Electric Circuit Theory

(1)

Electric Circuit Theory

Nam Ki Min

010-9419-2320

nkmin@korea.ac.kr

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Sinusoidal

Steady-State Analysis

Chapter 11 Nam Ki Min

010-9419-2320

nkmin@korea.ac.kr

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Contents and Objectives 3

Chapter Contents

11.1 The Sinusoidal Source 11.2 The Sinusoidal Response 11.3 The Phasor

11.4 The Passive Circuit Elements in the Frequency Domain 11.5 Kirchhoff’s Laws in the Frequency Domain

11.6 Series, Parallel, and Delta-to-Wye Simplifications

11.7 Source Transformations and Thevenin-Norton Equivalent Circuits 11.8 The Node-Voltage Method

11.9 The Mesh-Current Method 11.10 The Transformer

11.11 The Ideal Transformer 11.12 Phasor Diagrams

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Contents and Objectives 4

Chapter Objectives

1. Understand phasor concepts and be able to perform a phasor transform and an inverse phasor transform.

2. Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts.

3. Know how to use the following circuit analysis techniques to solve a circuit in the frequency domain:

· Kirchhoff’s laws;

· Series, parallel, and delta-to-wye simplifications;

· Voltage and current division;

· Thevenin and Norton equivalents;

· Node-voltage method; and · Mesh-current method.

4. Be able to analyze circuits containing linear transformers using phasor methods.

5. Understand the ideal transformer constraints and be able to analyze circuits containing ideal transformers using phasor methods.

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The Sinusoidal Source 5

 Sinusoidal Source

 Sinusoidal voltage source

• Produces a voltage that varies sinusoidally with time.

 Sinusoidal current source

 Sinusoidal source or sinusoidally time-varying excitation or sinusoid.

• Produce a signal that has the form of the sine or cosine function.

• Produces a current that varies sinusoidally with time.

𝑣

𝑒

𝑣

_𝑠

𝑖

_𝑠

𝑖

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 Alternating Circuit

 A sinusoidal current is usually referred to as alternating current (ac).

 Circuits driven by sinusoidal current or voltage sources are called ac circuits.

𝜋 2

− 𝜋 2

𝜋 0

𝜔 𝑣(𝑡)

 AC Waveforms

Sinusoidal wave

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 Why Sinusoids?

 Nature itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the economic fluctuations of the stock market, and the natural response of underdamped second-order systems’

 A sinusoidal signal is easy to generate and transmit.

It is the form of voltage generated throughout the world and supplied to homes, factories, laboratories, and so on.

It is the dominant form of signal in the communications and electric power industries.

 Through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals.

 A sinusoid is easy to handle mathematically.

𝜋 2

− 𝜋 2

𝜋 0

𝜔 𝑣(𝑡)

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 Sinusoidal Voltage

𝑣 𝑡 = 𝑉

_𝑚

sin 𝜔𝑡 𝑣 𝑡 = 𝑉

_𝑚

cos 𝜔𝑡

𝑉_𝑚cos 𝜔𝑡

𝑉_𝑚sin 𝜔𝑡 𝑣 𝑡

𝜙 = 90° 𝜔𝑡

𝑣 𝑡 = 𝑉

_𝑚

sin(𝜔𝑡 ± 90°) = ±𝑉

_𝑚

cos 𝜔𝑡 𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 ± 90°) = ∓𝑉

_𝑚

sin 𝜔𝑡

𝑉_𝑚

 A sinusoid can be expressed in either sine or cosine form.

 When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.

 This is achieved by using the following trigonometric identities:

sin(𝐴 ± 𝐵) = sin 𝐴 cos 𝐵 ± cos 𝐴 sin 𝐵

cos(𝐴 ± 𝐵) = cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵

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 Sinusoidal Voltage

𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙)

• 𝑉_𝑚: amplitude

• 𝜔 ∶ angular frequency in radians/s

• 𝜔𝑡: argument

• 𝑇 ∶ period in second

𝑇 = 2𝜋 𝜔

𝑣 𝑡 + 𝑇 = 𝑉

_𝑚

cos 𝜔 𝑡 + 𝑇 = 𝑉

_𝑚

cos 𝜔 𝑡 +

^2𝜋_𝜔

= 𝑉

_𝑚

cos 𝜔𝑡

= 𝑉

_𝑚

cos(𝜔𝑡 + 2𝜋)

= 𝑣(𝑡)

• 𝑓 ∶ frequency in Hz

𝑓 = 1

𝜔 = 2𝜋𝑓

 A more general expression:

• 𝜙: phase angle

(9.1)

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 Converting Sine to Cosine

𝑉_𝑚cos 𝜔𝑡

𝑉_𝑚sin 𝜔𝑡 𝑣 𝑡

𝜙 = 90° 𝜔𝑡 𝑉_𝑚

𝑣 𝑡 = 𝑉

_𝑚

sin(𝜔𝑡 + 90°) = 𝑉

_𝑚

cos 𝜔𝑡 𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 − 90°) = 𝑉

_𝑚

sin 𝜔𝑡

𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 ± 180°) = −𝑉

_𝑚

cos 𝜔𝑡 𝑣 𝑡 = 𝑉

_𝑚

sin(𝜔𝑡 ± 180°) = −𝑉

_𝑚

sin 𝜔𝑡 𝑣 𝑡 = 𝑉

_𝑚

cos(5𝑡 + 10°)

