ac circuits handout and tutorials

© Heriot-Watt University

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Heriot-Watt University Edinburgh Electrical Electronic and Computer Engineering

B3.8EB Circuits and Analysis

Part 2 1. Introduction The application of dc circuits/analysis in Electrical Engineering is very limited. However when we move into the realms of handling alternating waveforms things are completely different. These notes are an introduction into the real world of Electrical Engineering, Communications, Power Generation and much more. Part 2 Notes/Tutorials will cover the following topics:

• An introduction to alternating current and voltage • Sinusoids and phasors • Resistance, Inductance and Capacitance in ac circuits • Complex Impedance • Impedance and Admittance

2. Alternating Current (ac) If a current is varied in a repetitive manner then it is known as an alternating current, commonly abbreviated to ac. Current flows first in one direction and then in the other (electrons moving in one direction and then in the other), and the cycle of variation is exactly the same for each direction.

Sinusoidal wave

Square wave

Triangle wave

Figure 2.0.1 Alternating Current Waveforms

The curves relating current to time are known as waveforms. These can be of a variety of shapes as shown above or even more complicated as long as the shape above the zero line is identical to that below the zero line. However, the most common and most useful form is the sinusoid as: • This can be easily generated • It can form other interesting waveforms (Fourier) • It is a very naturally occurring shape • It is mathematically simple As ac waveforms appear to be so common we must first learn how to define the terms involved prior to being able to analyse circuits operating with ac rather than dc.

ECTE172 Introduction to Circuits and Devicesac circuits

© Heriot-Watt University 2

2.1 Generating ac Before looking at the generation of ac waveforms we should recall some basic principles. Consider the relationship: E = Blu where E is the voltage generated by a wire, length l (m) passing through a magnetic field of strength B at a velocity u (m/sec)

If we now consider a loop of conductor, AB, carried on a spindle, DD, rotated at a constant speed in an anticlockwise direction in a uniform magnetic field created by the poles of a magnet as shown.

Figure 2.1.1 Generating an alternating e.m.f. We may observe that when the loop is horizontal, no flux is being cut and no emf can be generated. Whereas, if the loop is vertical, then flux is being cut at a maximum rate. At any angle in between these limits, the flux is being cut at some rate less than the maximum but more than zero. It is not difficult to imagine that the signal that will emerge will be sinusoidal.

Figure 2.1.2 Emf in a rotating coil Figure 2.1.3 Instantaneous value generated emf

Taking the A part of the loop at an angle θ to the horizontal, we can see that if AL represents velocity u, then the horizontal component of this at angle θ is AM = AL sinθ.


Going back to the equation, E = Blu we now have the emf generated by one side of the loop as: Blu sin θ volts For both halves of the loop this doubles to:

e= 2Blu sin θ volts

The maximum value of voltage generated is when the sine function is unity and gives:

Em = 2Blu volts

If the loop has a breadth of b meters and has N turns and the rotational rate is n revs/s, then the circumference of the circle the loop makes is πb meters and this gives a speed of πbn meters/sec so,

e= 2Bl (πbn) sinθ volts and Emax = 2Bl πbn volts

Noting the area of the loop to be A = bl gives for an N turn coil:

e = 2πBAnN sinθ volts and Em = 2πBAnN volts

Note that lower case letters are conventionally used for instantaneous values and that upper case letters are used for definite values such as peak voltage. 3. Waveform Terms and Definitions The essential terms that define sinusoidal waveforms are: 1. Waveform: The variation of voltage or current against time as a graph. 2. Cycle: A complete repetition of the waveform assuming periodicity 3. Period: The duration of one cycle of the waveform 4. Instantaneous value: The magnitude at any given instant. This can be positive or

negative. 5. Peak Value: The maximum value the function can reach. 6. Peak-to-peak: The measure of the range between the maximum and minimum

values the function reaches. 7. Peak Amplitude: The maximum instantaneous value measured from the mean

value of the waveform 8. Frequency: Previously called cycles per second, which is self explanatory.

The correct name is Hertz (Hz) and this is the inverse of the period. If a waveform has a frequency of 50Hz the period is 1/50 = 0.02 or 20ms. In other words the waveform repeats every 20ms.


4. Sine and cosine Sine and cosine waveforms are universally accepted as the fundamental alternating waveforms associated with all aspects of electrical engineering theory. There are a number of reasons for this: many practical systems, including the National Grid and radio-communications systems, have essentially sinusoidal waveforms; sinusoids are readily handled mathematically (trigonometric formulae) and through the application of Fourier techniques all waveform shapes can be considered as composed of summations of sinusoids of varying frequency, amplitude and phase; the natural response of many electrical and non-electrical circuits is to generate a sinusoidal responses. 4.1 Phasors The sinusoidal and co sinusoidal waveforms can be best understood by considering them as being represented by the horizontal and vertical projection of a rotating phasor. Imagine a line rotating in an anticlockwise direction, so that its tip traces out a circle, as shown in Figure 4.1.1. The resultant waveform is the plot sin (θ) versusθ.

