC −  R ω L   

- real

By complex algebra and arithmetic 17 August 2005

Engineer M S Ayubi

i

θ

v i

iC + real

iL iR -j

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8.9 General Circuit Solution • We can solve ac networks with the same tools and methods used for dc networks – Must use complex representation of voltages, currents and impedances – Must choose one current or voltage as reference phasor and relate all others to it in terms of the phase angle θ

• Evaluation of power needs care (rms values help) • Evaluation of stored energy needs care

17 August 2005

Engineer M S Ayubi

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8.10 Instantaneous power, real power and reactive power Instantaneous power waveforms for a voltage of 2V peak and a current of 1.5A peak Flowing separately in a resistor, a capacitor and an inductor 3

v( t n)

4

2

i( t n) P ( t n)

1

v( t n)

2

2

i( t n)

0

P( t n)

0

1

−2

2

0 0

1

2

3

tn

4 2

Resistor case

v ( t n) i( t n)

Average power Pav =0.5Vm*Im Pav =vrms *irms

−2

4

2

0

1

0

tn

3

4 4

2

Inductor case

1

0

P ( t n)

Pav = 0

1

−2

2

2

0 0

1

2 tn

3

4 4

Capacitor case Pav = 0 17 August 2005

Engineer M S Ayubi

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i Z

v •

v=iZ

Consider a series connected resistor and inductor Z = R + jωL i=v

   R − jωL R − jωL  1  = v ⋅ =   2  =  2 Z  R + jωL   ( ) ( ) R + j ω L ⋅ R − j ω L    R + (ωL )

   

 R  i = i ⋅e where i = and θ = a cos 2 2  2 R 2 + (ωL )  R + (ωL ) and v = v ⋅ e j 0 if we take the voltage as reference − jθ

v

(

)

   

Thus

v ⋅ i = v ⋅ i ⋅ (e j 0 ⋅ e − jθ ) = v ⋅ i ⋅ e − jθ = v ⋅ i ⋅ ( cos (θ ) − j sin (θ ) ) We take only the real part as the power P = v ⋅ i ⋅ cos (θ ) And the imaginary part is the reactive power whic h defines the component of the voltage current product responsibl e for energy storage in reactive components . This is given the symbol Q. Q = − v ⋅ i ⋅ sin (θ ) 17 August 2005

Engineer M S Ayubi

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8.11 Solutions to Problems with Power Evaluation When Neither Voltage or Current Phasor is the Reference Phasor i = i ⋅e

jθ i

and v = v ⋅ e

jθ v

if we do not take the voltage as reference

The angle between th e voltage and current phasors is φ = θi − θ v now

(

P = v ⋅i = v ⋅ i ⋅ e

jθ v

⋅e

jθ i

) = v ⋅ i ⋅e (

j θ v +θ i

)

= v ⋅ i ⋅ ( cos (θ v + θ i ) − j sin (θ v + θ i ) ) We take only the real part as the power P = v ⋅ i ⋅ cos (θ v + θi )

Q = v ⋅ i ⋅ sin (θ v + θi )

The correct answer should be

P = v ⋅ i ⋅ cos (θi − θv ) or v ⋅ i ⋅ cos (θ v − θi ) since the sign of the angle does not change the sign of the cos function

But we must have Q = v ⋅ i ⋅ sin (θi − θv ) since this gives the correct sign for Q (try θ v = 0) 17 August 2005

Engineer M S Ayubi

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Solutions to Problems with Power Evaluation When Neither Voltage or Current Phasor is the Reference Phasor Thus to evaluate the power in the circuit we take the conjugate of the voltage or the conjugate of the current in the v ⋅ i product. jθ v − jθ i j (θ v −θ i ) * v ⋅i = v ⋅ i ⋅ e ⋅e = v ⋅ i ⋅e

(

)

= v ⋅ i ⋅ ( cos(θ v − θ i ) − j sin (θ v − θ i ) )

We take only the real part as the power P = v ⋅ i ⋅ cos(θ v − θ i )

If we want the correct sign for the reactive power Q, we must take the conjugate of the voltage only Thus reactive power Q is the imaginary part of v * ⋅ i

Q = v ⋅ i ⋅ sin (θ i − θ v ) 17 August 2005

Engineer M S Ayubi

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8.12 Power Dissipated in Resistors With Complex Currents and Voltages • In dc circuits the power dissipated in resistors was

v

i R

P =i 2 R

v=iR

v2 P = R

• In ac circuits with complex phasor notation the power dissipated in resistors is P =i P = 17 August 2005

2

v

R

2

R

Engineer M S Ayubi

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8.13 Power Factor in Power Systems •

The angle between the voltage and current (θ in our previous calculation) in a power network is called the power factor angle and is usually given the symbol φ .

•

The term cos(φ ) is called the power factor

•

Most power networks in factories and shops etc. have a lagging power factor. That is the load is inductive and resistive and φ is negative.

•

The charge for industrial electricity is based on the power consumed and the maximum VA product (Maximum demand).

•

This is because – The cables and transformers required to supply a load are rated according to their maximum current (power lost in device conductors is dependent on current squared). – The bigger the rated VA the greater the cost of the equipment required to supply it – Thus it is necessary to charge for maximum demand because otherwise you could have a high current demand with virtually no power and no revenue to cover your investment in expensive power supply equipment.

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Engineer M S Ayubi

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