_{= 𝑉}

𝑚

sin(5t + 90° + 10°)

= 𝑉

_𝑚

sin(5t + 100°) 𝑣(𝑡) = 𝑉

_𝑚

sin(5t − 260°)

= 𝑉

_𝑚

sin(5t − 180° − 80°)

= −𝑉

_𝑚

sin(5t + 90 − 170°)

= −𝑉

_𝑚

cos(5t − 180 + 10°)

= 𝑉

_𝑚

cos(5t + 10°)

= −𝑉

_𝑚

sin(5t − 80°)

= −𝑉

_𝑚

cos(5t − 170°)

(11)

 Phase Relation of a Sinusoidal Wave

(12)

 Phase Shift (difference)

 The two waves (A versus B) are of the same amplitude and frequency, but they are out of step with each other.

 In technical terms, this is called a phase shift (difference).

𝜙

_𝐴

𝜙

_𝐵

𝜙 = 0 𝑣 𝑡 = 𝑉

_𝑚

sin(𝜔𝑡 + 𝜙

_𝐴

)

𝑣 𝑡 = 𝑉

_𝑚

sin(𝜔𝑡 + 𝜙

_𝐵

)

𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙

_𝐴

) 𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙

_𝐵

)

𝜔

𝜙_𝐴 𝜙_𝐵

𝑣 𝑡 𝑣 𝑡

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 Phase Shift (difference)

 Examples of phase shifts

The sinusoids are said to be in phase.

The sinusoids are said to be out of phase.

B “lags” A

A “lags” B

Leading Lagging

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The Sinusoidal Response 14

 Sinusoidal Response

 A sinusoidal forcing function produces both a natural (or transient) response and a forced (or steady-state) response, much like the step function, which we studied in Chapters 7 and 8.

 The natural response of a circuit is dictated by the nature of the circuit, while the steady- state response always has a form similar to the forcing function.

 However, the natural response dies out with time so that only the steady-state response remains after a long time.

 When the natural response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state.

 It is this sinusoidal steady-state response that is of main interest to us in this chapter.

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15

𝑖 𝑡 = 1 𝑒

^𝑅^𝐿𝑡

𝑒

^𝑅^𝐿𝑡

𝑉

𝑚

𝐿 cos(𝜔𝑡 + 𝜙)𝑑𝑡 + 𝐾 𝑒

^𝑅^𝐿𝑡

The Sinusoidal Response

𝜇 𝑡 = 𝑒

^𝑅^𝐿𝑡

𝑔(𝑡) = 𝑉

_𝑚

𝐿 cos(𝜔𝑡 + 𝜙)

= 1 𝑒

^𝑅^𝐿𝑡

𝑉

_𝑚

𝐿 𝑒

^𝑅^𝐿𝑡

cos 𝜔𝑡 cos 𝜙 − sin 𝜔𝑡 sin 𝜙 𝑑𝑡 + 𝐾 𝑒

^𝑅^𝐿𝑡

= 1 𝑒

^𝑅^𝐿𝑡

𝑉

_𝑚

𝐿 cos 𝜙 𝑒

^𝑅^𝐿𝑡

cos 𝜔𝑡 𝑑𝑡 − sin 𝜙 𝑒

^𝑅^𝐿𝑡

sin 𝜔𝑡 𝑑𝑡 + 𝐾 𝑒

^𝑅^𝐿𝑡

cos(𝜔𝑡 + 𝜙) = cos 𝜔𝑡 cos 𝜙 − sin 𝜔𝑡 sin 𝜙

exp 𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 = 𝑒^𝑎𝑥

𝑎²+ 𝑏²(𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥) exp 𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 = 𝑒^𝑎𝑥

𝑎²+ 𝑏²(𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥)

(5)

(6)

(16)

The Passive Circuit Elements in the Frequency Domain 16

= 1 𝑒

^𝑅^𝐿𝑡

𝑉

_𝑚

𝐿 cos 𝜙 𝑒

^𝑅^𝐿𝑡

cos 𝜔𝑡 𝑑𝑡 − sin 𝜙 𝑒

^𝑅^𝐿𝑡

sin 𝜔𝑡 𝑑𝑡 + 𝐾 𝑒

^𝑅^𝐿𝑡

exp 𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 = 𝑒^𝑎𝑥

𝑎² + 𝑏²(𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥) exp 𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 = 𝑒^𝑎𝑥

𝑎²+ 𝑏²(𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥)

