SINUSOIDAL STEADY STATE ANALYSES USING PHASORS

ELECTRIC CIRCUITS

1.7 SINUSOIDAL STEADY STATE ANALYSES USING PHASORS

Steady state indicates the absence of transients and it is achieved after the five times of time constant (5t) of the switching action (Fig. 1.73).

V_mcos(wt + f)

R₂

L₂ L₁

R₁

i₂(t) i₁(t)

C

+ + +

+ +

+

− − −

− +

− −

−

Figure 1.73

|

Sinusoidal steady state analyses.

(ii) dV dt

V u t +2 = ( )

Here CF = ke^−2t and PI can be determined as PI =e^−2t u t e^2tdt=e ^2t 1 e^2tdt=

1 2

( )⋅ ⋅

∫

⁻

∫

Therefore, GS = V =ke^−2t+ 1 2 0

1 2

1 2

=ke0+ ⇒k=− or V = e ^t

1 2

1 ²

( − ⁻ )

1.7  SINUSOIDaL SteaDY State aNaLYSeS USING phaSOrS       63

Table 1.5

|

Operations and their phasor representa- tions

Operation Expression

Addition z

1 + z

2 = (x

1 + x

2) + j(y

1 + y

2)

Subtraction z

1− z

2 = (x

1− x

2) + j(y

1− y

2) Multiplication z

1z

2 = r

1r

2∠f1 + f2

Division

z z

r r

1 2

1 2

1 2

= ∠ −f f

Reciprocal

1 1 z r

= ∠ −f Square root z = r∠f/2 Complex

conjugate

z* =x−jy=r∠−f=re^−jf The relation between time domain and phasor domain are listed as follows in Table 1.6:

Table 1.6

|

Relation between time and phasor domain

Time Domain Phasor Domain

V t

mcos(w +f) V

m∠f

V t

msin(w +f) V

m∠f−90

I t

mcos(w +f) I

m∠f

I t

msin(w +f) I

m∠f−90° dv

dt

jwV

∫

vdt ^V

jw Transform the network into the phasor domain,

L j L

1= w 1, i

2(t) = I

2

  L

2 = jwL

2, i

1(t) = I

1

C

j C

= 1

w V

m∠ °f

V R I j L I I

m∠f= + w ( )

1 1 1 1− 2 (1.13)

0 1

2 2 2 2 2 1 2 1

= + + +

j C

I R I j L I j L I I

w . w w ( − ) (1.14)

I1

= 1

∆

∆ ,I

2

= 2

∆

∆ Therefore, i t I e^{j t}

1( )= RP[ 1 ^w ]A i t I e^{j t}

2( )= RP[ 2 ^w ]A

Problem 1.98: For the circuit shown in figure given below, determine the phasors I, I

1, I

2, E

2, E

0. j1Ω

1Ω

j2Ω E₂

I₁ I₂ I

−j1 0.5Ω

0.5Ω E₁ = 10∠10°

E₀ + +

−

−

Solution: Z V

I

= Impedance phasors,

Z = j1 + (1 - j1) || (1 + j2) = 1 61. ∠29 74. °

I E

Z

= ¹ = =

10 1 6

9 74 6 21 9 74 .

. . .

∠−1 ° ∠−1 °

I

I j

j j

1

1 2

1 1 1 2

=

+ + +

( )

−

=

+ + +

= 6 21 9 74 1 2

1 1 1 2

6 21 17 12

. . ( )

. .

∠ °

∠ °

−

−

j

j j

E2 = (1 - j1) . I

1

= 2 45 =8 78 27 88

∠− °I1 . ∠− . ° I

I j

j j

2

1 1

1 1 1 2

3 927 91 30

=

+ +

( )

. .

−

− = ∠−

Therefore, E

0 = -0.5I

2= +1.963Ð-89.7

Frequency of sinusoidal does not appear in its phasor representation.

z = x + jy is the rectangular form Z =r∠f is the polar form Z =re^jf is the exponential form

where r= x²+y², f= tan⁻¹y/x, x=rcosf and y =rsinf.

Addition and subtraction of complex numbers are better performed in the rectangular form. Multiplication and division are better done in polar form (Table 1.5).

1.7.1 Phasor

A phasor is a complex number that represents the ampli- tude and phase of a sinusoid. Phasor is a mathematical representative of an AC quantity in polar form.

Sinusoidal form Phasor notation

v(t) = 10sin(300t + 30°) 10∠30° V v(t) = 15sinwt 15∠0° V

5 − ° ∠ − °

Solution: We have

4 8

3 50 75

I I j

j I

+ =

w− w ∠ °

As w = 2, I(4−j4−j6)=50∠ °75 I

j

= = =

50 75 4 10

50 75 10 77 68 2

4 642 143 2

∠ ° ∠ °

∠ ° ∠ °

− . − .

