AC Circuit Analysis Notes

AC Circuit Analysis

Introduction

Alternating current (AC) is preferred for electricity generation, transmission, and distribution due to several reasons:
- AC is the natural form for electrical machines.
- AC voltages can be easily transformed to high voltages, reducing current and the cost/size of transmission lines.
- AC is easier to switch off.
AC circuit analysis involves using phasors to calculate steady-state voltages and currents.
Concepts such as filter circuits, active and reactive power, and power factor are crucial.
Phasors transform AC circuits described by differential equations into equivalent networks with complex impedances, which can be analyzed using algebraic equations.
Analysis involves complex numbers.

The AC Waveform

AC voltage supplied by electricity networks is sinusoidal.
Most electrical loads and generators draw or supply sinusoidal currents.
Non-sinusoidal currents and voltages increase losses and torque ripple.
Sinusoidal waveforms are characterized by:
- Peak value ( $V_p$ ).
- Frequency ( $f$ ).
- Phase angle ( $\theta$ ).
The period ( $T$ ) is the reciprocal of the frequency: $T = \frac{1}{f}$ .
Electrical engineers use the root mean square (rms) value: $V{rms} = \frac{Vp}{\sqrt{2}}$ .
Average power in an AC circuit is the product of rms voltage and rms current.
RMS value is equivalent to the DC voltage or current that produces the same power in a resistor.

AC Circuits

Peak value: $V_p$
Peak-to-peak value: $V{pp} = 2Vp$
Period: $T$ (seconds), Frequency: $f = \frac{1}{T}$ (Hz)
Angular frequency: $w = 2\pi f$ (rad/s)
Phase: $\theta$
Formula: $v = Vp \cos(2\pi ft - \theta) = Vp \cos(\omega t - \theta)$
The rms voltage or current is the equivalent DC voltage/current that generates the same mean amount of power loss $P$ in a resistor.
Power in DC circuit: $P = \frac{V^2}{R} = I^2 R$
Instantaneous power in AC circuit: $p = vi$
Mean power in AC circuit: $P = \frac{1}{T} \int_0^T p \, dt = \frac{V^2}{R}$
RMS voltage: $V{rms} = \sqrt{\frac{1}{T} \int0^T v^2 \, dt} = \frac{V_p}{\sqrt{2}}$
RMS current: $I{rms} = \sqrt{\frac{1}{T} \int0^T i^2 \, dt} = \frac{I_p}{\sqrt{2}}$

Circuit Elements and Kirchhoff’s Laws

Electric circuits are networks of electric devices connected by wires or cables.
At low frequencies, electrical energy is confined within wires and devices with negligible radiation.
The network can be represented by lumped parameters: resistors, inductors, and capacitors.
Circuit elements represent different energy manifestations:
- Resistance: Represents heat or increased molecular vibrations due to collisions with flowing charges.
- Capacitance: Measures the potential energy needed to store electric charges.
- Inductance: Measures magnetic flux produced by moving charge, representing kinetic energy.
Behavior is described by the relationship between current and voltage.
In equivalent circuits, elements are connected by wires with zero resistance, inductance, and capacitance.
Ideal switches have zero on-voltage and infinite off-resistance.
Charge and energy are conserved in lumped parameter circuits.
Kirchhoff’s Voltage Law (KVL): Conservation of energy.
Kirchhoff’s Current Law (KCL): Conservation of charge.
These laws are used to analyze circuits to determine voltage, current, and power distribution.
At high frequencies, lumped parameters are insufficient; electromagnetic field theory and Maxwell’s equations are needed.

Example – Analysis of an AC Circuit

Consider an RL circuit.
$v = V_p \cos(\omega t) = \sqrt{2} V \cos(\omega t)$
Applying KVL: $-v + iR + L \frac{di}{dt} = 0$
$L \frac{di}{dt} + iR = v$
The current will also be sinusoidal with the same frequency:
- $i = I_p \cos(\omega t + \phi) = \sqrt{2} I \cos(\omega t + \phi)$
Substituting, we get: $-L\omega \sqrt{2}I \sin(\omega t + \phi) + R\sqrt{2}I \cos(\omega t + \phi) = \sqrt{2}V \cos(\omega t)$
Solving this equation for $I$ and $\phi$ is complex.
Using complex functions and phasors simplifies the problem.

Phasors

There is a direct relationship between complex exponential functions and sinusoidal functions.
The projection of a rotating arrow on the real and imaginary axes traces sinusoidal waves.
Euler’s formula defines this relationship.
Complex exponential functions can represent sinusoidal functions.
This leads to a complex solution of the differential equation, from which the real sinusoidal solution is derived.

