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5 Phasor Relationships for R, L and C

# 5 Phasor Relationships for R, L and C

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11.24

Equality of the angles  and  is apparent, and the current and voltage are

thus in phase.

The voltage-current relationship in phasor form for a resistor has the same form

as the relationship between the time-domain voltage and current as illustrated

below:

i

I

v=Ri

R

time-domain

V=R I

R

frequency-domain

Figure 11.11

EXAMPLE 11.3 Phasor Analysis with a Resistor

Assume a voltage of 8 cos100t  50 across a 4  resistor. Working in the

time-domain, the current is:

it  

vt 

 2 cos100t  50

R

The phasor form of the same voltage is 8  50 , and therefore:

I

V

 2  50

R

If we transform back to the time-domain, we get the same expression for the

current.

No work is saved for a resistor by analysing in the frequency-domain – because

the resistor has a linear relationship between voltage and current.

Index

11 - The Phasor Concept

Phasor Relationships for R, L and C

PMcL

2015

11.25

11.5.2 Phasor Relationships for an Inductor

The defining time-domain equation is:

vt   L

dit 

dt

(11.68)

After applying the complex voltage and current equations, we obtain:

Vme j t    L

d

I me j t  

dt

(11.69)

Taking the indicated derivative:

Vme j t    jLI me j t  

(11.70)

By dividing throughout by e jt , we find:

Vme j  jLI me j

(11.71)

Thus the desired phasor relationship is:

V  jLI

(11.72)

Phasor V-I

relationship for an

inductor

The time-domain equation Eq. (11.68) has become an algebraic equation in the

frequency-domain. The angle of jL is exactly  90 and you can see from

Eq. (11.71) that   90   . I must therefore lag V by 90 in an inductor.

PMcL

2015

Phasor Relationships for R, L and C

Index

11 - The Phasor Concept

11.26

The phasor relationship for an inductor is indicated below:

i

I

v=L di

dt

L

time-domain

V=j L I

L

frequency-domain

Figure 11.12

EXAMPLE 11.4 Phasor Analysis with an Inductor

Assume a voltage of 8 cos100t  50 across a 4 H inductor. Working in the

time-domain, the current is:

i t   

vt 

dt

L

  2 cos 100t  50dt

 0.02 sin 100t  50

 0.02 cos 100t  140

The phasor form of the same voltage is 8  50 , and therefore:

I

V

8  50

 0.02  140

jL 100490

If we transform back to the time-domain, we get the same expression for the

current.

Index

11 - The Phasor Concept

Phasor Relationships for R, L and C

PMcL

2015

11.27

11.5.3 Phasor Relationships for a Capacitor

The defining time-domain equation is:

it   C

dvt 

dt

(11.73)

After applying the complex voltage and current equations, we obtain:

I me j t    C

d

Vme j t  

dt

(11.74)

Taking the indicated derivative:

I me j t    jCVme j t  

(11.75)

By dividing throughout by e jt , we find:

I me j  jCVme j

(11.76)

Thus the desired phasor relationship is:

I  jCV

(11.77)

Phasor V-I

relationship for a

capacitor

Thus I leads V by 90 in a capacitor.

PMcL

2015

Phasor Relationships for R, L and C

Index

11 - The Phasor Concept

11.28

The time-domain and frequency-domain representations are compared below:

dv

i=C dt

I =j C V

v

V

C

time-domain

C

frequency-domain

Figure 11.13

EXAMPLE 11.5 Phasor Analysis with a Capacitor

Assume a voltage of 8 cos100t  50 across a 4 F capacitor. Working in the

time-domain, the current is:

dvt 

dt

d

 4 8 cos 100t  50

dt

 3200 sin 100t  50

i t   C

 3200 cos 100t  40

The phasor form of the same voltage is 8  50 , and therefore:

I  jCV  100490  8  50  320040

If we transform back to the time-domain, we get the same expression for the

current.

Index

11 - The Phasor Concept

Phasor Relationships for R, L and C

PMcL

2015

11.29

11.5.4 Summary of Phasor Relationships for R, L and C

We have now obtained the phasor V  I relationships for the three passive

elements. These results are summarized in the table below:

Time-domain

i

Frequency-domain

v

v  Ri

V  RI

I

R

i

V

R

v

vL

L

di

dt

V  jLI

I

V

j L

v

i

Summary of phasor

V-I relationships for

the passive

elements

V

v

1

idt

C

V

1

I

jC

I

1 j C

C

All the phasor equations are algebraic. Each is also linear, and the equations

relating to inductance and capacitance bear a great similarity to Ohm’s Law.

Before we embark on using the phasor relationships in circuit analysis, we

need to verify that KVL and KCL work for phasors. KVL in the time-domain

is:

v1 t   v2 t     vn t   0

(11.78)

If all voltages are sinusoidal, we can now use Euler’s identity to replace each

real sinusoidal voltage by the complex voltage having the same real part,

divide by e jt throughout, and obtain:

V1  V2    Vn  0

(11.79)

KVL and KCL are

obeyed by phasors

Thus KVL holds. KCL also holds by a similar argument.