. . A

So, i(t) = 4.642 cos (2t + 143.2°) A

Problem 1.101: Using phasor approach, determine i(t).

4 8

3

50 2 75 i idt

di dt

t +

∫

− = cos( + °)

Problem 1.99: Transform the following sinusoids to phasors.

(a) i(t) = 6 cos (50t − 40°) I= 6∠−40°

(b) v = −4 sin (30t + 50°) sinv -sinA = cos (A + 90°)

v = 4 cos (30t + 50° + 90°)

= 4 cos (30t + 140°) v= 4∠140°

(c) v = −7 cos (2t + 40°) −cosA = cos (A + 180°)

v = 7 cos (2t + 40° + 180°)

= 7 cos (2t + 220°) v= 7∠220°

(d) i = 4 sin (10t + 10°) sinA = cos (90 - A)

= cos (A − 90°)

i = 4 cos (10t + 10°− 90°)

= 4 cos (10t − 80°) i= 4∠−80°

Problem 1.100: Find the sinusoids represented by the following phasors.

(a) I = −3 + j4

= 5∠126 87. °

i(t) = 5 cos (wt + 126.87°) A (b) v=j8e^−j20°

j= 1∠ °90

v=j8∠−20°=(1<90°)(8<−20°)

=8∠90−20=8∠ °70 V v t( )=8cos(wt+70°)V (c) v=−10∠ °30

= (j²)10∠ °30

= (1∠180°)(10∠ °30 )

= 10∠210°

v t( )=10cos(wt+210°) (d) I = j (5 - j12)

=12+j5=13∠22 62. ° i t( )=13cos(wt+22 62. °)

1.7.2 Phasor Relationship for Circuit Elements

The phasor relationship for circuit elements are listed as follows:

1. When current and voltage are in the same phase (Fig. 1.74)

i=I t+

mcos(w f) v=iR=RI t+

mcos(w f) V =RI

m∠f and I =I

m∠f V = RI

I

f 0 Im

Re V

Figure 1.74

|

Current and voltage in same phase.

2. When current lags the voltage by 90° (Fig. 1.75):

v Ldi

dt

LI t

= =−w msin(w +f) v=wI_mcos(wt+f+90°) V =wLI_me^j(f+90°)=wLI e^jfe^j

m 90°

=wLI fe^j =jwLI

m∠ ^90°

0 Im

Re I V

f

w

Figure 1.75

|

Current lags voltage by 90°.

3. When the current leads the voltage by 90° (Fig. 1.76).

V =V t+

mcos(w f) i C

dv dt

=

I j CV V

I j C

= w =

and w

1.7  SINUSOIDaL SteaDY State aNaLYSeS USING phaSOrS       65

Im

Re V

I

f

Figure 1.76

|

Current leads the voltage by 90°.

Problem 1.102: In the circuit shown in the figure given below, current I

1 leads I

2 by angle V

C

I₂ I₁

j2 V

V w = 2 rad/s

2Ω

(a) 30° (b) 60° (c) 90° (d) 135° Solution:

I I I

V Z

V Z

= + = +

1 2

C L

I

V Z

j V

C V X

1 = = = 90

C w C∠ °

I

V j

V

2

2 2 2 2 45

= +

= ∠− °

I₁

I₂ V 135° 90°

45°

Ans. (d) Problem 1.103: The currents i

1(t), −i

2(t) and i

3(t) are meeting at a junction

i₁ i₂ i₃ i1(t) = −6sinwt mA, i

2(t) = 9coswt mA, i

3(t) = ? Solution: Given that:

i1=6∠ °90 , i

2=8∠ °0 Applying KCL, we get −i

1− i

2− i

3 = 0 i3 = −(i

1 + i

2) = −6j − 8 =10∠180°+^tan−1(6 8^/ ) =10∠180°+36 86. °=10e^{j(p+36 86. °)}

i3(t) = 10 cos (wt + 36.86° + p)

w °

Problem 1.104: Determine the current I in the circuit shown in the figure given below.

3 ∠0° I_C = 4 ∠90° 1Ω I

I_L

ω = 3 rad/s

I 1 V

3 H

(a) 1∠ °90 (b) 3∠ °60 (c) 5∠−45° (d) 2∠ °45 Solution:

I e^j j

C =4∠ °90 =4 ^90°= 4 I

V j L

j

L = = = = j

w

3 0 3

1 3

3 90 3

∠ ° ∠ °

×

− −

I=I +I =j j =j=

L C 4− 3 1∠ °90 A

Ans. (a) 1.7.2.1 Phasor Diagrams

The pictorial representation of all the phase currents and phase voltages in a circuit is called the phasor diagram.

The phasor diagrams along with the characteristic equa- tions for different circuits are discussed as follows.