= 1 𝑒

^𝑅^𝐿𝑡

𝑉

_𝑚

𝐿 cos 𝜙 𝑒

^𝑎𝑥

𝑎

²

+ 𝑏

²

(𝑎 cos 𝑏𝑡 + 𝑏 sin 𝑏𝑡) − sin 𝜙 𝑒

^𝑎𝑥

𝑎

²

+ 𝑏

²

(𝑎 sin 𝑏𝑡 − 𝑏 cos 𝑏𝑡) + 𝐾 𝑒

^𝑅^𝐿𝑡

𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥 = 𝑎² + 𝑏² 𝑎

𝑎² + 𝑏²cos 𝑏𝑥 + 𝑏

𝑎²+ 𝑏²sin 𝑏𝑥 = 𝑎²+ 𝑏² cos 𝜃 cos 𝑏𝑥 + sin 𝜃 sin 𝑏𝑥

𝑎 = 𝑅 𝐿

𝑏 = 𝜔 𝑒

^𝑎𝑥

= 𝑒

^𝑅^𝐿𝑡

= 𝑉

_𝑚

𝐿 cos 𝜙 1

𝑎

²

+ 𝑏

²

(𝑎 cos 𝑏𝑡 + 𝑏 sin 𝑏𝑡) − sin 𝜙 1

𝑎

²

+ 𝑏

²

(𝑎 sin 𝑏𝑡 − 𝑏 cos 𝑏𝑡) + 𝐾 𝑒

^𝑅^𝐿𝑡

= 𝑎²+ 𝑏² cos(𝜃 − 𝑏𝑥) 𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥 = 𝑎²+ 𝑏² cos 𝜃 sin 𝑏𝑥 − sin 𝜃 cos 𝑏𝑥 = 𝑎²+ 𝑏² sin(𝑏𝑥 − 𝜃)

(7)

(8)

(17)

𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥 = 𝑎² + 𝑏² 𝑎

𝑎² + 𝑏²cos 𝑏𝑥 + 𝑏

𝑎²+ 𝑏²sin 𝑏𝑥 = 𝑎²+ 𝑏² cos 𝜃 cos 𝑏𝑥 + sin 𝜃 sin 𝑏𝑥

= 𝑎²+ 𝑏² cos(𝜃 − 𝑏𝑥) 𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥 = 𝑎²+ 𝑏² cos 𝜃 sin 𝑏𝑥 − sin 𝜃 cos 𝑏𝑥 = 𝑎²+ 𝑏² sin(𝑏𝑥 − 𝜃)

= 𝑉

_𝑚

𝐿 cos 𝜙 1

𝑎

²

+ 𝑏

²

𝑎

²

+ 𝑏

²

cos(𝜃 − 𝑏𝑡 − sin 𝜙 1

𝑎

²

+ 𝑏

²

𝑎

²

+ 𝑏

²

sin(𝑏𝑡 − 𝜃) + 𝐾 𝑒

^𝑅^𝐿𝑡

= 𝑉

_𝑚

𝐿

1 𝑎

²

+ 𝑏

²

cos 𝜙 cos(𝜃 − 𝜔𝑡) − sin 𝜙 sin(𝜔𝑡 − 𝜃) + 𝐾 𝑒

^𝑅^𝐿𝑡

= 𝑉

_𝑚

𝐿

1 𝑅 𝐿

2

+ 𝜔

²

cos 𝜙 cos(𝜔𝑡 − 𝜃) − sin 𝜙 sin(𝜔𝑡 − 𝜃) + 𝐾 𝑒

^𝑅^𝐿𝑡

𝑎 = 𝑅 𝐿 𝑏 = 𝜔

= 𝑉

_𝑚

𝐿 cos 𝜙 1

𝑎

²

+ 𝑏

²

(𝑎 cos 𝑏𝑡 + 𝑏 sin 𝑏𝑡) − sin 𝜙 1

𝑎

²

+ 𝑏

²

(𝑎 sin 𝑏𝑡 − 𝑏 cos 𝑏𝑡) + 𝐾 𝑒

^𝑅^𝐿𝑡

(9)

(10)

(18)

= 𝑉

_𝑚

𝐿

1 𝑅 𝐿

2

+ 𝜔

²

cos 𝜙 cos(𝜔𝑡 − 𝜃) − sin 𝜙 sin(𝜔𝑡 − 𝜃) + 𝐾 𝑒

^𝑅^𝐿𝑡

𝑖(𝑡) = 𝑉

_𝑚

1 𝑅

²

+ (𝜔𝐿)

²

cos(𝜔𝑡 + 𝜙 − 𝜃) + 𝐾 𝑒

^𝑅^𝐿𝑡

𝑎 = 𝑅 𝐿 𝑏 = 𝜔

0 = 𝑉

_𝑚

1 𝑅

²

+ (𝜔𝐿)

²

cos( 𝜙 − 𝜃) + 𝐾 𝐾 = −𝑉

_𝑚

1 𝑅

²

+ (𝜔𝐿)

²

cos( 𝜙 − 𝜃)

𝑖(𝑡) = 𝑉

_𝑚

𝑅

²

+ 𝜔𝐿

²

cos(𝜔𝑡 + 𝜙 − 𝜃) − 𝑉

_𝑚

𝑅

²

+ 𝜔𝐿

²

cos( 𝜙 − 𝜃) 𝑒

^−𝑅^𝐿𝑡

(10)

(11)

(12)

(13) (9.9)

(19)

𝑖 𝑡 = − 𝑉

_𝑚

𝑅

²

+ 𝜔𝐿

²

cos( 𝜙 − 𝜃) 𝑒

^−𝑅^𝐿𝑡

+ 𝑉

_𝑚

𝑅

²

+ 𝜔𝐿

²

cos(𝜔𝑡 + 𝜙 − 𝜃)

(9.9)

 Sinusoidal Response

Transient(Natural) Component

• The natural response of a circuit is dictated by the nature of the circuit.

