AC Circuits Analysis Summary

AC Circuits Analysis

Powers in AC Circuits

• References:

  • Lecture slides and notes

  • Reference book: HUGHES Electrical and Electronics Technology

    • Chapter 12: Power in AC Circuits

    • Chapter 13: Complex Notation (13.8-13.12)

Power in a Purely Resistive Load

• Instantaneous power $p = vi$. In a purely resistive circuit, voltage and current are in phase.

• If we assume $v = \sqrt{2}V \cos(\omega t)$, then $i = \frac{v}{R} = \frac{\sqrt{2}V}{R} \cos(\omega t) = \sqrt{2}I \cos(\omega t)$, where $I = \frac{V}{R}$.

• Instantaneous power can be expressed as $p = vi = (\sqrt{2}V \cos(\omega t))(\sqrt{2}I \cos(\omega t)) = 2VI \cos^2(\omega t) = VI(1 + \cos(2\omega t))$.

• The Active power of the resistor R is given by $P = I^2R$ [W].

• Average Power

  • The average power $P$ is calculated as $P = \frac{1}{T} \int<em>{0}^{T} p dt = \frac{1}{T} \int</em>{0}^{T} VI(1 + \cos(2\omega t)) dt = VI = \frac{V^2}{R} = I^2R$ [W].

  • Here, $T$ is the period of the AC signal.

Power in a Purely Inductive Load

• Let $v = \sqrt{2}V \cos(\omega t)$.

• Then, $i = \frac{1}{L} \int v dt = \frac{1}{L} \int \sqrt{2}V \cos(\omega t) dt = \frac{\sqrt{2}V}{\omega L} \sin(\omega t) = \sqrt{2}I \sin(\omega t)$, where $V = \omega L I = IX_L$.

• $X_L = \omega L$ is the inductive reactance.

• Instantaneous Power

  • $p = vi = (\sqrt{2}V \cos(\omega t))(\sqrt{2}I \sin(\omega t)) = VI \sin(2\omega t) = I^2 X<em>L \sin(2\omega t) = \frac{V^2}{X</em>L} \sin(2\omega t) = Q_L \sin(2\omega t)$

• Average power $P = 0$.

• Reactive Power of L

  • Defined as $Q<em>L = I^2 X</em>L = \frac{V^2}{X_L}$ [VAR].

  • Measured in Volt-Ampere Reactive (VAR).

  • Represents power used to magnetize the inductor or recovered when it is demagnetized. No energy is lost on average.

  • $Q_L$ is positive, meaning an inductor absorbs reactive power.

Power in a Purely Capacitive Load

• Let $v = \sqrt{2}V \cos(\omega t)$.

• Then, $i = C \frac{dv}{dt} = C \frac{d}{dt} (\sqrt{2}V \cos(\omega t)) = -\sqrt{2} \omega C V \sin(\omega t) = -\sqrt{2}I \sin(\omega t)$, and $V = - \frac{1}{\omega C} I = -X_C I$.

• $X_C = \frac{1}{\omega C}$ is the capacitive reactance.

• Instantaneous Power

  • $p = vi = (\sqrt{2}V \cos(\omega t))(-\sqrt{2}I \sin(\omega t)) = -VI \sin(2\omega t) = -I^2 X<em>C \sin(2\omega t) = -\frac{V^2}{X</em>C} \sin(2\omega t) = Q_C \sin(2\omega t)$

• Average or Active Power $P = 0$.

• Reactive Power of C

  • $Q<em>C = -I^2 X</em>C = -\frac{V^2}{X_C}$

  • The Reactive Power is measured in [VAR] and represents the power used for charging the capacitor or recovered during discharging. No energy is lost on average.

  • $Q_C$ is negative, indicating that a capacitor generates reactive power.

Summary of Active and Reactive Power

• Note: We can also use $\frac{V^2}{R}$, $\frac{V^2}{X<em>L}$, or $\frac{-V^2}{X</em>C}$ to calculate the active and reactive power for R, L, and C, respectively.

• Forreactive loads containing both R and L or C, both active power P and reactive power Q are involved.

• The conservation of energy holds:

  • The active power from the supply, $P<em>s$, equals the summation of active powers from each load: $P</em>s = P<em>1 + P</em>2 + \dots$

  • Similarly, the total reactive power from the supply $Q<em>s = Q</em>1 + Q_2 + \dots$

Practice Examples

• Example 1: A circuit with a current of $I = 10 \angle 0.5$ A flowing through a 4 Ω resistor.