Figure 4.1.1 The horizontal projection of a rotating line or phasor A sine wave can be expressed as a function of time by writing sin θ = sin ωt

Figure 4.1.2 Plot of the sine function versus θ = ωt

A phase angle φ can be added to the variable θ or ωt to cause the sine wave to shift to the left along the horizontal axis for positive phase angles. A negative phase angle causes the sine wave to shift to the right. See Figure 4.1.3.


Figure 4.1.3 Plot of sin θ versus θ showing the effect of the phase angle φ The sine wave is an example of a periodic waveform. If the value of a periodic sine wave is f (t1) at time t1 and is similar at times (t1 + nT), where n is an integer, then T is known as the period or periodic time of the function. The frequency, f, of an alternating waveform is the number of cycles that occur in 1 s. Frequency is inversely proportional to period.

Thus f =T

1. The units of frequency are cycles per second or in SI units Hz.

From Figure 1.0.2, it is clear that one cycle is the same as 2π radians. The number of radians produced in 1 s is (2π) times (f), or ω = 2πf rads-1. This is known as the angular frequency. Now sin ωt is sine (angle) and angle has units of radians. Remember to set your calculator to radian measurement! The phasor shown in Figure 4.1.1 is a convenient mathematical model for sinusoidal alternating waveforms and can be represented as v (t) = V exp. j (ωt + φ) where V is the peak voltage, ω is the angular frequency (rads-1) and φ is some reference phase angle. As (ωt) has units of radians, so the phase angle φ, must also have units of radians. In Figure 1 the magnitude of V was taken as unity. For convenience we refer to the horizontal (right-hand) x-axis as the reference, i.e. a phase of zero degrees. Now, v(t) = V exp. j(ωt + φ) expands to give v(t) = V cos (ωt + φ) + j sin (ωt + φ) hence a cosine wave can be considered as Rev(t) and a sine wave the Imv(t). The designations real (Re) and imaginary (Im) are simply there to differentiate between the two directional components. As alternating circuit theory is founded on phasors, i.e. a two dimensional co-ordinate axis system, mathematical manipulations must inevitably use complex number theory. 4.2 Phase Angles and the phasor representation In using the phasor as a mathematical tool to represent the sine and cosine functions we will restrict ourselves (for the moment) to dealing only with linear circuits that have multiple “input” and “output” phasors all of which rotate at the same angular frequency. Multiple phasors can be “frozen in time”; it is only the relative phase between one phasor and another that is of importance in ac theory.


Consider the “frozen” phasor shown in Figure 4.2.1

Re

Im

1

1

sin

cos

1

ejθ

θθ

θ

ejθ =1

<) ejθ =θ

Figure 4.2.1 Frozen Phasor

polar form exponential form rectangular form vector exponential functions complex numbers

θ≡ /VV θje.V θ+θ jVsin cosV

= ]jIm[e ]eRe[ jj θθ + Example 1:

vpvp /VV)tcos(Vv θ=⇒θ+ω=

Consider three phasors: v, i & i1

°=⇒°+ω= 45/3V)45tcos(3v volts

°−=⇒°−ω= 30/2I)30tcos(2i amps

°−=⇒°+ω−=ω= 90/10I)90tcos(10tsin10i 11 amps

V

II1

10

3

2

45o

-30o ref

As phasors with positive frequencies rotate anticlockwise: V leads I leads 1I

Example 2: Given that y(t) = 1 sin (3141.6)t Determine: a. the angular velocity b. the frequency c. the period of the waveform. a. Comparing y(t) = sin 3141.6t with sin ωt, we see that the angular velocity, ω, is 3141.6 rads-1. b. The frequency, f, is given by f = ω/2π = 3141.6/2π = 500 Hz. c. The period of the waveform is the inverse of frequency, thus T = 1/f = 1/500 Hz = 2 ms.


4.3 Phase relationships between sinusoids of the same frequency When we write V sin (ωt + φ) we understand from the mathematics of this equation that its waveform shape is sinusoidal, that it can be pictured as a phasor of magnitude V volts rotating anticlockwise at an angular rotation of ω rads-1. But what about the phase term (φ)? One interpretation is that the temporal waveform at t = 0 (?) has a value of sin (φ). But it is usually phase difference that matters in electrical engineering; i.e. our interest lies in the phase of one phasor relative to another. When two waveforms have different phase angles, the one shifted farthest to the left is said to lead the other. For example in Figure 4.3.1 we see two sinusoidal signals; the ac voltage is shifted left by 30 ° and the ac current is shifted right by 45 °. We can say that v (t) leads i (t) by 75 °. Alternatively it is equally correct to say that i (t) lags v (t) by 75 °.

Figure 4.3.1 A voltage 10 sin (ωt + 30°) volts, and a current 10 sin (ωt - 45°) amps.

v (t) leads i(t).

There is one other way of looking at Figure 4.3.1. Try to imagine the phasor equivalents at a set point in time, say t = 0. The voltage has a positive value in the first quadrant of the complex plane (+Re, +Im). The current has a negative value in the fourth quadrant (+Re, - Im). Try drawing the phasors for v(t) and i(t)!