• The natural response dies out with time.

Steady-state Component (Forced Response)

• The steady-state solution is a sinusoidal function.

• The frequency of the response signal is identical to the frequency of the source signal.

• The maximum amplitude of the steady-state response, in general, differs from the maximum amplitude of the source.

• The phase angle of the response signal, in general, differs from the phase angle of the source

.

𝑣

_𝑠

(𝑡) = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙)

When the natural response has become negligibly small compared with the steady- state response, we say that the circuit is operating at sinusoidal steady state.

It is this sinusoidal steady-state response that is of main interest to us in this chapter.

(20)

The Phasor 20

 Definition

 A phasor is a complex number that represents the amplitude and phase angle of a sinusoid.

𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙) → 𝐕 = 𝑉

_𝑚

𝑒

^𝑗𝜙

= 𝑉

_𝑚

∠𝜙

phasor representation

 When a phasor is used to describe an AC quantity,

the length of a phasor represents the amplitude of the wave while the angle of a phasor represents the phase angle of the wave relative to some other (reference) waveform.

(21)

21

 Complex Number

 A complex number z can be written in rectangular form as

𝑧 = 𝑥 + 𝑗𝑦 𝑗 = −1

Representation of a complex number z in the complex plane

𝑥:

the real part of z

the imaginary part of z

𝑦:

 The complex number z can also be written in polar or exponential form as

𝑧 = 𝑟𝑒

^𝑗𝜙

𝑟 = 𝑥

²

+ 𝑦

²

𝜙 = 𝑡𝑎𝑛

⁻¹

𝑦

𝑥 𝑧 = 𝑟∠𝜙

 The relationship between the rectangular form and the polar form is

𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos 𝜙 + 𝑗 sin 𝜙) = 𝑟∠𝜙

The Phasor

(22)

22

 Complex Number

 Addition and subtraction of complex numbers

𝑧

₁

= 𝑥

₁

+ 𝑗𝑦

₁

= 𝑟

₁

∠𝜙

₁

 Multiplication and division

𝑧

₁

𝑧

₂

= 𝑟

₁

𝑟

₂

∠𝜙

₁

+ 𝜙

₂

 Reciprocal

 Square Root

𝑧

₂

= 𝑥

₂

+ 𝑗𝑦

₂

= 𝑟

₂

∠𝜙

₂

𝑧

₁

+ 𝑧

₂

= (𝑥

₁

+ 𝑥

₂

) + 𝑗(𝑦

₁

+ 𝑦

₂

) 𝑧

₁

− 𝑧

₂

= (𝑥

₁

− 𝑥

₂

) + 𝑗(𝑦

₁

− 𝑦

₂

)

𝑧

₁

𝑧

₂

= 𝑟

₁

𝑟

₂

∠𝜙

₁

− 𝜙

₂

1 𝑧 = 1

𝑥 + 𝑗𝑦 = 1

𝑥

²

+ 𝑦

²

∠ − 𝜙 1

𝑧 = 1

𝑟∠𝜙 = 1

𝑟 ∠ − 𝜙

𝑧 = 𝑟∠𝜙/2

The Phasor

Representation of a complex number z in the complex plane

(23)

23

 Complex Conjugate

 Complex Conjugate

𝑧 = 𝑥 + 𝑗𝑦

𝑧

^∗

= 𝑥 − 𝑗𝑦 = 𝑟∠ − 𝜙 = 𝑟𝑒

^−𝜙

−𝑗 = 1

𝑗 ← 1

𝑗 = 1

1∠90° = 1

1 ∠ − 90° = −𝑗

The Phasor

Representation of a complex number z in the complex plane

(24)

24

 The Phasor Representation of the Sinusoid v(t)

 The idea of phasor representation is based on Euler’s identity. In general,

𝑒

^±𝑗𝜃

= cos 𝜃 + 𝑗 sin 𝜃

^(9.10)

 We can regard cos 𝜃 and sin 𝜃 as the real and imaginary parts of 𝑒^𝑗𝜃; we may write

cos 𝜃 = Re{𝑒

^𝑗𝜃

} sin 𝜃 = Im{𝑒

^𝑗𝜃

}

(9.11) (9.12)

 We write the sinusoidal voltage function given by Eq.(9.1) in the form suggested by Eq.(9.11)

(9.1)

𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙)

= Re{𝑉

_𝑚

𝑒

^{𝑗 𝜔𝑡+𝜙}

}

= Re{𝑉

_𝑚

𝑒

^𝑗𝜙

𝑒

^𝑗𝜔𝑡

}

𝑉

_𝑚

𝑒

^𝑗𝜙 : a complex number that carries the amplitude and phase angle of the given sinusoidal function.