  • Active power $P = 10^2 \times 4 = 400$ W.

• Example 2: A circuit with a current of $I = 10 \angle 0.5$ A flowing through an inductor with impedance $Z_L = j5$ Ω.

  • Reactive power $Q = I^2 X_L = 10^2 \times 5 = 500$ VAR.

• Example 3: A circuit with a current of $I = 10 \angle 0.5$ A flowing through a capacitor with impedance $Z_C = -j3$ Ω.

  • Reactive power $Q = -I^2 X_C = -10^2 \times 3 = -300$ VAR.

Complex Power

• Complex power S represents electrical power in the form of a complex number.

  • Real part: Active power (P)

  • Imaginary part: Reactive power (Q)

  • $S = P + jQ$

• Complex Conjugate

  • For a complex number $Z = x + jy = re^{j\theta}$, its complex conjugate is $Z^* = x - jy = re^{-j\theta}$.

  • For a complex current $I = Ie^{j\phi} = I \angle \phi$, its complex conjugate is $I^* = Ie^{-j\phi} = I \angle (-\phi)$.

• For a Load with Voltage V and Current I, Complex Power is $S = P + jQ = V \times I^*$.

• Power Triangle

  • Apparent Power $S = |S| = \sqrt{P^2 + Q^2}$ [VA].

  • $P = VI \cos(\phi)$

  • $Q = VI \sin(\phi)$

Example

• A circuit with R = 6 Ω and jXL = j11.31 Ω, and $V = 50 \angle 0°$ V.

• The total impedance of the load is $Z = R + jX_L = 6 + j11.31 = 12.8027 \angle 1.083$

• The complex current is $I = \frac{V}{Z} = \frac{50 \angle 0°}{12.8027 \angle 1.083} = 3.9054 \angle -1.083$ A.

• The complex power $S = V \times I^* = 50 \angle 0° \times 3.9054 \angle 1.083 = 195.27 \angle 1.083 = 91.52 + j172.5$

  • Active power $P = 91.52$ W.

  • Reactive power $Q = 172.5$ VAR.

  • Apparent power $S = |S| = 195.27$ VA.

• Alternatively:

  • Active power $P = I^2R = 3.9054^2 \times 6 = 91.52$ W.

  • Reactive power $Q = I^2X_L = 3.9054^2 \times 11.31 = 172.5$ VAR.

Power Factor (PF)

• Defined as $PF = \frac{P}{S} = \cos(\phi)$.

• Power factors near 1 are good; small power factors are poor.

• Utility charges for active power P, and customers pay per kWh because P is proportional to fuel used to generate the electricity.

• Poor power factor increases transmission losses or requires a larger generator (high V).

• Customers with poor power factors (less than 0.85) may have to pay a penalty or correct the power factor.

• As $\phi \rightarrow 0$, $PF \rightarrow 1$.

• As $\phi \rightarrow \pm \frac{\pi}{2}$, $PF \rightarrow 0$.

Power Factor Correction

• Most loads are inductive and absorb reactive power ($Q<em>L = I^2 X</em>L$).

• To correct the power factor, reactive power needs to be generated locally.

• A capacitor is a generator of reactive power ($Q<em>C = -I^2 X</em>C$)!

• A capacitor bank can be used for power factor correction.

Summary of AC Power

• AC circuits deliver power to resistive (contains only resistors) and reactive loads (contains a combination of resistor, inductor, and capacitor).

• For resistive loads, AC energy is dissipated like a DC current dissipates energy in a resistor. This kind of AC power is called Active power or Real power, $P = I^2R$.

• Most loads in AC circuits contain both resistive and reactive loads; the source delivers a mixture of active and reactive powers.

• Complex power $S = P + jQ = V \times I^*$ [VA] represents both powers.

• Apparent power is $S = |S| = \sqrt{P^2 + Q^2} = VI$.

• For a purely reactive load, AC energy is first delivered to the load and then returned to the source, like an unending rally in tennis. The power describing the rate of this energy moving in and out of the reactance is called Reactive power, Q.

  • For an inductor, $Q<em>L = I^2 X</em>L$.

  • For a capacitor, $Q<em>C = -I^2 X</em>C$.

• The ratio of active power and apparent power is called the Power Factor (PF), $PF = \frac{P}{S}$. Load power factor PF near 1 is good. Customers with poor power factors may have to pay a penalty or correct the power factor.