4.4 Complex Numbers: Revision The complex plane is a rectangular co-ordinate system in which real numbers are plotted along the horizontal (real) axis and imaginary numbers along the vertical axis.

Figure 4.4.1 Examples of complex numbers plotted in the complex plane.


The representation (a + jb) is called the rectangular form of a complex number. Every complex number can also be represented in polar form: y = M /θ where M is the magnitude of y and θ is its angle.

Figure 4.4.2 Converting between the polar and rectangular forms of the complex number.

Addition and Subtraction can only be performed in rectangular components. For example the sum of the two phasor voltages, v1(t) = 12/ -30° volts and v2(t) = 20 / 45° volts requires the conversion of the voltages from polar to rectangular format. Thus 12/ -30° volts plus 20 / 45° volts => (10.4 - j 6) + (14.1 + j 14.1) = (24.5 + j 8.1) volts. To get the answer back in polar notation, we do the reverse: (24.5 + j 8.1) => 25.8 / 18.3° volts. Multiplication and Division can only be performed in polar co-ordinates. For example 180 / 27° amps divided by 1.5 / 85 ° amps is straight forward as both currents

are in polar co-ordinates. Thus the answer is 5.1

180/(27 ° - 85 °) = 120 / - 58° amps. If (1 + j2)

was multiplied by (2 + j6) then we have to do the rectangular to polar conversion and then the multiplication. It is absolutely imperative that this sort of manipulation (on your calculator) becomes second nature. For everything we do in ac theory, and this will include complex impedance as well as voltages and currents will rely on complex manipulation and rectangular ⇔ polar conversions.


4.5 Spectra The range of frequencies used in engineering literally covers the frequency spectrum.

Figure 4.5.1 The Frequency Spectra


5 Tutorial: Sine and cosine (The Questions from this Tutorial are all drawn from Hughes Exercises 9 and 14) 1. An alternating current is represented by: i (t) = 10 sin 942 t amperes. Determine (a) the frequency (Hz), (b) the period (sec), (c) the time taken from t = 0 for the

current to reach a value of 6 A for the first and second time.

(150 Hz, 6.7 ms, 0.7 ms (and 2.65 ms), 7.4 ms) 2. In a certain circuit supplied from a 50 Hz mains supply, the potential difference has a maximum value of 500 volts and the current has a maximum value of 10 amps. At the instant t = 0, the instantaneous values of the p.d. and the current are 400 volts and 4 A respectively,

both increasing positively. Assuming sinusoidal variations, state trigonometric expressions for the instantaneous values of the p.d. and the current at time t. Calculate the instantaneous values at t = 0.015 s and determine the angle of phase difference between the p.d. and the current.

(500 sin (314t + 0.93) volts, 10 sin (314t + 0.412) A, - 300 volts, -9.15 A, phase = 29.6°)

3. Express in rectangular and polar notation the phasors for the following quantities: (a) i (t) = 10 sin ωt amps; (b) i (t) = 5 sin (ωt - π/3) amps; (c) v (t) = 40 sin (ωt + π/6) volts

(10 +j0, 10/0°, 2.5 - j4.33, 5/-60°, 34.64 + j20, 40/30°) 4. Express each of the following phasors in polar notation: (a) 10 + j5; (b) 3 - j8. Now add the two phasors (a) and (b). Express in both rectangular and polar notations. Now subtract (b) from (a). Express in both rectangular and polar notations.

(11.2/26.5°, 8.54/-69.4°, 13 - j3, 13.34/-13°, 7 +j13, 14.76/61.7°) 5. Derive expressions in rectangular and polar notation for: (a) 1/ (10 + j15); (b) 1/ (20 - j10); (c) 1/ (50/20°);

(0.0555/-56.3°, 0.0308 -j0.0462, 0.0447/26.56°, 0.04 +j0.02, etc…) 6. Determine the resultant of the following four voltages: v1 (t) = 25 sin ωt; v2 (t) = 30 sin (ωt + π/6); v3 (t) = 30 cos ωt; v4 (t) = 20 sin (ωt - π/4) volts. Express your answer in a similar form.

(72 sin (ωt + 0.4424) volts)


7. Two sinusoidal e.m.f.’s of peak values 50 V and 20 V respectively but differing in phase by 30° are induced in the same circuit. Draw the phasor diagram and find the peak value of the resultant e.m.f.

(68 volts (8.45°) 8. Two impedance's are connected in parallel to the mains supply. The first impedance takes a current of 40A at a lagging phase angle of 30°, the second takes a current of 30 A at a leading phase angle of 45°. Determine the total current taken from this supply.

(55.9 A, 1.2°) 9. Two circuits connected in parallel take alternating currents that can be expressed trigonometrically as I1(t) = 13 sin 314t amperes and I2 (t) = 12 sin (314t + π/4) amperes. Sketch the waveforms of these currents. Determine the resultant of these currents.