Series RL Circuit

In the RL series circuit shown in Fig. 1.77(a), we have VR = IR and V

L = IZ

L = IjwL = IjX

L=IX

L∠ °90 In any RL circuit due to the inductances, the cur- rent lags the voltage, therefore, the phasor diagram is as shown in Fig. 1.77(b).

R

L V

V_R

V_L +

+

−

+

−

−

(a)

V

I V_L = IX_L

f

(b)

Figure 1.77

|

RL series (a) circuit and (b) phasor diagram.

L R

I

C V

V_R V_C +

+

−

+

− V_R −

+ −

Figure 1.79

|

Series RLC circuit.

Three cases are possible depending on the capacitor and inductor voltages, that is, V

L > V

C; V

L< V

C; and V

L = VC. The corresponding phasor diagrams are illustrated in Fig. 1.80 (a), (b) and (c), respectively and the expres- sions for voltage, impedance angle and power factor given in Table 1.7.

V

90°

90° I

(V_L − V_C) V_L

V_C V_R f

(a)

V

90°

90° I

(V_C − V_L) V_L

V_C V_R f

(b)

I

I V_L

V_C V_R

V_R

(c)

Figure 1.80

|

Phasor diagrams for RLC series circuit.

(a) V

L > V

C; (b) V

L< V

C and (c) V_L = V_C.

From Fig. 1.77(b), we have Voltage V = V +V

R L

2 2

Impedance angle is f = tan ^L

R

−1 V V













Power factor = cosf (lagging) Series RC Circuit

In the series RC circuit shown in Fig. 1.78(a), we have VR = IR, so

V IZ I

j C

C = C =

1

w =-IjwC=IX_C∠-90° As in any RC circuit due to the capacitive nature, the current leads the voltage, so the phasor diagram is as shown in Fig. 1.78(b).

R

C V

V_R V_C +

+

−

+

−

−

(a)

90˚

I

V

V_C = IX_C V_R = IR f

(b)

Figure 1.78

|

RC series (a) circuit and (b) phasor diagram.

From Fig. 1.78(b), we have V = V +V

C R

2 2

f= tan ^C

R

−1 V V













Power factor = cos f (leading) Series RLC Circuit

In the series RLC circuit shown in Fig. 1.79, we have VR = IR

V IX

L = L∠ °90

V IX

C = C∠−90°

1.7  SINUSOIDaL SteaDY State aNaLYSeS USING phaSOrS       67

Table 1.7

|

Parameters for series RLC circuit conditions Condition Voltage Impedance Angle Power Factor

VL> V

C V = V + V V

R L C

2 2

( − ) f= tan ^{L C}

R

−₁ V −V V











 cosf (lagging) VL< V

C

V = V + V V

R C L

2 2

( − ) f= 











tan⁻

1 V −V V

C L

R

cosf (leading) VL = V

C V = V

R f= 0 cos f = 1 or power factor is unity. Hence the

circuit is resistive in nature and the j terms in y or z is 0

Parallel RL Circuit

Consider the parallel RL circuit given in Fig. 1.81(a).

Here, the urrent through resistor and inductor are:

I V

R V R

R

= R = I

V Z

V j L

V X

L L L

L L

= = =

w ∠−90°

So, the corresponding phasor diagram is as shown in Fig. 1.81(b).

I

L

V V_R R V_L V

I_R I_L

+ +

−

+

− − (a)

I

V I_R =

I_L = f

V R

V X_L (b)

Figure 1.81

|

RL parallel (a) circuit and (b) phasor diagram.

From Fig. 1.81(b), we have I= I +I

R L

2 2

f= tan ^L

R

−1 I I













cosf= Power factor (lagging)

Parallel RC Circuit

From the parallel RC circuit shown in Fig. 1.82(a), we have I

V R

R = I

V Z

V j L V X

C C C

C

C C

= = w = ∠ °90

I V X

C C

C

= ∠ °90

Therefore, the phasor diagram is as shown in Fig. 1.82(b) I

R C

V V_R V_L V

I_R I_C

+ +

−

+

−

−

(a) I

V I_C

I_R f

(b)

Figure 1.82

|

RL parallel (a) circuit and (b) phasor diagram.

From Fig. 1.82(b), we have I= I +I

R C

2 2

f= tan ^C

R

−1 I I













cosf= Power factor (leading) Parallel RLC Circuit

From the RLC parallel cicuits as shown in Fig. 1.83, we have

I

90°

V

(I_C − I_L)

I_L IC

I_R f

(b)

V I_C

I_L I = I_R

(c)

Figure 1.84

|

Phasor diagrams for RLC parallel circuits. (a) I

C> I

L (b) I

C< I

L (c) I

C = I

L.

I

R V

V_R V_L

I_R I_L I_C

+ +

−

V_C +

−

−

Figure 1.83

|

Parallel RLC circuit.