PMcL

2015

Phasor Relationships for R, L and C

Index

11 - The Phasor Concept

11.30

11.5.5 Analysis Using Phasor Relationships

We now return to the series RL circuit that we considered several times before,

shown as (a) in the figure below. We draw the circuit in the frequency-domain,

as shown in (b):

A circuit and its

frequency-domain

equivalent

R

R

i( t )

I

VR

vs (t ) =

Vm cos( t )

L

VL

Vs

(a)

j L

(b)

Figure 11.14

From KVL in the frequency-domain:

VR  VL  Vs

(11.80)

We now insert the recently obtained V  I relationships for the elements:

RI  jLI  Vs

(11.81)

The phasor current is then found:

I

Vs

R  jL

(11.82)

The source has a magnitude of Vm and a phase of 0 (it is the reference by

which all other phase angles are measured). Thus:

The response of the

circuit in the

frequency-domain

Index

11 - The Phasor Concept

I

Vm0

R  jL

Phasor Relationships for R, L and C

(11.83)

PMcL

2015

11.31

The current may be transformed to the time-domain by first writing it in polar

form:

Vm

I

R 2   2 L2

 I m

  tan 1 L R 

(11.84)

Transforming back to the time-domain we get:

it   I m cost   

L 

cos t  tan 1

2

2 2

R

R  L

The response of the

circuit in the timedomain

Vm

(11.85)

which is the same result as we obtained before the “hard way”.

11.6 Impedance

The voltage-current relationships for the three passive elements in the

frequency-domain are:

V  RI

V  jLI

V

I

jC

(11.86)

If these equations are written as phasor-voltage phasor-current ratios, we get:

V

R

I

V

 jL

I

V

1

I

jC

(11.87)

Phasor V-I

relationships for the

passive elements

These ratios are simple functions of the element values, and in the case of the

inductor and capacitor, frequency. We treat these ratios in the same manner we

treat resistances, with the exception that they are complex quantities and all

algebraic manipulations must be those appropriate for complex numbers.

PMcL

2015

Impedance

Index

11 - The Phasor Concept

11.32

We define the ratio of the phasor voltage to the phasor current as impedance,

symbolized by the letter Z :

Z

Impedance defined

V

I

(11.88)

The impedance is a complex quantity having the dimensions of ohms.

Impedance is not a phasor and cannot be transformed to the time-domain by

multiplying by e jt and taking the real part.

In the table below, we show how we can represent a resistor, inductor or

capacitor in the time-domain with its frequency-domain impedance:

Impedances of the

three passive

elements

Time-domain

Frequency-domain

R

R

L

j L

C

1 j C

Impedances may be combined in series and parallel by the same rules we use

for resistances.

In a circuit diagram, a general impedance is represented by a rectangle:

Z

I

V = ZI

Figure 11.15

Index

11 - The Phasor Concept

Impedance

PMcL

2015

11.33

EXAMPLE 11.6 Impedance of an Inductor and Capacitor in Series

We have an inductor and capacitor in series:

100 F

5 mH

At   104 rads -1 , the impedance of the inductor is Z L  jL  j50  and the

impedance of the capacitor is

ZC  1 jC   j1  . Thus the series

combination is equivalent to Zeq  Z L  ZC  j50  j1  j 49  :

j 49 

The impedance of inductors and capacitors is a function of frequency, and this

equivalent impedance is only valid at   104 rads -1 . For example, if

  5000 rads -1 , then the impedance would be Zeq  j 23  .

Impedance may be expressed in either polar or rectangular form.

In polar form an impedance is represented by:

Z  Z 

(11.89)

No special names or symbols are assigned to the magnitude and angle. For

example, an impedance of 100  60  is described as having an impedance

magnitude of 100  and an angle of  60 .

PMcL

2015

Impedance

Index

11 - The Phasor Concept

11.34

In rectangular form an impedance is represented by:

Impedance is

composed of a

resistance (real part)

and a reactance

(imaginary part)

Z  R  jX

(11.90)

The real part, R, is termed the resistive component, or resistance. The

imaginary component, X, including sign, but excluding j, is termed the reactive

component, or reactance. The impedance 100  60  in rectangular form is

50  j86.6  . Thus, its resistance is 50  and its reactance is  86.6  .

It is important to note that the resistive component of the impedance is not

necessarily equal to the resistance of the resistor which is present in the circuit.

EXAMPLE 11.7 Impedance of a Resistor and Inductor in Series

Consider a resistor and an inductor in series:

20 

5H

At   4 rads -1 , the equivalent impedance is Zeq  20  j 20  . In this case the

resistive component of the impedance is equal to the resistance of the resistor

because the network is a simple series network. Now consider the same

elements placed in parallel:

20 

5H

The equivalent impedance is:

Zeq 

20 j 20

 10  j10 

20  j 20

The resistive component of the impedance is now 10  .

Index

11 - The Phasor Concept

Impedance

PMcL

2015

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