(23.1 sin (314t +0.374) A)


6.0 The Resistor, Inductor and Capacitor in ac circuits There are only three passive components, the R, L and C. We need to have an in depth understanding of all three; after all they are all we have! We will start our analysis by assuming that they are ideal devices. 6.1 The Resistor 6.1.1 ac voltage and current in the Resistor Ohm's Law can be applied to an ac circuit containing a resistance to determine the ac current in the resistance when an ac voltage is connected across it, see Figure 6.1.1. At every instance of time, the current in the resistor is the voltage at that instant divided by the resistance. Thus given a resistance (R) and a voltage v(t) = V sin ωt, the current is given by,

i(t) = R

Vsin ωt.

i(t) = V/R sin ωt

v(t) = V sin ωt

R

Figure 6.1.1.1(a) ac voltage across

and current through a Resistor

Figure 6.1.1.1(b) In a resistor the voltage

and current are in phase 6.1.2 ac Power in a Resistor You will recall that in a dc circuit power can be calculated using any of the three

relationships: VI, I2R and R

V 2

Watts. In ac circuits both voltage and current are time-varying

quantities, and so therefore is power. The power at any instant, the instantaneous power, can be computed using instantaneous values of voltage and/or current.

ac Power, p = v(t).i(t) = (V sin ωt)(I sin ωt) = VI sin2 ωt Watts Now the sin2 ωt term can be viewed in two ways. Firstly, although sin ωt goes both positive and negative, in the function sin2 ωt all such values are squared, so the function is always positive. This is illustrated by Figure 1.1.1, which shows the square of the current sine wave, I sin ωt. The waveform is positive going, has a sin2 shape versus time and clearly has an average value. This average value for power is best explained by taking the second approach to the sin2 term.


Figure 6.1.2.1 The square of a sinusoidal current, I sin ωt

Secondly, we can use standard trigonometric formulae to expand sin2 ωt:

sin2 ωt = 0.5 (1 - cos 2ωt).

Thus ac power = VI sin2 ωt = 0.5VI (1 - cos 2ωt) Watts. Now, consider the cos 2ωt term. The average value of cos 2ωt over some period of time (t>> a period) is zero (equal positive and negative areas). Thus average ac power = 0.5VI. This relates to rms. (root mean square values) of voltage and current:

Vrms = 2

Vvolts and Irms =

2

I amps.

Thus ac power, p = 0.5 VI = Vrms.Irms. = R

Vrms2

etc.

6.2 Inductor 6.2.1 ac voltage and current in an Inductor Whereas in the resistor the Voltage and Current are related by the linear relationship we call Resistance (Ω), for the Inductor (and the Capacitor) the situation is not so straightforward.

For an Inductor the voltage to current relationship is: V = Ldt

diwhere V is the voltage across

the inductor and i is the current through the inductor. So let us return to our basic circuit with a voltage v(t) = V sin ωt volts applied across an inductor (L), Figure 2.0.1.

Both sides of the equation, V = Ldt

di must agree. If the voltage is sinusoidal then the current

must be I sin (ωt - π/2), i.e.

V sin ωt = L dt

d-I cos ωt


Now dt

d-I cos ωt = ω I sin ωt so substituting in the above equation gives:

i(t)

v(t) = V sin ωt

L

Figure 6.2.1.1(a) ac voltage across and current through an

Inductor

Figure 6.2.1.1(b) In an Inductor the voltage

LEADS the current by 90°

V sin ωt = L dt

d-I cos ωt = L ω I sin ωt

So the fundamental relationship between Voltage and Current for an inductor is ωL which naturally has units ofΩ. ωL is known as the Inductive Reactance and is denoted XL. So the key features for the Inductor are: • In an Inductor the Voltage LEADS the Current by 90° (It is perhaps worthy of note that

by convention for Inductive circuits we usually refer to the inductor as a LAGGING circuit element i.e. we say Current LAGS Voltage. Please be content with either definition.

• The relationship between Voltage and Current is the inductive Reactance, XL = (ωL) Ω • As XL = (ωL) Ω the value of Inductive Reactance is directly proportional to Frequency 6.2.2 Power in an Inductor The instantaneous power delivered from the supply, v(t), is given by: ac power p(t) = v(t).i(t) = (V sin ωt). (- I cos ωt) Watts Refer to the trigonometric identity: SinA cosB = 0.5[sin (A - B) + sin (A + B)] Thus, p(t) = (V sin ωt). (-I cos ωt) = - 0.5VI [sin 2ωt] Watts


Figure 6.2.2.1 Voltage, current and power in an Inductor

Now this is an interesting result as we have already proved that a sine or cosine term has an average value of zero. Thus the perfect Inductor dissipates no power. This is entirely due to the 90° relationship between Voltage and Current. This is not surprising as 90° infers that the two quantities are orthogonal. Note: • During the time that the voltage and current are both positive the power p(t) is positive

and power and energy is delivered from the source to the Inductor and stored in the magnetic field.

• During the time that the voltage and current have opposite signs the power p(t) is negative the stored energy is returned from the Inductance back to the source.