I V R

R = I

V X

L L

= ∠−90° I V X

C C

= ∠ °90 There are three phasor diagrams possible, depending on the relation between capacitor and inductor current, that isIC > I

L; I

C < I

L and I

C= I

L, as shown in Fig. 1.84 (a), (b) and (c), respectively. The expressions for character- istic parameters are listed in Table 1.8.

I

90°

90° V

(I_C − I_L) I_C

I_L I_R f

(a)

Table 1.8

|

Parameters for parallel RLC circuit conditions Condition Voltage Impedance Angle Power Factor

IC> I

L I = I + I I

R² ( C− L)² f= tan ^{C L}

R

−₁ I −I I











 cosf (leading)

IC< I

L I= I + I I

R² ( L− C)² f = tan ^{L C}

R

−₁ I −I I













cosf (lagging)

IC= I

L I = I

R f= 0 cosf= 1 or power factor is unity. Hence the

circuit is resistive in nature.

Problem 1.105: In the circuit shown in the figure given below, if I I

1 = 2 =10A then

8A

C R

L I₁ = 10A

120∠0°

I_L I_R= 6 I₂ = 10A

(a) I

1 will be lead by tan⁻¹ 8 6



 

, I

2 will lag by tan⁻¹ 8 6



 



(b) I

1 will lead by tan⁻¹ 6 8



 

, I

2 will lag by tan⁻¹ 6 8



 



(c) I

1 will lag by tan⁻¹ 8 6



 

, I

2 will lead by tan⁻¹ 8 6



 



(d) I

1 will lag by tan⁻¹ 6 8



 

, I

2 will lead by tan⁻¹ 6 8



 



1.7  SINUSOIDaL SteaDY State aNaLYSeS USING phaSOrS       69

Solution: From the given circuity diagram I= I + I I

R² ( C− L)² Given that I

C = 8 A, I = 10 A and I

R = 6 A. Then current through the inductor is

10= 36+( 8)² = 16 IL− ⇒I A The phasor diagram is

I_L−I_C = 8A I_L = 16A I2

= 10A

I

1 = 10A

I_C = 8A

I_R=

120∠0° 90°

6A f1

f2

I2 leads by tan⁻¹ 8 6



 

. I1 lags by tan⁻¹

8 6



 

.

Power factor cos f= = = 6

10

0 6 I

I

R . (lags) The power delivered by source,

P =V I

rms rmscosf

= =

120 2

10 2

0 6 360 . .( . )

Ans. (c)

|I

C| = wCV = 300 × 2p× 50 × 159.23 × 10⁻⁶ = 15 A

I I I

R = ²− C² = (25)²−(15)² =20A R

V I

= ^R = =

R

300 20

15W With 360 Volts main, the circuit is

C R = 15Ω

25A

360 f₁

IR = = 360

15

24 A IC = 25²−24² =7A

I V C

C = w

7 = 360 × 2p× f

1× 159.23 × 10⁻⁶ f1 = 19.4 Hz

Ans. (19.4)

Problem 1.106: A 159.23 μF capacitor is in parallel with a resistance R, draws current of 25 A, from a 300 V, 50 Hz mains. Using phasor relations, find the frequency (f  ) at which this combination draws the same current from a 360 Volts mains.

C R

I_C 25A

300V 50HZ

I_R

Solution: From the circuit, we have I

V Z

C C

=

Problem 1.107: In the circuit shown in figure given below, the reading of the ammeter (A

1) is 6 A, and ammeter (A

2) is 8 A. The reading of the ammeter (A) is

V R L

A

A₁ A₂

+

−

Solution: For the given circuit, the phasor diagram is 10

f 6

8

I= I +I

R L

2 2

= 36+64=10 A

f= 

 

 ° tan⁻¹ .

8 6

25 32

= cosf= = . ( )

6 10

0 6 lags

Ans. (10)

The dot convetion and equations for voltage V

o for different circuits are shown in Fig. 1.86.

L₁ i

L₂ V_o

− + M

V M

di dt

o =

(a)

L₁ i

L₂ V_o

− + M

V M

di dt

o =−

(b)

L₁ i

L₂ V_o

− + M

V M

di dt

o =−

(c)

L₁ i

L₂ V_o

− + M

V M

di dt

o =

(d)

Figure 1.86

|

Circuits with dot convention and voltage equations.

1.8.2 Series Connection of Coupled Inductors

Figures 1.87(a) and (b) show two circuits for series - connected coupled inductors.

L₁ V₁

L₂ M

V₂ i₂

i₁ V₁ L₁

L₂ M

V₂ i₂ i₁

(a) (b)

Figure 1.87

|

Series connection of coupled inductors.

1.8 MAGNETICALLY COUPLED