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AC CIRCUIT ANALYSIS PROFESSOR SULEIMAN SHARKH

INTRODUCTION Electricity is mostly generated, transmitted and distribute in alternating current (AC) for a number of reasons: electric machines are AC in nature; AC can be readily transformed to a high voltage so as to reduce the current and the size and cost of the transmission lines and cables; and it is easier to switch off. In this chapter you will learn how to analyse AC circuits, using phasors, to calculate steady state voltages and currents. We will also learn about active and reactive power and power factor, and their physical and practical significance. Using phasors an AC real circuit, that is described by differential equations, is transformed into an equivalent network of complex impedances that can be described by algebraic equations and analysed in the same way we analyse circuits containing resistive elements (without capacitors or inductors). The difference is that the analysis of the phasor equivalent circuits involves the use of complex numbers. We will therefore briefly review complex number arithmetic, but you are strongly encouraged to review the material you have learnt in the mathematics module.

THE AC WAVEFORM The AC voltage supplied by the electricity network are sinusoidal, as shown in Figure 1. Most electricity loads and generators (e.g., appliances, machinery, UPS systems, solar farms wind generator etc.) draw or supply sinusoidal currents from or into the network – they are in fact obliged to do so by regulations. Non-sinusoidal currents and voltage increase losses in electrical devices, and increase the torque ripple of electric machines, which can cause vibrations and increase acoustic noise. The sinusoidal waveform is characterised by its peak (or maximum) value, frequency and phase angle. The period is the time difference between two peaks or two troughs; it is the reciprocal of the frequency. Electrical engineers normally use the root mean square value or rms, which is equal to the peak value divided by the square root of 2, as shown in Figure 2. The reason for this is that the average power in an AC circuit is given by the product of the rms voltage times the rms current, as we will see later. The rms value of AC voltage or current is in fact the equivalent DC voltage or current that produces the same power in a resistor, as can be seem in Figure 2.

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Figure 1– Characteristics of the sinusoidal waveform

Figure 2– Definition of the root mean square

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CIRCUIT ELEMENTS AND KIRCHOFF’S LAWS Electric circuits, as the name suggests, are circuits or networks of electric devices connected by wires or cables. When the frequency is low, most of the electrical energy will be confined to flow within wires and devices, with negligible radiated electromagnetic energy in the form of electromagnetic fields (e.g. radio waves). In this case the network of electrical devices can be represented by equivalent circuit elements (lumped parameters): resistors; inductors; and capacitors 1. These circuit elements are conceptual electric devices that represent the different energy manifestation of electricity: resistance which represents heat or increased vibrations of the molecules of a conductor as a result of collisions with the flowing electrical charges (usually electrons); capacitance which is a measure of the amount of the potential energy needed to store electric charges; and inductance which is a measure of the magnetic flux produced due to the movement of electric charge, hence it can be thought of as a measure of the kinetic energy of the electric charge. The behaviour of these circuit elements is described by the relationship between the current flowing through them as shown in Figure 3. In equivalent circuits the elements are connected to each other and to electricity sources using ‘wires’ of zero resistance inductance and capacitance. Electricity flow is permitted or interrupted using ideal switches with have zero on voltage when on and infinite resistance when off. In lumped parameter equivalent circuits Charge and energy are conserved by definition (as we assume no energy leakage due to radiation). Kirchhoff’s voltage law (KVL) – see Figure 4 – is a statement of the law of conservation of energy and Kirchhoff’s current law (KCL) is a statement of the conservation of charge – see Figure 5. Using these laws we can analyse electric circuit to determine the distribution of voltage and current and hence the power in the different circuit elements.

Figure 3 – Circuit elements .