6.3 Capacitor 6.3.1 ac voltage and current in a Capacitor

For the Capacitor the voltage to current relationship is: v = ∫ idtC

1 where V is the voltage

across the Capacitor and i is the Current through the Capacitor. If we return to our basic circuit with a voltage v(t) = V sin ωt volts applied across a Capacitor (C), Figure C1.

i(t)

v(t) = V sin ωt

C

Figure 6.3.1.1(a) ac voltage across

and current through a Capacitor

Figure 6.3.1.1(b)

In a Capacitor the voltage LAGS the current by 90°


Both sides of the equation must agree. If the voltage is sinusoidal the current must be co sinusoidal, i.e.

V sin ωt = ∫ idtC

1 = ∫ . cos I

1dtt

Cω

Now ∫ . cos I dttω = ω1

I sin ωt, so substituting in the above equation gives:

V sin ωt = ∫ idtC

1 = ∫ . cos I

1dtt

Cω =

C

I

ωsin ωt

So the fundamental relationship between Voltage and Current for a Capacitor is Cω1

which is

the Capacitive reactance, XC. So the key features for the Capacitor are: • In a Capacitor the Voltage LAGS the Current by 90°

• The relationship between Voltage and Current is the Capacitive Reactance, XC = C

I

ω Ω

• As C

I

ω Ω the value of Capacitive Reactance is inversely proportional to Frequency

(See Figure 3.0.2)

Figure 6.3.1.2 Plot of Capacitive Reactance, XC, versus frequency


6.3.2 Power in a Capacitor The instantaneous power delivered from the supply, v(t), is given by: ac power p(t) = v(t).i(t) = (V sin ωt). (I cos ωt) Watts Refer to the trigonometric identity: sinA cosB = 0.5[sin (A - B) + sin (A + B)] Thus, p(t) = (V sin ωt). (I cos ωt) = 0.5VI [sin 2ωt] Watts

Figure 6.3.2.1 Voltage, current and power in a Capacitor Now this is the same result (except for the change in sign) as for the Inductor. Again this is not surprising as 90° infers the two quantities are orthogonal. Thus the perfect Capacitor dissipates no power, taking energy from the power supply during part of the ac cycle and returning it back to the source during another part of the cycle. 6.4 What about this (j) operator thing? AC Theory has revealed the distinct differences between the three passive components. The key facts are summarised in the following table.

Property Resistor Inductor Capacitor v versus i relationship V = I r

V = Ldt

di v = ∫ idt

C

1

Average Power dissipated

vrmsirms 0 0

v versus i phase relationship

v in phase with i v LEADS i by 90 ° (π/2) v LAGS i by 90 ° (π/2)

Reactance (Ω) - XL = (ωL) XC =

C

I

ω

Reactance versus angular frequency

- proportional inversely proportional

This (j) thing?

? ? ?

Table 6.4.1

Clearly in ac theory we need to take into account the phase angles of 90° between the voltage and currents in the reactive devices and the zero phase shift in the resistor. The obvious way to present this information is by using the complex plane. All we need is a reference datum!


The universally accepted approach is to use the Real axis of the complex plane to represent the current axis. This is an important fact to remember as in order to establish any rule-base you must always remember to what you are referring. By making the real axis a current reference we see that for the resistor (voltage in phase with current) we can represent the phasors for voltage and current as both on the real axis. Now for the Inductor, the voltage leads the current by 90°. The (j) operator is of course just a mathematical representation of a 90° anticlockwise rotation. Reference to current, the inductor voltage is 90° ahead or + j. Thus we display Inductor voltage on the + j axis of the complex plane. Thus Inductive reactance becomes:

XL = j ωL For the Capacitor, the voltage lags the current by 90°. We thus display the Capacitor voltage on the - j axis of the complex plane. Thus Capacitive Reactance becomes:

XC = - jC

I

ω

The complex plane will from now on always be our means of handling currents and voltages and resistance's and reactance's. The ac world is a world of complex mathematics (though not necessarily computationally complexity!). This is the move that opens the way to explore the whole big world of Electrical Engineering! 6.5 The complex Planes

- j|XC|

j|X L|

R

VR

VL

VC

i

The Impedance Plane The Voltage Plane (series circuits)

Figure 6.5.1 The complex plane

6.6 Aide-memoire to voltage/currents/lead/lag/Inductor/Capacitor One memory trick is to remember "CIVIL". Reading from left to right:

"In a Capacitor I (current) leads Voltage - (V)oltage leads I in an L (Inductor)"


7.0 Impedance (Z) 7.1 Introduction In the previous section we saw that the Inductor and Capacitor have Inductive and Capacitive

Reactance’s of XL = j ωL and XC = - jCω1

respectively. This led us to the concept of

complex impedances and the use of the complex plane. Now very few circuits are pure R, L or C. Most circuits are combinations of these R, L and C and either Series or Parallel or combinations of Series/Parallel circuitry. The rules for series and parallel circuits are identical to those established in the dc circuits part of your course. However the difference in the ac circuits is that all voltages and currents must be treated as phasors. 7.2 Series ac circuits The total impedance Z of a series circuit containing the three passive components, resistance R, inductive reactance XL = jωL, and capacitive reactance XC = -j (1/ ωC) can be found by combining the phasor forms of each of these components on an impedance diagram, see Figure 7.2.1.

