Chapter 5 Steady-State Sinusoidal Analysis Notes/Chapter 05.pdf · 14.14 14.14 8.660 5 20 45 10 30...

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Chapter 5Steady-State Sinusoidal Analysis


1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal.

2. Solve steady-state ac circuits using phasors and complex impedances.

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4. Find Thévenin and Norton equivalent circuits. Lightly.

5. Determine load impedances for maximum power transfer.

3. Compute power for steady-state ac circuits.

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SINUSOIDAL CURRENTS AND VOLTAGES

Vt = Vm cos(ωt +θ)

Vm is the peak value

ω is the angular frequency in radians per second

θ is the phase angle

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Tπω 2=

fπω 2=

( ) ( )o90cossin −= zz

Frequency T

f 1=

Angular frequency

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Root-Mean-Square Values

( )dttvT

VT

2

0rms

1∫=

RVP

2rms

avg =

( )dttiT

IT

2

0rms

1∫=

RIP 2rmsavg =

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RMS Value of a Sinusoid

2rmsmVV =

The rms value for a sinusoid is the peak value divided by the square root of two. This is not true for other periodic waveforms such as square waves or triangular waves.

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Phasor Definition

( ) ( )111 cos :function Time θtωVtv +=

111 :Phasor θV ∠=V

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Adding Sinusoids Using Phasors

Step 1: Determine the phasor for each term.

Step 2: Add the phasors using complex arithmetic.

Step 3: Convert the sum to polar form.

Step 4: Write the result as a time function.

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Using Phasors to Add Sinusoids

( ) ( )o45cos201 −= ttv ω

( ) ( )o60cos102 += ttv ω

o45201 −∠=V

o30102 −∠=V

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o

oo

7.3997.2914.1906.23

5660.814.1414.1430104520

21s

−∠=−=

−+−=−∠+−∠=

+=

jjj

VVV

( ) ( )o7.39cos97.29 −= ttvs ω

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Sinusoids can be visualized as the real-axis projection of vectors rotating in the complex plane. The phasor for a sinusoid is a snapshot of the corresponding rotating vector at t = 0.

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Phase Relationships

To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. Then when standing at afixed point, if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ . Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)

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To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval tp between positive peaks of the two waveforms. Then, the phase angle isθ = (tp/T )× 360°. If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t) lags v1(t).

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COMPLEX IMPEDANCES

LL Lj IV ×= ω

o90∠== LLjZL ωω

LLL Z IV =

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CCC Z IV =

o90111 −∠==−=CCjC

jZC ωωω

RR RIV =

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Kirchhoff’s Laws in Phasor Form

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.

The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

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Circuit Analysis Using Phasors and Impedances

1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)

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2. Replace inductances by their complex impedances ZL = jωL. Replacecapacitances by their complex impedances ZC = 1/(jωC). Resistances have impedances equal to their resistances.

3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.

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AC Power Calculations

( )θcosrmsrms IVP =

( )θcosPF =

iv θθθ −=

( )θsinrmsrmsIVQ =

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rmsrmspower apparent IV=

( )2rmsrms

22 IVQP =+

RIP 2rms=

XIQ 2rms=

RVP R

2rms=

XVQ X

2rms=

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THÉVENIN EQUIVALENT CIRCUITS

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The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.

ocVV =t

We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.

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The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

scsc

oc

IV

IV t

tZ ==

scII =n

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Maximum Power TransferIf the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.

If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.

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BALANCED THREE-PHASE CIRCUITS

Much of the power used by business and industry is supplied by three-phase distribution systems. Plant engineers need to be familiar with three-phase power.

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Phase Sequence

Three-phase sources can have either a positive or negative phase sequence.

The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.

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Wye–Wye ConnectionThree-phase sources and loads can be connected either in a wye configuration or in a delta configuration.

The key to understanding the various three-phaseconfigurations is a careful examination of the wye–wye circuit.

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( ) ( )θcos3 rmsrmsavg LY IVtpP ==

( ) ( )θθ sin3sin2

3 rmsrms LYLY IVIVQ ==

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YZZ 3=∆

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Chapter 5 Steady-State Sinusoidal Analysis Notes/Chapter 05.pdf · 14.14 14.14 8.660 5 20 45 10 30...

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Transcript of Chapter 5 Steady-State Sinusoidal Analysis Notes/Chapter 05.pdf · 14.14 14.14 8.660 5 20 45 10 30...