 We define the phasor representation or phasor transform of the given sinusoidal function as

𝑣 𝑡 = Re{𝐕𝑒

^𝑗𝜔𝑡

} 𝐕 = 𝑉

_𝑚

𝑒

^𝑗𝜙 ^(9.15)

(9.14)

The Phasor

(25)

25

 The Phasor Representation of the Sinusoid v(t)

 Phasor representation

𝐕 = 𝑉

_𝑚

𝑒

^𝑗𝜙 _(9.15)

𝐕 = 𝑉

_𝑚

∠𝜙

𝐕 = 𝑉

_𝑚

(cos 𝜙 + 𝑗 sin 𝜙)

^(9.16)

 One way of looking at Eqs. (9.15) and (9.16) is to consider the plot of the 𝐕𝑒^𝑗𝜔𝑡 on the complex plane.

 As time increases, the 𝐕𝑒^𝑗𝜔𝑡 rotates on a circle of radius

𝑉

_𝑚 at an angular velocity ω in the counterclockwise direction, as shown in Fig. (a). In other words, the entire complex plane is rotating at an angular velocity of ω.

 We may regard 𝑣(𝑡) as the projection of the 𝐕𝑒^𝑗𝜔𝑡 on the real axis, as shown in Fig.(b).

 The value of the 𝐕𝑒^𝑗𝜔𝑡 at time t = 0 is the phasor 𝐕 of the sinusoid 𝑣(𝑡). The 𝐕𝑒^𝑗𝜔𝑡 may be regarded as a rotating phasor.

Thus, whenever a sinusoid is expressed as a phasor, the term 𝒆^𝒋𝝎𝒕 is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency ω of the phasor; otherwise we can make serious mistakes.

𝐕

𝐕𝑒^𝑗𝜔𝑡 The Phasor

(26)

26

 Phasor Transform : Summary

 Eqs.(9.14) through (9.16) reveal that to get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we take out the time factor 𝑒^𝑗𝜔𝑡, and whatever is left is the phasor corresponding to the sinusoid.

 By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:

= Re{𝑉

_𝑚

𝑒

^𝑗𝜙

𝑒

^𝑗𝜔𝑡

} 𝑣 𝑡 = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙)

= Re{𝐕𝑒

^𝑗𝜔𝑡

}

𝐕 = 𝑉

_𝑚

𝑒

^𝑗𝜙

= 𝑉

_𝑚

∠𝜙

⁽¹⁾

Time-domain representation Phasor or frequency-domain representation

𝑉

_𝑚

sin(𝜔𝑡 + 𝜙) = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙 − 90°)

The Phasor

(27)

27

 Inverse Phasor Transform

 Equation (1) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time factor 𝑒^𝑗𝜔𝑡 and take the real part.

𝐕 = 115∠ − 45°

𝐕 = 𝑉

_𝑚

∠𝜙

𝜔 = 500 rad/s

= 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙) 𝑣 𝑡 = 115 cos(500𝑡 − 45°)

𝑣(𝑡) = Rm{𝑉

_𝑚

𝑒

^𝑗𝜙

𝑒

^𝑗𝜔𝑡

}

= 𝑉

_𝑚

𝑒

^𝑗𝜙

𝐕 = 𝑗8𝑒

^−𝑗20°

= (1∠90°)(8∠ − 20°)

= 8∠90° − 20°

= 8∠70°° 𝑣 𝑡 = 8 cos(𝜔𝑡 + 70°)

𝐕 = 𝑗 5 − 𝑗12 = 12 + 𝑗5

= 12

²

+ 5

²

∠𝜙

^{𝜙 = tan}⁻¹₁₂⁵ ^{= 22.62°}

𝑣 𝑡 = 13 cos(𝜔𝑡 + 22.62°)

= 13∠22.62°

The Phasor

(28)

28

 Phasor Diagram

 Since a phasor has magnitude and phase (“direction”), it behaves as a vector and is printed in boldface.

For example, phasors

𝐕 = 𝑉

_𝑚

∠𝜙

and

𝐈 = 𝐼

_𝑚

∠ − 𝜃

are graphically represented in Figure.

 Such a graphical representation of phasors is known as a phasor diagram.

The Phasor

(29)

29

 Phasor Diagram

 The relationship between two phasors at the same frequency remains constant as they rotate; hence the phase angle is constant.

 Consequently, we can usually drop the reference to rotation in the phasor diagrams and study the relationship between phasors simply by plotting them as vectors having a common origin and separated by the appropriate angles.

Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of

Phasor Diagram 𝝓

𝜙

𝐈 𝐕

𝐕 𝜙 𝐈

𝑣(𝑡) = 𝑉_𝑚cos(𝜔𝑡 + 𝜙₂)

𝑖(𝑡) = 𝐼_𝑚 cos(𝜔𝑡 + 𝜙₁) 𝐕 = 𝑽_𝒎

∠𝜙

₂

𝐈 = 𝑰_𝒎

∠𝜙

₁

𝝓 = 𝝓_𝟐 − 𝝓_𝟏

The Phasor

(30)

< The V- I Relationship for a Resistor>

 Time Domain

• Current

𝑖(𝑡) = 𝐼

_𝑚

cos(𝜔𝑡 + 𝜙)

• Voltage

𝑣 𝑡 = 𝑖𝑅 = 𝑅𝐼

_𝑚

cos(𝜔𝑡 + 𝜙)

 Phasor form 𝐈 = 𝐼

_𝑚

∠𝜙

𝐕 = 𝑅𝐈 = 𝑅𝐼

_𝑚

∠𝜙

Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.

This shows that the voltage-current relation for the resistor in the phasor domain continues to be Ohm’s law, as in the time domain.

(31)

< The V- I Relationship for a Resistor>

 Phasor Diagram

𝐈 = 𝐼

_𝑚

∠𝜙

𝐕 = 𝑅𝐈 = 𝑅𝐼

_𝑚

∠𝜙

𝐈 = 𝐼

_𝑚

∠0°

𝐕 = 𝑅𝐈 = 𝑅𝐼

_𝑚

∠0°

(32)

< The V- I Relationship for an Inductor>

 Time Domain

𝑖(𝑡) = 𝐼

_𝑚

cos(𝜔𝑡 + 𝜙) 𝑣 𝑡 = 𝐿 𝑑𝑖

𝑑𝑡 = −𝜔𝐿𝐼

_𝑚

sin(𝜔𝑡 + 𝜙) 𝑣

𝑖

= −𝜔𝐿𝐼

_𝑚

cos(𝜔𝑡 + 𝜙 − 90°)

= 𝜔𝐿𝐼

_𝑚

cos(𝜔𝑡 + 𝜙 + 90°)

(33)

 Phasor form 𝐈 = 𝐼

_𝑚

∠𝜙

𝐕 = 𝜔𝐿𝐼

_𝑚

∠𝜙 + 90°

𝑣(𝑡) = 𝜔𝐿𝐼

_𝑚

cos(𝜔𝑡 + 𝜙 + 90°)

= (1∠90°)(𝜔𝐿𝐼

_𝑚

∠𝜙)

𝑗 = 1∠90°

= 𝑗𝜔𝐿𝐼

_𝑚

∠𝜙

= 𝑗𝜔𝐿𝐈

Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.

(34)

𝑑𝑖

𝑑𝑡 = 𝜔𝐼

_𝑚

cos(𝜔𝑡 + 𝜙 + 90°)

= Re{𝜔𝐼

_𝑚

𝑒

^𝑗𝜔𝑡

𝑒

^𝑗𝜙

𝑒

^𝑗90°

}

= Re{𝑗𝜔𝐼

_𝑚

𝑒

^𝑗𝜔𝑡

𝑒

^𝑗𝜙

}

= Re{𝑗𝜔𝐈 𝑒

^𝑗𝜔𝑡

}

𝑑𝑖 𝑑𝑡

𝐈 = 𝐼

_𝑚

∠𝜙 𝑖(𝑡) = 𝐼

_𝑚

cos(𝜔𝑡 + 𝜙)

𝑗𝜔𝐈

Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by

𝑗𝜔

^.

𝑣 𝑡 = 𝐿 𝑑𝑖

𝑑𝑡 = 𝐿𝑗𝜔𝐈 = 𝑗𝜔𝐿𝐈

(35)

35

 Phasor Diagram

The Passive Circuit Elements in the Frequency Domain

𝐈 = 𝐼

_𝑚

∠0°

= 𝑗𝜔𝐿𝐈

Voltage leads current by 90°in an inductor Current lags voltage by 90° in an inductor

𝐕 = 𝜔𝐿𝐼_𝑚∠90°

𝐈 = 𝐼_𝑚∠𝜙 𝐕 = 𝜔𝐿𝐼_𝑚∠𝜙 + 90°

(36)

< The V- I Relationship for a Capacitor>

 Time Domain

𝑣(𝑡) = 𝑉

_𝑚

cos(𝜔𝑡 + 𝜙) 𝑖 𝑡 = 𝐶 𝑑𝑣

𝑑𝑡 = −𝜔𝐶𝑉

_𝑚

sin(𝜔𝑡 + 𝜙)

= −𝜔𝐶𝑉

_𝑚

cos(𝜔𝑡 + 𝜙 − 90°)

= 𝜔𝐶𝑉

_𝑚

cos(𝜔𝑡 + 𝜙 + 90°)

The current leads the voltage by 90◦.

𝑣

𝑖

Voltage lags current by 90^o in a pure capacitive circuit.