R

j|X L|

- j|X C|

Z

O

O

- j|XC|

j|X L|

R

- j|XC|j|X L|

R

φ

Z = R + j( - |XC|) |X L| -

Figure 7.2.1 Series RLC circuit impedance diagram

If there is only one reactive term then the diagram is modified accordingly. Like series d.c. circuits, the current in a series a.c. circuit is the same through each series connected component, the significant difference for the a.c. circuit being that the current is a phasor. When computing Kirchhoff’s voltage law remember to treat all voltages as phasors. Voltage phasors can be combined using a phasor diagram in much the same way that impedance phasors are combined, see Figure 7.2.2.


R

LC

VR

VL

VC

e

+

VR

VL

VC

i

φ

(VL - VC)

VRi

e

Figure 7.2.2 Voltage relationships in series RLC circuit

From Figure 7.2.2 the following points should be noted: 1. The voltage across the inductor vL leads the current i through it by 90 degrees 2. The current i in the capacitor leads the voltage vC across it by 90 degrees 3. The current i in a series circuit is in phase with the voltage across the resistor Vr . 4. The applied voltage e is the phasor sum of all the voltage drops in the circuit. 5. The angle θ between the applied voltage e and the current i is the same as the angle of the total impedance of the circuit. 7.2.1 Series Resonance The circuit, Figure 7.2.1 is a series R, L and C circuit. Consider this circuit connected to a sinusoidal signal generator whose frequency can be varied over a wide range of frequencies. As the frequency is increased, the magnitude of the inductive reactance increases according to |XL| = ωL, whereas the magnitude of the capacitive reactance term decreases according to |XC| = 1/ ωC. At some frequency, which we shall call fS, |XL| will have exactly the same magnitude as |XC|. The frequency at which |XL| = |XC| is called the resonant frequency and at this frequency (only) the circuit will have no reactance term and will thus be purely resistive. This will be a situation where the circuit will have a minimum impedance (resistance) and thus a maximum current. Since the voltage across the resistor is vr = iR, it also follows that the voltage across the resistor will have a maximum value at resonance. We will return to resonance in more detail later in the course.


Example: Series circuit

In the series circuit shown a 100/0º volt sinewave connected across a series RLC circuit. Show that the sum of the voltage drops over the three components is equal to the applied voltage. ZTOTAL = Z1 + Z2 + Z3 = (4 + j3 - j6)Ω = (4 - j3) = 5 / -36.9º Ω

Example: Series circuit

Total current I = AmpsZ

V

TOTAL

°∠=°−∠

°∠= 9.36209.365

0100

Voltage across the resistor = IR = (20 / 36.9º) (4) = 80 / 36.9º volts Voltage across the inductor = IXL = (20 / 36.9º) (3 /90º) = 60 / 126.9º volts Voltage across the capacitor = IXC = (20 / 36.9º) (6 /-90º) = 80 / -53.1º volts Thus the sum of the voltage drops across the three components: VR + VL + VC = (64 + j48) + (-36 +j48) + (72 - j96) = (100 + j0)


7.3 Parallel ac circuits The fundamental difference between parallel a.c. circuits and series a.c. circuit theory is that for the parallel circuit the voltage across each branch of the parallel circuit is identical. In parallel a.c. circuit theory we have to obey Kirchhoff’s current law, but once again, in phasor format. Figure 7.3.1 illustrates the phasor current addition for a parallel RLC circuit.

R LC

e

+iR iCiL

i

iC

iLe iR

φ

e

(iC - iL)

iR

i

Figure 7.3.1 Parallel RLC circuit and current phasors One very important thing to notice about Figure 7.3.1 is that in the parallel circuit, it is the voltage that is the common term. We thus take the voltage across the Resistor as the reference. In the capacitor the current is leading the voltage by 90° and thus must appear as a (+ j) term relative to the real axis. Likewise the Inductor has a current that lags voltage, hence the (- j) term. This nomenclature is contrary to the previously adopted system used with series circuits. For the parallel circuit configurations however it is sensible to adopt the above phasor diagram when describing the current flow. 7.3.1 Parallel Resonance In a series circuit we said that at a certain frequency the capacitive and inductive reactance’s cancelled leaving a circuit with only a resistive term. In the case of the parallel circuit we will have a minimum current as the capacitive and inductive phasor currents cancel. Thus at resonance a parallel circuit will have a maximum impedance (resistance). We will return to resonance in more detail later in the course.


