Sinusoidal Excitation of Circuits Chapter 6 Sinusoidal Excitation of Circuits

6.1 The Sinusoidal Source Adding a phase angle to argument shift it backward

Example 6.1 Convert the cosine function to a sine function and give its representation in terms of sine and cosine function that have no phase angle.

6.1.1 Representation of General Waveforms via the Fourier Series A general periodic function satisfies the property that Where T is the period of the waveform

It is possible to represent any periodic waveform as an Infinite sum of sinusoids: a Fourier series where

For example: The sum of the First seven terms

Once we replace the waveform that has arbitrary time variation with the sum of sinusoidal sources representing the Fourier components of the waveform we may apply superposition to determine the response to the original waveform as the sum of the responses to the individual sinusoidal components of that waveform

6.1.2 Response of Circuits to Sinusoidal Sources Will be discussed with 6.3 The Phasor (Frequency Domain) Circuit

6.2 Complex Numbers, Complex Algebra, and Euler’s Identity The complex number can be represented as A vector in two-dimensional complex plan The length of the vector is the magnitude of the complex number The angle is measured counterclockwise

The polar representation of complex numbers in terms of its magnitude and angle

Euler’s Identity From Euler’s identity we can prove that

Addition and Subtraction of Complex Numbers For two numbers

Multiplication and Division of Complex Numbers Multiplication and division is best accomplished in polar form Multiplication and division can also be accomplished in rectangular form

Define conjugate of a complex number In polar form

Division in rectangular form Multiply numerator and denominator by conjugate of denominator Other identities that can be proven

6.3 The Phasor (Frequency- Domain) Circuit We will represent inductors and capacitors as complex valued resistors and use all previous analysis techniques to solve problems

Response of Linear Circuit to Sinusoidal Source The response will be the same trigonometric functions as the source with the same frequency Differs from the source in magnitude and phase Our task is to determine the magnitude and phase angle of currents and voltages

6.3.1 Representation of Sinusoidal Sources with Euler’s Identity

The key to the Phasor method: To determine the response to either a sine or a cosine source, replace the source with the complex exponential source and use superposition

The response to source is the real part to the source Similarly, the response to source is the imaginary part to the source

6.3.2 The Phasor Circuit (Frequency-domain circuit) Replace the source or with exponential source Represent all response voltages and currents as having the same form Cancel the factor in source and all response variables Replace the resistors, inductors, and capacitors with their impedances

Resistor Impedance For resistor, the impedance is

Inductor Impedance The pahsor current and voltage of the inductor are related by Impedance of inductor is

Capacitor Impedance

6.4 Applications of Resistive-Circuit Analysis Techniques in the Phasor Circuit Ex 6.4:Determine the voltage v(t) in the circuit Replace: source with desired voltage v(t) with Common factor is removed from all Impedance of capacitor is

A single-node pair circuit Hence time-domain voltage becomes Since the source was a cosine we take the real part

Ex 6.5 Determine the current i(t) and voltage v(t) Single loop phasor circuit The current By voltage division The time-domain

Ex 6.6 Determine the current i(t) The phasor circuit is Combine resistor and inductor

Use current division to obtain capacitor current Hence time-domain current is:

Ex 6.7 Determine i(t) using source transformation Phasor circuit Transformed source Voltage of source: Hence the current In time-domain

Ex 6.9 Find voltage v(t) by reducing the phasor circuit at terminals a and b to a Thevenin equivalent Phasor circuit

The Thevenin impedance can be modeled as 1.19 resistor in series with a capacitor with value or

6.5 Circuits Containing More than One Sinusoidal Source Apply superposition in the time domain to yield time-domain circuits with one sinusoidal source. Then solve each of these one-source circuits by reducing them to phasor circuit.

Ex 6.10 Determine i(t)

Cannot simplify further because radian frequencies are different

6.5.1 Sources of the Same Frequency Sinusoidal sources of the same radian frequency can be combined into one phasor circuit. However, all sources must be either sine or cosine form

Ex 6.11 Determine i(t) by placing both sinusoidal sources in the same phasor circuit Superposition in frequency domain In time-domain:

If use superposition in time-domain

6.6 Power The instantaneous power delivered to an element is The average power is

Average power is real power dissipated by the element usually in the form of heat or converted to useful work as in an electric motor. It can be computed also through the complex power

The complex power The magnitude of the complex power 1/2VI is referred to as apparent power and its units are VA. The real part of the complex power is the average power: Watts The imaginary part of the complex power is the reactive power: Reactive power represents energy stored in reactive elements (inductors and capacitors). Its unit is Volt-Amperes Reactive VAR

This implies that in any circuit, conservation of average power and In any circuit, conservation of complex power is achieved This implies that in any circuit, conservation of average power and Conservation of reactive power are achieved However, the apparent power (the magnitude of the complex power) is not conserved

Ex:6.13 Determine the average and reactive power delivered by the source. The phasor current leaving the source is The average power delivered by the source is:

The reactive power delivered by the source is: And the complex power delivered by the source is

Determine the average power and reactive power delivered to each element The voltage across the elements are: Thus the complex power delivered to each element is

show that conservation of complex power, average power, and reactive power is achieved.