Complex Power - Additional Details

• Complex power $S$ represents electrical power in the form of a complex number, with its real part as active power and imaginary part as reactive power.

• $S = P + jQ = V \times I^*$

• For a complex number $Z = x + jy = re^{j\theta}$, its complex conjugate is $Z^* = x - jy = re^{-j\theta}$.

• For a complex current $I = Ie^{j\phi} = I \angle \phi$, its complex conjugate is $I^* = Ie^{-j\phi} = I\angle(-\phi)$.

Complex Power for R, L, and C Elements

Power Factor Correction Example

• An inductive load with a power factor of 0.5 draws 80 kVA at 400 V, 50 Hz.

• Find active 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:

  • Original apparent power $S = VI = 80 kVA$.

  • $PF = \frac{P}{S} = \cos(\phi) = 0.5$, $\phi = \arccos(0.5) = 60°$.

  • Active power $P = S \cos(\phi) = 80 \times 0.5 = 40 kW$. (This remains constant before and after the capacitor connection).

  • Reactive power before the capacitor: $Q = S \sin(\phi) = 80 \times 0.866 = 69 kVAR$.

  • Reactive power generated by the capacitor: $Q<em>c = \frac{-V^2}{X</em>c} = -V^2\omega C = -V^2 2\pi f C = -50 kVAR$.

  • Total reactive power after adding the capacitor: $Q_{new} = 69 - 50 = 19 kVAR$.

  • New apparent power $S<em>{new} = \sqrt{P^2 + Q</em>{new}^2} = 44 kVA$.

  • New power factor: $\cos(\phi<em>{new}) = \frac{P}{S</em>{new}} = \frac{40}{44} = 0.9$ lagging.

Low-Pass and High-Pass Filters

• References:

  • Lecture notes

  • Reference book: HUGHES Electrical and Electronics Technology

    • Chapter 15: Network theorems applied to AC networks

    • Chapter 17: Passive filters

Impedances and Frequency

• Inductor

  • Low frequencies: Low impedance (Short circuit)

  • High frequencies: High impedance (Open circuit)

• Capacitor

  • Low frequencies (like DC): High impedance (Open circuit)

  • High frequencies: Low impedance (Short circuit)

Impedance Summary

Tidal Turbine Generator

• Example of a System Including: Turbine blade, DC generator, Gear boxes, Filter capacitor

AC Filters

• Low-pass filters attenuate (reduce in amplitude) high-frequency signals and pass low-frequency signals.

• High-pass filters attenuate low-frequency signals and pass high-frequency signals.

Low Pass Filters (RC)