Complex Function Equivalents
$L \frac{d\mathbf{i}}{dt} + \mathbf{i}R = \mathbf{v} = \sqrt{2}Ve^{j\omega t}$
$\mathbf{i} = \sqrt{2}Ie^{j(\omega t + \phi)}$

Substituting, canceling terms:
$j\omega LIe^{j(\omega t + \phi)} + RIe^{j(\omega t + \phi)} = Ve^{j\omega t}$
$j\omega LIe^{j\omega t}e^{j\phi} + RIe^{j\omega t}e^{j\phi} = Ve^{j\omega t}$
$Ie^{j\phi}(j\omega L + R) = Ve^{j0}$
Define the following:
- $\mathbf{V} = Ve^{j0}$
- $\mathbf{I} = Ie^{j\phi}$
- $\mathbf{Z} = R + j\omega L$
Then we get:
$\mathbf{V} = \mathbf{Z}\mathbf{I}$
This is analogous to Ohm’s Law, $V = RI$ , and is known as the Generalized Ohm’s law.
$V$ and $I$ are phasors.
Phasors are stationary arrows in the complex plane, not functions of time.
$e^{j\phi}$ is commonly written as $\angle \phi$
$\mathbf{Z}$ is impedance.
- The real part of impedance is resistance.
- The imaginary part is reactance: $\omega L = 2\pi fL$
The problem reduces to analyzing a complex phasor equivalent circuit.
The circuit can be analyzed like DC circuits using algebraic equations with complex numbers.
The phasor current $I$ can be calculated as:
$\mathbf{I} = I\angle \phi = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{V\angle 0}{R + j\omega L} = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \angle \left(-\tan^{-1} \frac{\omega L}{R} \right)$

Impedance

The impedance of a resistor is R, and the impedance of an inductor is $j\omega L$ .
The impedance of a capacitor is $\frac{1}{j\omega C} = -\frac{j}{\omega C}$ .

Impedances of Circuit Elements

Element	Time Domain	Phasor Domain
Resistor	$v = iR$	$V = RI$
Inductor	$v = L\frac{di}{dt}$	$V = j\omega LI$
Capacitor	$i = C\frac{dv}{dt}$	$V = \frac{1}{j\omega C} I$


Inductor:
	$i(t) = \sqrt{2}I cos(\omega t + \phi)$
	$v(t) = L\frac{di}{dt} = -\sqrt{2}\omega LIsin(\omega t + \phi)$
	$Z<em>L = \frac{V}{I} = j\omega L = jX</em>L$

| Capacitor:
| | $i(t) = \sqrt{2}I cos(\omega t + \phi)$
| | $v(t) = \frac{1}{C}\int i\,dt = \frac{\sqrt{2}I}{\omega C} sin(\omega t + \phi)$
| $Zc = \frac{V}{I} = \frac{1}{j\omega C} = \frac{-j}{\omega C} = -jXc$

Phase Relationships Between the Voltages and Currents of Circuit Elements

Resistor: Voltage and current are in phase.
Inductor: Current lags the voltage by 90 degrees.
Capacitor: Current leads the voltage by 90 degrees.
CIVIL: Capacitor-current leads voltage; inductor-voltage leads current.

Capacitive and Inductive Loads

Capacitive load: Has a negative imaginary impedance, current leads voltage.
Inductive load: Has a positive imaginary impedance, current lags voltage.

Example 1 – Analysis of an RLC Circuit

Find the current in the RLC circuit.
Convert to phasor domain using $\omega = 4000$ rad/s.
$\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{140 \angle (-10^\circ)}{3.6 + j4.8 - j6.25} = \frac{140 \angle (-10^\circ)}{3.88 \angle (-21.9^\circ)} = 36.1 \angle (11.9^\circ) \text{ A}$
$\mathbf{V}_R = 36.1 \angle (11.9^\circ) \times 3.6 = 130 \angle (11.9^\circ) \text{ V}$
$\mathbf{V}_L = 36.1 \angle (11.9^\circ) \times j4.8 = 36.1 \angle (11.9^\circ) \times 4.8 \angle (90^\circ) = 173 \angle (102^\circ) \text{ V}$
$\mathbf{V}_C = 36.1 \angle (11.9^\circ) \times (-j6.25) = 36.1 \angle (11.9^\circ) \times 6.25 \angle (-90^\circ) = 225 \angle (-78.1^\circ)$
Convert back to sinusoidal:
- $i = 36.1\sqrt{2} \cos(4000t + 11.9^\circ) \text{ A}$
- $v_R = 130\sqrt{2} \cos(4000t + 11.9^\circ) \text{ V}$
- $v_L = 173\sqrt{2} \cos(4000t + 102^\circ) \text{ V}$
- $v_C = 225\sqrt{2} \cos(4000t - 78.1^\circ) \text{ V}$