6.6.1 Power Relations for the Resistor The voltage and current are in phase so Average power is: Reactive power is zero for resistor

6.6.1 Power Relations for the Inductor The voltage leads the current by 90 so that

6.6.1 Power Relations for the Capacitor The current leads the voltage by 90 so that

6.6.2 Power Factor Since

EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load Only through the resistor

This could also be calculated from the complex power delivered to the load The power factor of the load is: The load is lagging because the current lags the voltage

A typical power distribution circuit The consumer is charged for the average power consumed by the load The load requires a certain total apparent power

The load current is obtained from Ex 6.16 Suppose that the load voltage in the previous figure is 170V, the line resistance is 0.1 ohm and the load requires 10KW of average power. Examine the line losses for a load power factor of unity and for a power factor of 0.7 lagging. The load current is obtained from For unity power factor this is For power factor of 0.7 The powers consumed in the line losses The power needed to be generated 720 W extra power to be generated if pf is 0.7 to supply the load

Power Factor Correction Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct the power factor to unity if power frequency is 60Hz. (Negative because lagging) The angle of the current is: Thus the current into the load is: The current through the added capacitor is: Hence the total current Solving this to cancel out the reactive component gives: C=120.02/(2*3.14*60*170)=0.001873F

6.6.3 Maximum Power Transfer Source-load Configuration Determine the load impedance so that maximum average power is delivered to that load. Represent the source and the load impedances with real and imaginary parts: The load current is:

The average power delivered to the load is: Since the reactance can be negative and to max value, we choose leaving Differentiate with respect to RL and set to zero to determine required RL which is RL= RS Hence: In this case the load is matched to the source. The max power delivered to the load becomes:

6.6.4 Superposition of Average Power Average power computation when circuit contains more than one source

The instantaneous power delivered to the element is Substituting Using the identity

The instantaneous power becomes Average powers delivered individually by the sources Suppose that the two frequencies are integer multiples of some frequency as The instantaneous power becomes

Averaging the instantaneous over the common period THUS: we may superimpose the average powers delivered by sources of different frequencies, but we may not, in general, apply superposition to average power if the sources are of the same frequency. where

Ex 6.18: Determine the average power delivered by the two sources of the circuit Hence the average power delivered by the voltage source is This can be confirmed from average powers delivered to the two resistors

By current division: The voltage across the current source is Hence the average power delivered by the current source is This may be again confirmed by computing the average power delivered to the Two resistors: Since frequencies are not the same, total average power delivered is the sum of average powers delivered individually by each source

EX 6.19: Determine the average power delivered by the two sources Since both sources have the same frequency, we can’t use superposition. So we include both sources in one phasor circuit. The total average power delivered by the sources is equal to the average power delivered to the resistor We use superposition on the phasor circuit to find the current across the resistor

The phasor current is: Hence the average power delivered to the resistor is Note that we may not superimpose average powers delivered to the resistors by the individual sources We can compute this total average power by directly computing the average power delivered by the sources from the phasor circuit The voltage across the current source is The average power delivered by voltage source is The average power delivered by the current source is The total average power delivered by the sources is

6.6.5 Effective (RMS) Values of Periodic Waveforms Sinusoidal waveform is one of more general periodic waveforms Apply a periodic current source with period T on resistor R The instantaneous power delivered to the resistor is The average power delivered to the resistor is Hence the average power delivered to the resistor by this periodic waveform can be viewed as equivalent to that produced by a DC waveform whose value is This is called the effective value of the waveform or the root-mean-square RMS value of the waveform

Ex 6.20 Determine the RMS value of the current waveform and the average power this would deliver to resistor The RMS value of the waveform is Hence the average power delivered to the resistor is

RMS voltages and currents in phasor circuits The sinusoid has a RMS value of Hence the average power delivered to a resistor by a sinusoidal voltage or current waveform is In general, the average power delivered to an element is Therefore, if sinusoidal voltages and currents are specified in their RMS values rather than their peak values, the factor ½ is removed from all average-power expressions. However, the time-domain expressions require a magnitude multiplied by square root of 2 Since X is the peak value of the waveform. Common household voltage are specified as 120V. This is the RMS value of the peak of 170V.

Ex 6.21 Determine the average power delivered by the source and the time-domain current i(t) Phasor circuit with rms rather than peak The phasor current is Hence the average power delivered by the source is The time-domain current is

6.9 Commercial Power Distribution is rms value of the voltage The peak is Time domain representation

The line-to-line voltage between conductors of the transmission line are the phasor sums of the phase voltages Thus the line-to-line voltages are larger than the phase voltages:

6.9.1 Wye-Connected Loads Transmission of power from generator to the load The three impedances making up the load are identical (Balanced load) Hence the current returning through the neutral wire is zero, and the neutral may be removed

For a balanced wye-connected load, the voltages across the individual loads are the respective phase voltages whether the neutral is connected or not. The power delivered to the individual loads is three times the power delivered to an individual load, because the individual loads are identical. No ½ factor in power expression because values are rms Power in terms of line-to-line voltages and load current gives

Example 6.24 Consider a balanced, wye-connected load where each load impedance is and the phase voltages are 120 V. Determine the total average power delivered to the load. The line currents are Hence, the average power delivered to each load is The total average power delivered to the load is

Example 6.25 If the line voltage of a balanced, wye-connected load is 208 V and the total average power delivered to the load is 900 W, determine each load if their power factors are 0.8 leading. Thus The phase voltage is Thus the magnitude of the individual load impedance is Since the power factor is 0.8 leading (current leads voltage; voltage lags current) Thus the individual loads are