• Transfer function: $H(\omega) = \frac{V<em>{out}}{V</em>{in}} = \frac{1}{1 + j\omega RC}$ = $G =$\frac{V{out}}{V{in}}$=$\frac{1}{\sqrt {1 + \omega RC ^2}}$.</p></li><li><p>Phase of$H(\omega)$:$\phi = - \arctan(\omega RC)$.</p></li></ul><h4 id="59baf1cf-d1ee-49aa-a8c5-c18b18928844" data-toc-id="59baf1cf-d1ee-49aa-a8c5-c18b18928844" collapsed="false" seolevelmigrated="true">Analysis of Low Pass RC Filter</h4><ul><li><p>Using a potential divider:$\frac{V{out}}{V{in}} = \frac{ZC}{ZR + Z_C} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} = \frac{1}{1 + j\omega RC}$.</p></li><li><p>Gain:$G = \frac{V{out}}{V{in}} = \frac{1}{\sqrt{1 + (\omega RC)^2}}$.</p></li><li><p>Phase:$\phi = -\arctan(\omega RC)$.</p></li><li><p>As$\omega \rightarrow 0$,$Gain \rightarrow 1$.</p></li><li><p>As$\omega \rightarrow \infty$,$Gain \rightarrow 0$.</p></li><li><p>Cut-off frequency$(\omegaC)$: Corner frequency, Half-power frequency;$\omegaC = \frac{1}{RC}$.</p></li><li><p>At half-power frequency, the power of an AC voltage is proportional to$V^2$, therefore,$\omega_C = 1/RC$.</p></li></ul><h4 id="56632a7d-9e65-474a-af5a-8f40fb9fb18c" data-toc-id="56632a7d-9e65-474a-af5a-8f40fb9fb18c" collapsed="false" seolevelmigrated="true">Decibel Scale</h4><ul><li><p>Bel (B) scale: A logarithmic scale used for expressing gains in terms of signal amplitude or power.</p></li><li><p>$P{Gain} (B) = \log{10} \frac{P{OUT}}{P{IN}}$.</p></li><li><p>Decibel (dB): One-tenth of a bel, i.e., 1dB = 0.1B.</p></li><li><p>$P{Gain} (dB) = 10 \cdot \log{10} \frac{P{OUT}}{P{IN}}$.</p></li><li><p>For signal amplitude gain, as$P = \frac{V^2}{R}$,$V{Gain} (B) = \log{10} (\frac{V{OUT}}{V{IN}})^2 = 2 \cdot \log{10} \frac{V{OUT}}{V_{IN}}$.</p></li><li><p>$V{Gain} (dB) = 10 \cdot \log{10} (\frac{V{OUT}}{V{IN}})^2 = 20 \cdot \log{10} \frac{V{OUT}}{V_{IN}}$.</p></li><li><p>A positive Gain on the decibel scale corresponds to an amplification ($V{OUT} > V{IN}$), whereas a negative Gain corresponds to an attenuation ($V{OUT} < V{IN}$).</p></li></ul><h4 id="40aec217-37b8-4375-9439-150ba2603e43" data-toc-id="40aec217-37b8-4375-9439-150ba2603e43" collapsed="false" seolevelmigrated="true">Bode Plot</h4><ul><li><p>Plots frequency response (Gain (dB) vs.$\omega$). Bode plots cover a wider frequency range and give linear approximations in the graph.</p></li><li><p>For the RC filter (1st order filter), the slope of the Gain for$\omega$above$\omega_c$: -20dB per decade (factor-of-10 increase in frequency).</p></li></ul><h4 id="3189e2db-3979-41b2-b8b0-c5ff5165042d" data-toc-id="3189e2db-3979-41b2-b8b0-c5ff5165042d" collapsed="false" seolevelmigrated="true">Low-Pass Filter Design Example</h4><ul><li><p>Using a 1KΩ resistor, design an RC low-pass filter that would attenuate the 50 Hz noise by 20 dB.</p></li><li><p>Calculate the cut-off frequency of this filter.</p></li><li><p>Solution:</p><ul><li><p>$Gain{dB} = 20 \log(Gain) = 20 \log(\frac{V{out}}{V_{in}}) = -20$dB.</p></li><li><p>$Gain = \frac{1}{\sqrt {1 + \omega RC ^2}}$.</p></li><li><p>Therefore,$20\log_{10} \frac{1}{\sqrt{1 + (\omega RC)^2}} = -20$<br>$ \frac{1}{\sqrt {1 + \omega RC ^2}} = \frac{1}{10}$</p></li></ul></li><li><p>Given f = 50 Hz,$\omega = 2\pi f = 100\pi$.</p></li><li><p>By taking the square of both sides we get:</p><ul><li><p>$C = \frac{99}{\omega^2 R^2} = \frac{99}{(100\pi)^2 \times 1000} = 31.67\mu F$</p></li></ul></li><li><p>The cut-off frequency is$\omega_C = \frac{1}{RC} = \frac{1}{1000 \times 31.67 \times 10^{-6}} = 31.57 rad/sec$</p></li><li><p>$ fc = \frac{\omegac}{2\pi} = 5.03 Hz$</p></li></ul><h4 id="a07780fe-468f-4977-a504-96bcaf40420a" data-toc-id="a07780fe-468f-4977-a504-96bcaf40420a" collapsed="false" seolevelmigrated="true">High Pass RC Filter</h4><ul><li><p>$\frac{V{out}}{V{in}} = \frac{ZR}{ZR + Z_C} = \frac{R}{R + \frac{1}{j\omega C}} = \frac{1}{1 + \frac{1}{j\omega RC}}$.</p></li><li><p>Transfer function:$H(\omega) = \frac{V{out}}{V{in}} = \frac{1}{1 + \frac{1}{j\omega RC}}$.</p></li><li><p>$ = \frac{1}{\sqrt {1 + (\frac{1}{\omega RC})^2}}$.</p></li><li><p>Phase:$\phi = arctan(\frac{1}{\omega RC})$.</p></li><li><p>As$\omega \rightarrow 0$,$Gain \rightarrow 0 $.</p></li><li><p>As$\omega \rightarrow \infty $,$Gain \rightarrow 1$.</p></li><li><p>Cut-off frequency:$\omega_c = \frac{1}{RC}$.</p></li></ul><h4 id="f988aab8-6a8c-4f2f-b44a-ab18a0e19de4" data-toc-id="f988aab8-6a8c-4f2f-b44a-ab18a0e19de4" collapsed="false" seolevelmigrated="true">High Pass Filter Design Example</h4><ul><li><p>Calculate the size of resistor necessary to give it a cut-off frequency of 3kHz.</p></li></ul><p>*$ \frac{V{out}}{V{in}} = \frac{j \omega L}{R + j \omega L}$</p><ul><li><p>Transfer function,$ H( \omega)= \frac{V{out}}{V{in}} = \frac{1}{1 + \frac{R}{ j \omega L}}$</p></li></ul><p>*$Gain = \frac{V{out}}{V{in}} = \frac{1}{\sqrt {1 + (\frac{R}{\omega L})^2}}$</p><p>*$\omega \rightarrow \infty Gain \rightarrow 1$<br>$\omega \rightarrow 0 , Gain \rightarrow 0$</p><p>Cut-off frequency$\omega_C = \frac{R}{L}$</p><p><em>solution</em><br>$ \frac{ \omega_C }{2 \pi } = \frac{R}{2 \pi L}$</p><ul><li><p>Then the natural frequency of cut-off frequency$f = \frac{ \omega_C }{2 \pi } = \frac{R}{2 \pi L} = 3000Hz$</p></li><li><p>Therefore$ R = 2 \pi f L = 5.65K \Omega$</p></li></ul><h4 id="c27dee0c-90f2-407b-a534-9494a61a843c" data-toc-id="c27dee0c-90f2-407b-a534-9494a61a843c" collapsed="false" seolevelmigrated="true">Filter Types</h4><ol><li><p>A low-pass filter permits the passage of low-frequency signals while rejecting those at higher frequencies. (Y)</p></li><li><p>A filter circuit changes the amplitude of an AC voltage but keeps its frequency and phase unchanged. (N)</p></li><li><p>The frequency response of a filter circuit describes how the output of the circuit varies with the frequency of the input signal. (Y)</p></li><li><p>If the maximum output voltage of a filter circuit, at any frequency, is$Vm$, the frequency at which the output voltage is$Vm/\sqrt{2}$is called the cut-off frequency. (Y)</p></li></ol><h4 id="b3eb66af-b3b4-4c9b-8df2-526c225bafe6" data-toc-id="b3eb66af-b3b4-4c9b-8df2-526c225bafe6" collapsed="false" seolevelmigrated="true">Additional topics on Filters</h4><ul><li><p>Bandpass filter.</p></li><li><p>Active filter.</p></li></ul><h4 id="aba3615e-519a-4488-8a33-dae0139c4c0c" data-toc-id="aba3615e-519a-4488-8a33-dae0139c4c0c" collapsed="false" seolevelmigrated="true">Revision Notes</h4><p>*Derive Transfer Function, Frequency Response and$\omega_c$$ of a filter