1

If the frequency is high equivalent and radiated energy is significant then circuits using lumped parameters (resistors, capacitors and inductors) are not an accurate representation. In this case we need to use electromagnetic field theory to solve Maxwell’s equations, which is outside the scope of this module. 3

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Figure 4 – Kirchhoff’s Voltage Law (KVL)

Figure 5 – Kirchhoff’s Current Law (KCL)

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EXAMPLE – ANALYSIS OF AN AC CIRCUIT Consider the circuit in Figure 6. Let 𝑣 = 𝑉𝑝 cos 𝜔𝑡 = √2 𝑉 cos 𝜔𝑡

(1)

Applying KVL we obtain the following equation: −𝑣 + 𝑖𝑅 + 𝐿 or 𝐿

𝑑𝑖

𝑑𝑡

𝑑𝑖 =0 𝑑𝑡

+ 𝑖𝑅 = 𝑣

(2)

(3)

As the voltage is sinusoidal, the current will be also sinusoidal of the same frequency (due to the fact that electric circuits are linear, described by linear ordinary differential equation). The current will have the general form: 𝑖 = 𝐼𝑝 cos(𝜔𝑡 + 𝜙) = √2 𝐼 cos(𝜔𝑡 + 𝜙)

(4)

Figure 6 – RL circuit using real functions.

Substituting (4) into (3) we obtain −𝐿𝜔 √2𝐼 sin(𝜔𝑡 + 𝜙) + 𝑅√2𝐼 cos(𝜔𝑡 + 𝜙) = √2𝑉 cos(𝜔𝑡)

(5)

The objective is to solve the above equation to find 𝐼 and 𝜙. That is possible, but it is not easy to do! Fortunately, there is an easier way to solve the problem using complex function and phasors.

PHASORS As can be seen in Figure 7, there is a direct relationship between the complex exponential function, which represents a rotating arrow in the complex plane, and the sinusoidal functions; the projection of the arrow on the real and imaginary axes trace sinusoidal waves as the arrow rotates. The relationship is defined by Euler’s formula. In other words, we can use complex exponential functions to represent sinusoidal functions as illustrated in Figure 8 and use that to obtain a complex solution of the differential equation from which we can derive the real sinusoidal solution.

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Figure 7 – Relationship between sinusoidal and complex function

2

Figure 8 –RL Circuit using complex functions.

The complex function equivalents of equations (3) and (4) are 𝐿

𝑑𝐢 + 𝐢𝑅 = 𝐯 = √2𝑉𝑒 𝑗𝜔𝑡 𝑑𝑡 𝐢 = √2𝐼𝑒 𝑗(𝜔𝑡+𝜙)

(6) (7)

Substituting (7) into (6) and cancelling the √2 we obtain 𝑗𝜔𝐿𝐼𝑒 𝑗(𝜔𝑡+𝜙) + 𝑅𝐼𝑒 𝑗(𝜔𝑡+𝜙) = 𝑉𝑒 𝑗𝜔𝑡

(8)

Expanding the exponentials we obtain 𝑗𝜔𝐿𝐼𝑒 𝑗𝜔𝑡 𝑒 𝑗𝜙 + 𝑅𝐼 𝑒 𝑗𝜔𝑡 𝑒 𝑗𝜙 = 𝑉𝑒 𝑗𝜔𝑡

(9)

Cancelling the 𝑒 𝑗𝜔𝑡 terms in the above equation we get, 𝐼𝑒 𝑗𝜙(𝑗𝜔𝐿 + 𝑅) = 𝑉𝑒 𝑗0

(10)

𝐕 = 𝑉𝑒 𝑗0

(11)

Define

Note that in electrical engineering, for obvious reasons, the 𝑗 rather than 𝑖 represents the imaginary number √−1.

2

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(12) (13)

Substituting (11), (12) and (13) into (10) we get 𝐕 = 𝐙𝐈

(14)

The above equation as analogous to Ohms Law, 𝑉 = 𝑅𝐼, and it is known as the Generalised Ohm’s law. The complex variables 𝐕 and 𝐈 are called phasors. Note that they are not functions of time and are represented as stationary arrows in the complex plane as shown in Figure 9 - complex variables that are functions of as shown in Figure 7. Also note that it is common to write 𝑒 𝑗𝜙 as ∠𝜙

Figure 9 –Phasor representation in the complex plane

𝐙 is knows as impedance. Impedance is a complex number with the real part representing the resistance. The imaginary part is called reactance, which in this example is equal to 𝜔𝐿, the product of the angular frequency, 𝜔 = 2𝜋𝑓, times the inductance. Based on equation (14) we the problem reduces to analysing a complex phasor equivalent circuit as shown in Figure 10. We can now analyse this circuit in the same way we analyse circuits with resistors only, or dc circuits to write and solve algebraic equations. The difference is that we now have complex numbers instead of real numbers. The phasor current 𝐈 in Figure 10 can be calculated directly from the generalised Ohms law as 𝐈 = 𝐼∠𝜙 =

𝐕 𝑉 𝑉∠0 𝜔𝐿 ) = = ∠ (−tan−1 2 2 𝑅 𝐙 𝑅 + 𝑗𝜔𝐿 �𝑅 + (𝜔𝐿)

(15)

See a summary of complex number arithmetic in the appendix.