Example – Analysis Using Phasor Diagrams

Using the graphical method, find the source voltage $V_s$ in the circuit, which consists of an ac source supplying an inductive load through a cable.
Treat phasors as vectors for addition and subtraction.
$V_s = (R + jX)I + V = RI + jXI + V = 2I + 5I\angle 90^\circ + V = 4 \angle (-30^\circ) + 10 \angle (60^\circ) + V$
$Vs = VR + V_L + V$
The supply voltage is the sum of the voltages across the resistor, inductor, and the load.
The resistor’s voltage $V_R = 2I = 4 \angle (-30^\circ)$ is proportional to the current and in phase with it.
The inductor’s voltage $V_L = 5I \angle 90^\circ = 10 \angle 60^\circ$ is perpendicular to the current.

Example – AC Circuit Analysis

Problem: Find $I$ , $V1$ , $V2$ in the AC circuit with a voltage source, a resistor, an inductor and a capacitor.
Solution:
- The current is found by dividing the voltage by the total impedance.
- The impedance of the right-hand loop (resistance in series with the capacitance) is $15 - j30 \Omega = 33.5 \angle (-63.4^\circ) \Omega$ .
- This is in parallel with the $j20 \Omega$ of the inductor:
 $\mathbf{Z}_x = \frac{j20 \times 33.5 \angle (-63.4^\circ)}{j20 + 15 - j30} = \frac{671 \angle (26.6^\circ)}{18 \angle (-33.7^\circ)} = 37.2 \angle (60.3^\circ) = 18.5 + j32.3 \Omega$
- The total impedance is:
 $\mathbf{Z} = 10 + 18.5 + j32.3 = 28.5 + j32.3 = 43.1 \angle (48.6^\circ) \Omega$
- The current $I$ is:
 $\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{100 \angle (20^\circ)}{43.1 \angle (48.6^\circ)} = 2.32 \angle (-28.6^\circ) \text{ A}$
- Using the voltage divider rule:
 $\mathbf{V}1 = \frac{\mathbf{Z}x}{\mathbf{Z}} \mathbf{V} = \frac{37.2 \angle (60.3^\circ)}{28.5 + j32.3} \times 100 \angle (20^\circ) = \frac{3720 \angle (80.3^\circ)}{43.1 \angle (48.6^\circ)} = 86.4 \angle 32^\circ \text{ V}$
 $\mathbf{V}_2 = \frac{-j30}{15 - j30} \times 86.4 \angle 32^\circ = \frac{2590 \angle (-58^\circ)}{33.5 \angle (-63^\circ)} = 77.3 \angle (5^\circ) \text{ V}$

RC Filters

RC filters are analyzed using phasors to derive gain plots.

Low Pass Filter

The output voltage is:
$\mathbf{V}{out} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} \times \mathbf{V}{in} = \frac{\mathbf{V}{in}}{1 + jR\omega C} = \frac{V{in}}{\sqrt{1 + (RC\omega)^2}} \angle (-\tan^{-1}(RC\omega))$
The voltage gain is:
$V{GAIN} = \left| \frac{\mathbf{V}{out}}{\mathbf{V}{in}} \right| = \frac{V{out}}{V_{in}} = \frac{1}{\sqrt{1 + (RC\omega)^2}}$
At very low frequencies, the gain is approximately unity.
At high frequencies, the gain is approximately $\frac{1}{RC\omega}$ , inversely proportional to frequency.
This filter passes low-frequency components and attenuates high-frequency signals.
The corner frequency (cut-off frequency or bandwidth) $\omega_c$ is defined as the frequency at which the power is halved.
$\omega_c = \frac{1}{RC}$
$fc = \frac{\omegac}{2\pi} = \frac{1}{2\pi RC}$

High Pass Filter

The resistor and capacitor positions are swapped compared to the low pass filter.
$V_{GAIN} = \frac{RC\omega}{\sqrt{(RC\omega)^2 + 1}}$
When $\omega$ is large, the voltage gain approaches unity.
At low frequencies, the gain is approximately $RC\omega$ , proportional to frequency.
Low-frequency signals are attenuated.
$fc = \frac{\omegac}{2\pi} = \frac{1}{2\pi RC}$

Expressing Gain in Decibels (dB)

The bel (B) scale is a logarithmic scale for expressing gains in terms of a signal amplitude or power
$Gain{P}(B) = log(\frac{P{OUT}}{P_{IN}})$
A decibel (dB) is one-tenth of a bel, i.e., 1 dB = 0.1 B
Power gain in decibels:
$Gain{P}(dB) = 10log(\frac{P{OUT}}{P_{IN}})$
Amplitude gain in decibels:
$Gain{V}(dB) = 20log(\frac{V{OUT}}{V_{IN}})$

Power in AC Circuits

Instantaneous power fluctuates periodically at double the frequency for a purely resistive load.