(37)

 Phasor form 𝐕 = 𝑉

_𝑚

∠𝜙

𝐈 = 𝜔𝐶𝑉

_𝑚

∠𝜙 + 90°

𝑖(𝑡) = 𝜔𝐶𝑉

_𝑚

cos(𝜔𝑡 + 𝜙 + 90°)

= (1∠90°)(𝜔𝐶𝑉

_𝑚

∠𝜙)

𝑗 = 1∠90°

= 𝑗𝜔𝐶𝑉

_𝑚

∠𝜙

= 𝑗𝜔𝐶𝐕

𝐕 = 1 𝑗𝜔𝐶 𝐈

Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.

(38)

 Phasor Diagram

𝐕 = 𝑉_𝑚∠𝜙 𝐈 = 𝜔𝐶𝑉_𝑚∠𝜙 + 90°

𝐕 = 𝑉_𝑚∠0°

𝐈 = 𝜔𝐶𝑉_𝑚∠90°

= 𝑗𝜔𝐶𝐕

(39)

The Passive Circuit Elements in the Frequency Domain 39 Summary

Summary of voltage-current relationships

(40)

< Impedance, Reactance, and Admittance>

 Impedance

 We obtained the voltage-current relations for the three passive elements as

𝐕 = 𝑅𝐈

𝐕 = 1 𝑗𝜔𝐶 𝐈 𝐕 = 𝑗𝜔𝐿𝐈

𝐕

𝐈 = 𝑅

𝐕

𝐈 = 𝑗𝜔𝐿

𝐕

𝐈 = 1 𝑗𝜔𝐶

From these three expressions, we obtain Ohm’s law in phasor form for any type of element as

𝐙 = 𝐕

𝐈

^or

𝐕 = 𝐙𝐈

Z is a frequency-dependent quantity known as impedance, measured in ohms.

(41)

 Impedance

 The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.

 The impedance represents the opposition which the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.

 The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39).

Table 9.1 summarizes their impedances.

(42)

 Resistance and Reactance

 As a complex quantity, the impedance may be expressed in rectangular form as

𝐙 = 𝐕

𝐈 = 𝑅 + 𝑗𝑋

𝑅 : Resistance 𝑋 : Reactance

 Inductive and capacitive reactance

𝐙 = 𝑅 + 𝑗𝑋 𝑋<0 :

𝑋>0 :

𝐙 = 𝑅 − 𝑗𝑋

Inductive reactance Capacitive reactance

𝐙 = 𝑅 𝐙 = 𝑗𝜔𝐿

𝐙 = 1

𝑗𝜔𝐶 = −𝑗 1 𝜔𝐶

Resistor : Inductor : Capacitor:

(43)

 Impedance in Polar Form

𝐙 = 𝑅 + 𝑗𝑋 𝐙 = 𝑍

∠𝜃

𝑍 = 𝑅

²

+ 𝑋

² _{𝜃 = tan}⁻¹^𝑋

𝑅

𝐙

𝑅 𝑗𝑋

𝜃

𝑅 = 𝑧 cos 𝜃

𝑋 = 𝑍 sin 𝜃

 Admittance

 It is sometimes convenient to work with the reciprocal of impedance, known as admittance.

 The admittance Y is the reciprocal of impedance, measured in siemens (S).

𝐘 = 1 𝐙 = 𝐈

𝐕

 As a complex quantity, we may write Y as

𝐘 = 𝐺 + 𝑗𝐵

𝐺 : Conductance 𝐵 : Susceptance

𝐘 = 1

𝑅 + 𝑗𝑋 = 𝑅 − 𝑗𝑋 𝑅

²

+ 𝑋

²

𝐺 = 𝑅

𝑅

²

+ 𝑋

²

𝐵 = − 𝑋

𝑅

²

+ 𝑋

²

(44)

𝐘 = 𝐺 + 𝑗𝐵

𝐺 : Conductance 𝐵 : Susceptance

𝐘 = 1

𝑅 + 𝑗𝑋 = 𝑅 − 𝑗𝑋 𝑅

²

+ 𝑋

²

𝐺 = 𝑅

𝑅

²

+ 𝑋

²

𝐵 = − 𝑋

𝑅

²

+ 𝑋

²

(45)

Kirchhoff’s Laws in the Frequency Domain 45

 KVL in the frequency Domain

 For KVL, let 𝑣₁, 𝑣₂, ⋯ , 𝑣_𝑛 be the voltages around a closed loop. Then

𝑣

₁

+ 𝑣

₂

+ ⋯ + 𝑣

_𝑛

= 0

 In the sinusoidal steady state, each voltage may be written in cosine form, so that Eq.