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Figure 10 – Phasor equivalent circuit

In summary, the procedure for analysing AC circuits using phasors as as follows: 1.

2.

Draw the phasor equivalent circuit a. Replace the real sinusoidal 𝑣 and 𝑖 with their equivalent phasors. b. Replace the circuit elements with their equivalent impedances. Analyse the circuit as you would analyse a dc circuit (or a circuit with resistors only). The difference is that we now need to deal with complex numbers.

IMPEDANCE In the previous section we observed that the impedance of a resistor is simply 𝑅 and that of an inductor is 𝑗𝜔𝐿. The detailed proof of this is shown in Figure 11 and Figure 12. It can be also readily shown that the impedance of a capacitor is

1

𝑗𝜔𝐶

=−

𝑗

𝜔𝐶

as shown in Figure 13. A summary is shown in Figure 14.

Figure 11 – Impedance of a resistor

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Figure 12 – Impedance of an inductor

Figure 13 – Impedance of a capacitor

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Figure 14 – Impedances of circuit elements.

PHASE RELATIONSHIPS BETWEEN THE VOLTAGES AND CURRENTS OF CIRCUIT ELEMENTS It is useful to consider the phase relationship between the voltages and currents of circuit elements. For a resistor the current and voltage are in phase, i.e. their peaks, troughs and zero crossings occur at the same time and their phasors will be parallel to each other, i.e. they will have the same angle as illustrated in Figure 15.

Figure 15 – Waveforms and phasor diagrams of voltage and current of a resistor – the voltage and current are in phase.

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For an inductor, there is a 90 degrees phase difference between the voltage and the current; when one is at its peak, the other will be crossing zero. In Figure 16, moving in the positive direction of the time axis, one will see the peak of the voltage waveform first, followed by the peak of the current waveforms; in other words, the current lags the voltage. One could also see this if one imagines the phasors of the voltage and current rotating in the counter clockwise direction; one would see the tip of the current phasor 90 degrees later after seeing the tip of the voltage phasor. For a capacitor, the current leads the voltage as shown in Figure 17. To help you remember this think of the word ‘CIVIL’: starting from the left or start of the word you see that for a capacitor C the current I lead the voltage V; starting from the right end of the word you will see I occurring after or lagging V.

Figure 16 – Waveforms and phasor diagram of voltage and current of an inductor - the current lags the voltage.

Figure 17 – Waforms and phasor diagram of voltage and current of a capacitor – the current leads the voltage.

CAPACITIVE AND INDUCTIVE LOADS If the impedance of the load has a negative imaginary part, then its current will be leading the voltage, and it will be a capacitive load, even though it may include an inductance, as illustrated in Figure 18. An inductive load has an impedance with a positive imaginary part and its current lags the voltage as shown in Figure 19.

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Figure 18 – An example of a capacitive load – it has a negative imaginary part and the its current will lead the voltage.

Figure 19 – An example of an inductive load – it has a net positive imaginary part and its current lags the voltage.

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EXAMPLE 1 – ANALYSIS OF AN RLC CIRCUIT Find the current 𝐼 in the circuit shown in Figure 20.

Figure 20 – RLC Circuit.