Active Power, P [W]

Average power is also known as active power.
In a purely resistive load: $P = VI = I^2R = \frac{V^2}{R}$
The units of active power are Watts (W).
In a purely inductive load, average power is zero.
In a purely capacitive load, average power is zero.

Reactive Power, Q [VAR]

The alternating magnetizing current flowing into/out of an inductor and the alternating flow of charge into/out of a capacitor does not consume average power.
They do matter because they cause losses in transmission lines and voltage drops.
Reactive power formula: $Q = I^2X = \frac{V^2}{X}$
Reactive power of a capacitor is negative.
The unit of reactive power is Volt Amp Reactive (VAR).

The Power Factor

Power factor is defined as $cos \phi$ where $\phi$ is the angle between voltage and current.
Active power: $P = I^2R = VIcos\phi$
Reactive power in an inductor: $QL = I^2XL = VIsin\phi$
Reactive power of a capacitor: $QC = -I^2XC = -VIsin\phi$

Complex and Apparent Powers [VA], and the Power Triangle

Complex Power is defined as $S = P +jQ$ with a magnitude called Apparent Power and measured in Volt-Amps [VA].

Effect of the Power Factor

Transmission losses increase if the power factor is less than unity.
Inductive loads have high voltage regulation, which can cause the voltage to dip.
Capacitive loads increase the load voltage above the source voltage.
Non-unity power factors are not desirable.

Power Factor Correction

Electricity distribution network operators charge for active energy (kWh).
A poor power factor increases transmission losses and thus incurs costs.
Customers with a poor power factor load have to pay a penalty unless they correct it.
Most loads are inductive and have a lagging power factor so they absorb reactive power.
To correct the power factor, reactive power needs to be supplied locally by connecting a capacitor in parallel with the inductive load.

Power Factor Correction Example

Problem: An inductive load with a power factor of 0.5 draws 80 kVA at 400 V, 50 Hz.
Determine the power and reactive power and draw the power triangle.
If a power factor correction capacitor of 1mF is connected across the load, determine the new power factor.
Solution:
- Active power: $P = S \cos \phi = 80 \times 0.5 = 40 \text{ kW}$
- Reactive power before the capacitor: $Q = S \sin \phi = 80 \times 0.886 = 69 \text{ kVAR}$
- The reactive power supplied by the capacitor is:
 $Qc = - \frac{V^2}{Xc} = -V^2 \omega C = -2\pi fCV^2 = -50 \text{ kVAR}$
- Total reactive power of the combined capacitor and load:
 $Q_{new} = 69 - 50 = 19 \text{ kVAR}$
- New apparent power:
 $S_{new} = \sqrt{P^2 + Q^2} = \sqrt{40^2 + 19^2} = 44 \text{ VA}$
- New power factor:
 $\cos \phi{new} = \frac{P}{S{new}} = \frac{40}{44} = 0.9 \text{ lagging}$

Appendix: Complex Numbers

Complex number: $z = x + jy$ (Cartesian form)
- $x$ is the real part.
- $y$ is the imaginary part.
Polar form: $z = re^{j\theta}$
- $r = \sqrt{x^2 + y^2}$
- $\theta = \tan^{-1} \frac{y}{x}$
- $x = r \cos \theta$
- $y = r \sin \theta$
Let
- $\mathbf{Z}1 = x1 + jy1 = r1 \angle \theta_1$
- $\mathbf{Z}2 = x2 + jy2 = r2 \angle \theta_2$

Summation and Substraction

$\mathbf{Z}1 + \mathbf{Z}2 = (x1 + jy1) + (x2 + jy2) = (x1 + x2) + j(y1 + y2)$
$\mathbf{Z}1 + \mathbf{Z}2 = \sqrt{((x1 + x2)^2 + (y1 + y2)^2}\angle tan^{-1}(\frac{y1 + y2}{x1 + x2})$

Multiplication and Division

$\mathbf{Z}1 \mathbf{Z}2 = r1r2\angle(\theta1 + \theta2)$
$\frac{\mathbf{Z}1}{\mathbf{Z}2} = \frac{r1}{r2} \angle (\theta1 - \theta2)$