(9.36) becomes

(9.36)

𝑉

_𝑚1

cos(𝜔𝑡 + 𝜃

₁

) + 𝑉

_𝑚2

cos(𝜔𝑡 + 𝜃

₂

) + ⋯ + 𝑉

_𝑚𝑛

cos(𝜔𝑡 + 𝜃

_𝑛

) = 0

^(9.37)

 We now Euler’s identity to write Eq.(9.37) as

Re 𝑉

_𝑚1

𝑒

^𝑗𝜃¹

𝑒

^𝑗𝜔𝑡

} + Re{𝑉

_𝑚2

𝑒

^𝑗𝜃²

𝑒

^𝑗𝜔𝑡

} + ⋯ + Re{𝑉

_𝑚𝑛

𝑒

^𝑗𝜃^𝑛

𝑒

^𝑗𝜔𝑡

} = 0

^(9.38)

or

Re (𝑉

_𝑚1

𝑒

^𝑗𝜃¹

+ 𝑉

_𝑚2

𝑒

^𝑗𝜃²

+ ⋯ + 𝑉

_𝑚𝑛

𝑒

^𝑗𝜃^𝑛

)𝑒

^𝑗𝜔𝑡

} = 0

 If we let

𝐕

_𝑘

= 𝑉

_𝑚𝑘

𝑒

^𝑗𝜃^𝑘^{, then}

Re (𝐕

₁

+ 𝐕

_𝟐

+ ⋯ + 𝐕

_𝒏

)𝑒

^𝑗𝜔𝑡

} = 0

 Since

𝑒

^𝑗𝜔𝑡

≠ 0

^,

𝐕

₁

+ 𝐕

_𝟐

+ ⋯ + 𝐕

_𝒏

= 0

(9.39)

(9.40)

(9.41)

indicating that Kirchhoff’s voltage law holds for phasors.

+ 𝑣

₁

− + 𝑣

₂

−

+ 𝑣

_𝑛

−

(46)

Kirchhoff’s Laws in the Frequency Domain 46

 KCL in the frequency Domain

 By following a similar procedure, we can show that Kirchhoff’s current law holds for phasors.

If we let 𝑖₁, 𝑖₂, ⋯ , 𝑖_𝑛 , in be the current leaving or entering a closed surface in a network at time t, then

𝑖

₁

+ 𝑖

₂

+ ⋯ + 𝑖

_𝑛

= 0

 If

𝐈

₁

, 𝐈

_𝟐

, ⋯ , 𝐈

_𝑛, are the phasor forms of the sinusoids 𝑖₁, 𝑖₂, ⋯ , 𝑖_𝑛 , then

(9.42)

(9.43)

𝐈

₁

+ 𝐈

_𝟐

+ ⋯ + 𝐈

_𝒏

= 0

which is Kirchhoff’s current law in the frequency domain.

𝑖

₁

𝑖

₂

𝑖

_𝑛

(47)

Series, Parallel, and Delta-to-Wye Simplifications 47

 Series Connection

showing that the total or equivalent impedance of series-connected impedances is the sum of the individual impedances.

This is similar to the series connection of resistances.

𝐕

_ab

= 𝐕

_𝟏

+ 𝐕

_𝟐

+ ⋯ + 𝐕

_𝒏

= 𝐈𝐙

₁

+ 𝐈𝐙

_𝟐

+ ⋯ + 𝐈𝐙

_𝐧

= 𝐈(𝐙

₁

+ 𝐙

_𝟐

+ ⋯ + 𝐙

_𝐧

) 𝐙

_ab

= 𝐕

_ab

𝐈 = 𝐙

₁

+ 𝐙

_𝟐

+ ⋯ + 𝐙

_𝐧

(48)

 Parallel Connection

This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.

𝐈 = 𝐈

_𝟏

+ 𝐈

_𝟐

+ ⋯ + 𝐈

_𝒏

= 𝐕 1

𝐙

_𝟏

+ 1

𝐙

₂

+ ⋯ + 1

𝐙

_n

= 𝐕 𝒁

_𝑎𝑏

1 𝒁

_𝑎𝑏

= 1

𝐙

_𝟏

+ 1

𝐙

₂

+ ⋯ + 1 𝐙

_n

𝒀

_𝑎𝑏

= 𝐘

₁

+ 𝐘

₂

+ ⋯ + 𝐘

_𝑛

(49)

 Delta-to-Wye Transformation Z

₁

= 𝑍

_𝑏

𝑍

_𝑐

𝑍

_𝑎

+ 𝑍

_𝑏

+ 𝑍

_𝑐

, (9.51) Z

₂

= 𝑍

_𝑐

𝑍

_𝑎

𝑍

_𝑎

+ 𝑍𝑏 + 𝑍𝑐 , (9.52) Z

₃

= 𝑍

_𝑎

𝑍

_𝑏

𝑍

_𝑎

+ 𝑍𝑏 + 𝑍𝑐 , (9.53)

Z

_a

= 𝑍

₁

𝑍

₂

+ 𝑍

₂

𝑍

₃

+ 𝑍

₃

𝑍

₁

𝑍

₁

, (9.54) Z

_b

= 𝑍

₁

𝑍

₂

+ 𝑍

₂

𝑍

₃

+ 𝑍

₃

𝑍

₁

𝑍

₂

, 9.55 Z

_c

= 𝑍

₁

𝑍

₂

+ 𝑍

₂

𝑍

₃

+ 𝑍

₃

𝑍

₁

𝑍

₃

, (9.56)