SOLUTION The first step is to draw the corresponding phasor-domain circuit shown in Figure 21 using 𝜔 = 4000 rad/s of the source. The current is 𝐕

𝐈=𝐙=

140 ∠�−10° �

3.6+𝑗4.8−𝑗6.25

=

140 ∠�−10° �

3.88 ∠�−21.9 °�

= 36.1 ∠(11.9° ) A

This current can be used to obtain the voltage phasors: 𝐕𝐑 = 36.1 ∠(11.9° ) × 3.6 = 130 ∠(11.9° ) V 𝐕𝐋 = 36.1 ∠(11.9° ) × 𝑗4.8 = 36.1 ∠(11.9° ) × 4.8 ∠(90° ) = 173 ∠(102° ) V 𝐕𝐂 = 36.1 ∠(11.9° ) × (−𝑗6.25) = 36.1 ∠(11.9° ) × 6.25 ∠(−90° ) = 225 ∠(−78.1)°

4.8 Ω

3.6 Ω 𝐼

+

-j6.25Ω

𝑉 = 140 ∠(−10) V

Figure 21 – Phasor-domain RLC circuit.

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The corresponding sinusoidal quantities are 𝑖 = 36.1 √2 cos(4000𝑡 + 11.9° ) A, 𝑣𝑅 = 130 √2 cos(4000𝑡 + 11.9° ) V, 𝑣𝐿 = 173 √2 cos(4000𝑡 + 102° ) V, 𝑣𝑐 = 225 √2 cos(4000𝑡 − 78.1° ) V.

EXAMPLE – ANALYSIS USING PHASOR DIAGRAMS Find the source voltage 𝐕 in the circuit shown in Figure 22 using the phasor diagram graphical method.

Figure 22 – An AC source supplying an inductive load through a cable.

SOLUTION We can treat phasors in the same way we treat vectors; they can be added or subtracted graphically in the same way we add or subtract vectors. To add two phasors for example we draw them connected top to tail and the resultant will be the phasor closing the triangle. In the above circuit we can write 𝐕𝐬 = (𝑅 + 𝑗𝑋)𝐈 + 𝐕 = 𝑅𝐈 + 𝑗𝑋𝐈 + 𝐕 = 2𝐈 + 5𝐈 ∠(90° ) + 𝐕 = 4 ∠(−30° ) + 10 ∠(60° ) + 𝐕 The above equation can be written as 𝐕𝐬 = 𝐕𝑅 + 𝐕𝐿 + 𝐕

(16)

In other words, the supply voltage is the sum of the voltages across the resistor, inductor and the load. The resistor’s voltage phasor, 𝐕𝑅 = 2𝐈 = 4 ∠(−30° ), is proportional to that of the current phasor, and hence the two are in parallel. But the inductor’s voltage phasor, 𝐕𝐿 = 5𝐈 ∠90° = 10 ∠(60° ), is perpendicular to that of the current. The phasor diagram representing equation (16) is shown in Figure 23.

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20

15

10

𝐕𝐬

5

0

𝐕

𝐈 -5 -5

0

𝑗𝑋𝐈

5

𝑅𝐈 10

15

20

Figure 23 – Phasor diagram

EXAMPLE – AC CIRCUIT ANALYSIS Find the current 𝐈, 𝐕𝟏 , 𝐕𝟐 in the circuit shown in Figure 24.

𝐈

10 Ω

15 Ω

+ 𝐕 = 100 ∠(20° ) V

𝐕1

𝑗20 Ω

𝐕2

−𝑗30 Ω

Figure 24 – AC circuit

SOLUTION The current can be found by dividing the voltage by the total impedance. The impedance of the right hand loop resistance in series with the capacitance, 15 − 𝑗30 Ω = 33.5 ∠(−63.4° ) Ω are connected in parallel with the 𝑗20 Ω of the inductor: 𝐙𝐱 =

𝑗20 × 33.5 ∠(−63.4° ) 671 ∠(26.6° ) = 37.2 ∠(60.3° ) = 18.5 + 32.3 Ω = 18 ∠(−33.7° ) 𝑗20 + 15 − 𝑗30

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Note that in the above expression we used the polar form when doing multiplication or division and the Cartesian form when doing addition or subtraction. The resultant impedance is inductive as the imaginary part is positive. The above value of 𝐙𝐱 plus the 10 Ω of the series resistor gives the total impedance: 𝐙 = 10 + 18.5 + 𝑗32.3 = 28.5 + 𝑗32.3 = 43.1 ∠(48.6° )Ω. The current 𝐈 can now be calculated as 𝐈=

𝐕 100∠(20° ) = 2.32 ∠(−28.6° ) A. = 𝐙 43.1 ∠(48.6° )