  • Be able to recognise low and high-pass filter configurations

  • Know and be able to define all the quantities, including stop band and pass band.

  • Be able to design a filter by choosing suitable components to meet a specification (based on transfer function).

  • Know what a Bode plot is, and what the slope of the linear approximations are that are often superimposed on the plot.

  Impedance Triangle & Inductive and Capacitive Loads

  Inductive Loads

  • Inductive load has a positive imaginary part with the form R+jX, where X>0

  • Current lags the voltage.

  • Inductive load can be made of a combination of R, L and C!

  Equivalent impedance Z=1+j5-j2=1+j3 (Ω)
  𝐕 = 10 (V)
  𝐈 = 𝐕/𝐙 = 10/1+j3 = 3.16∠(−71. 60) (A)

  The total impedance Z=𝑅 + j𝑋𝐿
  𝑽 = 𝑽𝑅 + 𝑽𝐿 = 𝐼∠0°𝑅 + 𝐼∠0°𝑋𝐿∠90°= 𝐼𝑅 + 𝑗𝐼𝑋𝐿 assume 𝑰 = 𝐼∠0°

  Capacitive Loads

  • Capacitive load has a negative imaginary part of the form R-jX, where X>0

  • Current leads the voltage.

  • They can be made of a combination of R, L and C!
    Equivalent impedance Z=2+j1-j6=2-j5
    𝐕 = 10 V

  𝑰 = 𝑽/𝒁 = 10/2 − j5 = 1.86∠(+68. 20)
  The total impedance Z=𝑅 − j𝑋𝐶