Example: Parallel circuit

In the parallel circuit shown a 50/0º volt sinewave connected across a parallel RLC combination circuit. Show that the sum of the currents in each branch equals the total current flowing out of the generator. Determine the total impedance for the circuit. IT = I1 + I2 + I3 amps

10050

1

°∠=I ( )°∠°∠=

535

0502I

°−∠°∠=9.3610

0503I

I1 = 5/0º I2 = 10/-53º I3 = 5/36.9º

Example: Parallel circuit

(3 + j4) = 5 /53º Ω

(8 - j6) = 10/-36.9º Ω IT = (I1 + I2 + I3) = 5 + (6 - j 8) + (4 + j3) = (15 - j5) amps IT = (15 - j5) = 15.9/-18.45º amps Total Impedance:

( ) ( ) °∠+°−∠+=−

++

+=++= 9.361.0532.01.068

143

11011111

321 jjZZZZT

)1.03.0()06.008.0()16.012.0(1.01

jjjZT

−=++−+=

( ) ( ) Ω+=°∠=°−∠

=−

=∴ )13(43.18164.343.18316.0

1

1.03.0

1j

jZT

Thus the total current: ( ) ampsj

IT °−∠=°∠

°∠=+

°∠= 43.1889.1543.18146.3

050

13

050


7.4 Thevenin and Norton Norton and Thevenin circuit analysis is central to much future work in ac analysis. The follow example is designed to remind the reader of how to handle these theorems in ac circuits. Determine the Thevenin equivalent for the network below. (Hughes: Example 15.8)

(a) ac network with load resistor (b) Thevenin Voltage: remove the load

resistor & determine the open circuit voltage

(c) Thevenin Impedance: voltage source

short circuited (d) Thevenin Equivalent Circuit

Working (excuse the mix of rectangular/polar units): (b) Potential division of 100/0º volts by (6 +j8) and (-j3)

VT = ( )°−∠=

°∠°−∠=

+°−∠°∠=

++−−∠ °

8.1294.38)8.3981.7()90300

)56(903)(0100(

)86()30()3)(0100(

jjj

jvolts

(or (-24.58 – j-29.5) volts in rectangular units)! (c) ZT = (6 +j8) in parallel with (-j3) =

Ω°−∠=

°∠°−∠=

°∠°−∠°∠=

−++−+

8.7684.3)8.3981.7(

)3730()8.3981.7(

)903)(5310()3()86(

)3)(86(jj

jj

(or (0.876 –j3.73)Ω, which is a more correct units for ohms)


8.0 Tutorial: Series and Parallel circuits 1. The current through a 47 Ω resistor is i(t) = 1.5 sin (ωt + 70° ) A. What is the voltage across the resistor? Sketch a phasor diagram showing the voltage and current relationship.

(70.5 sin (ωt + 70° A). 2. The voltage across a 0.015 µF capacitor is v(t) = 16.4 sin (3400 t - 63° ) volts. What is the current through the capacitor? Sketch a phasor diagram showing the voltage and current relationship.

(0.836 sin (3400 t + 27°) mA)

3. The current through a 0.2 H inductor is i(t) = 0.06 sin (2π x 103 t - 80°) A. What is the voltage across the inductor? Sketch a phasor diagram showing the voltage and current relationship.

(75.4 sin (2π x 103t + 10°) volts) 4. The current in an R - L series circuit of R = 5 Ω and L = 30 mH lags the applied voltage

by 80°. Determine the supply frequency and the circuit impedance.

(150.4 Hz, (5 + j28.4) Ω) 5. A 2 A rms. current source of frequency 100 Hz is connected to a series circuit consisting

of a 35 Ω resistor in series with a 0.1 H inductor. Determine the circuit impedance in both rectangular and polar form and establish the

voltage developed across this circuit. Sketch the phasor diagram for this question.

((35 + j62.8) Ω, 71.9 / 61° Ω, 144 / 61° volts) 6. A parallel circuit consists of a 1 kΩ resistor and a 100 µF capacitor.

Determine the component values for an equivalent series circuit when a UK ac. mains voltage (240 / 0° V AT 50 Hz) is applied. Determine the supply current.

(1Ω in series with 100 µF, 7.55 / 88° A)

7. A series circuit consists of a 510 Ω resistor and a 0.3 µF capacitor.

Determine the component values for the equivalent parallel circuit operating at a frequency of 1 kHz

(1060 Ω in parallel with 0.156 µ F)

8. An ac circuit consists of a 200 Ω resistor connected in series with the parallel combination

of a 1 H inductor and a 4 µF capacitor. Determine the circuit impedance and the total current that will flow when this circuit is connected to the mains supply (240 / 0° V ).

((200 + j518) Ω, 0.43 / -70° A)


9. For the network shown in Figure Q9 determine the current in XL. Use both superposition and nodal analysis.

Figure Q9

(5.7 /- 58.6º A)

10. Determine the voltage V across the 10 Ω resistor for the circuit of Figure Q10.

(20 +J20) V(2 + j1)A R1 =

10Ω

R2 = 4Ω

XL = + j 6Ω

V

Figure Q10

(27.7 /35.3º volts) 11. Determine the Thevenin equivalent for Figure Q11.