We can use the voltage divider rule twice to calculate 𝐕𝟏 and 𝐕𝟐 : 𝐕𝟏 =

𝒁𝒙 37.2 ∠(60.3° ) 3720 ∠(80.3° ) = 86.4 ∠32° V. 𝐕= × 100 ∠(20° ) = 43.1 ∠(48.6° ) 𝒁 28.5 + 𝑗32.3 𝐕𝟐 =

−𝑗30 2590 ∠(−58° ) = 77.3 ∠(5° ) V. × 86.4 ∠32° = 33.5 ∠(−63° ) 15 − 𝑗30

RC FILTERS You have done experiments to measures the frequency response of simple first order RC filter circuits: low pass and high pass filters. Using phasors we can now analyse these circuits and derive their gain plots that you measured in the experiments.

LOW PASS FILTER Consider the circuit shown in Figure 25. The output voltage can be calculated using the voltage divider rule as:

𝐕𝐨𝐮𝐭

1 𝑉𝑖𝑛 ∠(− tan−1 (𝑅𝐶 𝜔)) 𝐕𝐢𝐧 𝑗 𝜔𝐶 = × 𝐕𝐢𝐧 = = 1 1 + 𝑗𝑅 𝜔𝐶 �1 + (𝑅𝐶 𝜔)2 𝑅 + �𝑗 𝜔𝐶�

The above assumes that the phase angle of 𝐕𝐢𝐧 is zero and its magnitude is 𝑉𝑖𝑛 . The voltage gain is defined as the ratio of the magnitude of the output to the magnitude of the input: 𝑉𝐺𝐴𝐼𝑁 = �

𝐕𝐨𝐮𝐭 1 𝑉𝑜𝑢𝑡 = �= 𝑉𝑖𝑛 𝐕𝐢𝐧 �1 + (𝑅𝐶 𝜔)2

At very low frequencies, the gain is approximately unity. At high frequencies the gain is approximately

1

𝑅𝐶 𝜔

,

i.e. it is inversely proportional to frequency. Hence such a filter will let the low frequency components of a signal pass through, but attenuates the high frequency signals, and hence this circuit is known as a low pass filter. It is used to remove high frequency noise from a signal. The corner frequency (also known as the bandwidth) is defined as the frequency at which the power of the corresponding components of the signal is halved. Recalling that power is proportional to the square of the voltage, then the corner frequency will be that at which the gain equals 𝑓𝑐 = 16

𝜔𝑐 1 = 2 𝜋 𝑅𝐶

1

√2

or 𝑅𝐶 𝜔𝑐 = 1. Hence, (17)

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Figure 25 – Low pass RC filter.

Expressing gain using the decibel (dB) scale: The bel (B) scale is a logarithmic scale used for expressing gains, in terms of a signal amplitude, or the power contained in the signal. A decibel (dB) is just one-tenth of a bel, i.e. 1dB = 0.1B (to yield more convenient numerical values). In the case of signal power, the gain in bels is simply:

P PGAIN (B ) = log OUT  PIN

  

 POUT  PIN

Therefore, in decibels, PGAIN (dB ) = 10 ⋅ log

  

For the signal amplitude gain, note that the power is proportional to the square of the amplitude (e.g. in the electrical case P = V2 ÷ R). Using the laws of logarithms, voltage gain can therefore be expressed as:

 V  2  V VGAIN (B ) = log  OUT   = 2 ⋅ log  OUT   V IN    VIN  

  

 V  2  V VGAIN (dB ) = 10 ⋅ log  OUT   = 20 ⋅ log OUT   VIN  V IN   

  

Note that a positive value on the bel scale corresponds to an amplification, whereas a negative value corresponds to an attenuation. It is useful to recognise several commonly occurring gains expressed in decibels – see the table below. Signal Gain √2 (~1.4) 10 100

Gain (dB) 3 20 40

Signal Gain 1/√2 (~0.7) 0.1 0.01

Gain (dB) -3 -20 -40

The corner frequency of a low pass filter would therefore correspond to the point when the dB gain reduces to approximately -3 dB. 17

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HIGH PASS FILTER Basically, the circuit of the high pass filter is similar to that that of the low pass filter, but with the resistor and capacitor swopping place...