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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

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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.

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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

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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.

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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

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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.
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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)

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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.

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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

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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 .

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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

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2015