Figure Q11

(22.4 /26.6º volts, (3.2 – j4.4)Ω)


12. Obtain the Thevenin and Norton equivalent circuits (a b) for Figure Q12a and Figure Q12b.

Figure Q12a

Figure Q12b

(a) (V th = 5.59/26.4° volts, Z th = 6.76/68° Ω, I th = 0.83/- 41° A)

(b) (V th = 11.4/- 85° volts, Z th = 8.26/- 15° Ω, I th = 1.38/- 70° A)

13. For the circuit of Figure Q13 determine the current I.

1mA

R1 = 6.8kΩ

Xc = - j3.3kΩ

XL =

j2.2kΩ

R2 = 10kΩ

R3 = 4.7kΩ

I

Figure Q13

(0.18 /9.8º A)


14. For the circuit shown in Figure Q14 determine the current I flowing in the 60 Ω resistor.

Figure Q14

(0.33 /- 22.3º A)

15. Obtain the Thevenin equivalent at ba for the bridge circuit shown in Figure Q15.

Consider one end of the voltage source as a reference point.

Figure Q15

(V th = 0.33/169°volts, Z th = 47.3/26.7° Ω)


9.0 Impedance and Admittance Impedance relates voltage to current. If we only have resistance to deal with then impedance would revert to being the simple case of Ohm’s law: V = IR.

However as we enter the ac complex impedance world impedance consists of both real and imaginary parts (we are dealing with phasors). Remember j is just a 90º operator.

Impedance = Resistance + Reactance or Z = R + jX

When we have two parallel connected components having impedances Z1 and Z2, the total equivalent impedance of such a circuit is given by Z total, where

Z total = )(

.

21

21

ZZ

ZZ

+ Ω

All the arithmetic operations in the above equation must be performed in phasor format. An alternative approach often used in parallel circuits is to use not impedance terms but admittance. The reciprocal of resistance R is called conductance G, and it has the units of Siemens (S). Similarly, the reciprocal of reactance is called susceptance, B = 1/ X. There is naturally inductive susceptance and capacitive susceptance.

Admittance (Y) = Conductance (G) + Susceptance (B)

The general term for conductance, susceptance and combinations of these is admittance, which has the symbol Y. The relationship between the impedance of a circuit and its admittance is Z = 1/ Y. The most important fact about admittance is that the total admittance of parallel connected components is the sum of the admittance’s of the individual components.

Since i = Z

eand Z =

Y

1, the relationship between current, voltage and admittance is i = e Y

The Relationship between Impedance and Admittance: Z = 1/Y

Thus: jBG

jXR+

=+ 1 now remembering that we have a complex number to invert:

22

1

BG

jBG

jBGjXR

+−=

+=+

Consider a parallel RLC circuit. If we work in admittance: For the resistor: YR = 1/R For the inductor YL = 1/jωL For the capacitor YC = jωC

Figure 9.0.1 Parallel circuit


The total admittance is Y = (1/R + 1/jωL + jωC) which can be written as a complex number G + jB where B in this case is (ωC – 1/ωL). (Note the negative sign.) I = V/Z or VY so I = V (1/R + j (ωC – 1/ωL)) I1 = V/jωL I2 = V/R I3 = VjωC The relationships between current and voltage in the components are exactly as before. Example If ω = 1000 rad/s and R = 100Ω, L = 100mH and C = 20µF in the parallel circuit as above find the supply current for an input voltage of 1000 /30º volts. R = 100Ω YR = 0.01 S XL = jωL = j100 Ω YL = - j0.01 S XC = -j(1/ωC) = -j50 Ω YC = +j0.02 S Y = (1/R + 1/jωL + jωC) = (0.01 - j 0.01 + j 0.02) = (0.01 + j0.01) = 0.0141 /45º S I = VY = 1000 /30º x 0.0141 /45º = 14.1 /75º Amps


10.0 Tutorial: Admittance 1. Find the admittance of each set of series elements.

Figure Q1 (a) Figure Q1 (b)

(0.06 + j0.08 S, 0.087 – j0.034 S) 2. Three impedances: 10 /- 30º Ω , 20 /60º Ω , 40 /0º Ω are connected in parallel. Using

admittance methods calculate the equivalent total impedance. (7.32 /- 3º Ω)

3. The admittance of a circuit is (0.05 –j0.08) S. Find the values of the resistance and the

inductive reactance of the circuit if they are: (i) In parallel and (ii) In Series

(20Ω, +j12.5 Ω) and (5.62 Ω, +j8.99 Ω) 4. Find the total admittance and impedance of the circuit of Figure 5. Identify the value of

conductance and susceptance. Draw the admittance diagram.

(0.333 –j0.56 S, 2.92 +j0.49 Ω)

Figure Q4

5. Determine the equivalent impedance Z and admittance Y for the four branch circuit shown

in Figure Q5: Z1 = + j5ΩΩΩΩ, Z2 = (5 + j 8.66) ΩΩΩΩ, Z3 = 15 ΩΩΩΩ, Z4 = - j 10 ΩΩΩΩ.

Figure Q5

(Y = 0.22 / -58º , Z = 4.55 / 58º )


6. If the total current I in Figure Q5 is 33 / -58° A, determine the supply voltage V and the branch current in the 15 Ω resistor.

(150 / 0 o V, 10 A.) 7. For the the circuit of Figure Q7 find a parallel circuit that will have the same total

impedance ZT.

(4kΩ, - j4kΩ)

Figure Q7

ac circuits handout and tutorials

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