Comprehensive Study Guide: Alternating Current Circuits

Fundamentals of Alternating Current (AC) Signals

• Household electrical appliances operate on alternating current (AC) circuits, where voltage and current periodically reverse direction.
• An AC circuit consists of a combination of circuit elements (resistors, inductors, capacitors) connected to a power source providing alternating voltage $Δv$.
• Standard Mathematical Notation in AC Circuit Analysis:
  • Lowercase symbols (such as $v$, $i$, $q$) represent instantaneous values that vary with time.
  • Capital letters (such as $V_{max}$, $I_{max}$, $R$, $L$, $C$) represent fixed values, amplitudes, or maximum peak quantities.
• Mathematical Definition of AC Voltage:
  • The output of an AC power source is sinusoidal and varies with time according to the equation:     $Δv = ΔV_{max} \sin(\omega t)$
  • $Δv$ is the instantaneous voltage at time $t$.
  • $ΔV_{max}$ is the maximum output voltage of the source, also known as the voltage amplitude.
  • $\omega$ is the angular frequency of the AC voltage source.
• Angular Frequency and Periodicity Equations:
  • The angular frequency $\omega$ is related to the frequency $f$ and period $T$ of the source by:     $\omega = 2\pi f$$\omega = \frac{2\pi}{T}$
  • $f$ is the frequency of the source in hertz ($Hz$).
  • $T$ is the period of the source in seconds ($s$).
  • The voltage is positive during one half of the cycle and negative during the other half.
• Commercial Power Standards:
  • Commercial electric power plants in the United States operate at a frequency of $f = 60.0\,Hz$.
  • This frequency corresponds to an angular frequency of:     $\omega = 2\pi (60.0\,Hz) \approx 377\,rad/s$
  • The current in any circuit driven by an AC source is an alternating current that varies sinusoidally with time at the driving frequency.

Resistors in AC Circuits

• Circuit Configuration:
  • A purely resistive AC circuit consists of an AC source connected across a resistor of resistance R$.\n- Instantaneous Voltage Across a Resistor:\n - Symbolized by \u0394v_R, the instantaneous voltage across the resistor equals:\n    \u0394v_R = \u0394V_{max} \sin(\omega t) = V_{max} \sin(\omega t)\n- Instantaneous Current in a Resistor:\n - The instantaneous current i_R in the resistor is given by Ohm's law:\n    i_R = \frac{\u0394v_R}{R} = \frac{\u0394V_{max}}{R} \sin(\omega t) = I_{max} \sin(\omega t)\n - I_{max} is the maximum current in the resistor:\n    I_{max} = \frac{\u0394V_{max}}{R}\n - The instantaneous voltage across the resistor can also be written as:\n    \u0394v_R = I_{max} R \sin(\omega t)\n- Phase Relationship in Resistors:\n - The current and voltage reach their maximum values at the exact same time.\n - The current and voltage are in phase (\phi = 0).\n - For a sinusoidal applied voltage, the current in a resistor is always in phase with the voltage across the resistor.\n - The direction of the current has no effect on the behavior of the resistor; resistors behave essentially the same in both DC and AC circuits.\n\n# Phasor Diagrams and Representations\n\n- Graphical Construction:\n - To simplify the analysis of AC circuits, a graphical constructor called a phasor diagram is used.\n- Definition of a Phasor:\n - A phasor is a vector whose length is proportional to the maximum value (amplitude) of the variable it represents (such as V_{max}$or$I_{max}).\n - The vector rotates counterclockwise at an angular speed equal to the angular frequency \omega associated with the variable.\n - The projection of the phasor tip onto the vertical axis represents the instantaneous value of the quantity it represents.\n- Alternative Graphical Representations:\n - Rectangular Coordinates:\n - Voltage or current is on the vertical axis.\n - Time is on the horizontal axis.\n - Polar Coordinates / Phase Space:\n - The radial coordinate represents the amplitude of the voltage or current.\n - The angular coordinate is the phase angle (\omega t + \phi).\n - The vertical axis coordinate of the tip of the phasor represents the instantaneous value.\n - The horizontal coordinate does not represent any physical quantity in this context.\n - Alternating currents are represented by phasors in the same manner as alternating voltages.\n\n# RMS Values and Power Dissipation in AC Circuits\n\n- Time-Average Current in AC Circuits:\n - The average current over one complete cycle is zero (I_{avg} = 0) because current flows equally in both directions.\n- Thermal Energy and Current Magnitude:\n - Resistors experience a temperature increase that depends on the magnitude of the current, regardless of current direction.\n - The rate of energy dissipation (power) is related to the square of the instantaneous current:\n    P = i^2 R\n- Root-Mean-Square (RMS) Definitions:\n - The rms current I_{rms} is the average value of importance in AC circuit power calculations:\n    I_{rms} = \frac{I_{max}}{\sqrt{2}} \approx 0.707 I_{max}\n - Alternating voltages are similarly defined in terms of rms values:\n    \u0394V_{rms} = \frac{\u0394V_{max}}{\sqrt{2}} \approx 0.707 \u0394V_{max}\n- Power Calculations:\n - The rate at which electrical energy is delivered to a resistor is given by P = i^2 R.\n - The heating effect produced by an AC current with a maximum value of I_{max} is not the same as that of a DC current of the same value because maximum current occurs only instantaneously.\n - The average power delivered to a resistor carrying an alternating current is:\n    P_{avg} = I_{rms}^2 R\n- Reasons for Using RMS Values:\n - AC ammeters and voltmeters are designed and calibrated to read rms values.\n - Equations for AC power and voltage take the exact same algebraic forms as their DC counterparts.\n\n# Inductors in AC Circuits\n\n- Circuit Configuration and Governing Loop Rule:\n - An AC circuit consisting of an AC source connected across an inductor of inductance L\n - Applying Kirchhoff's loop rule:\n    \u0394v + \u0394v_L = 0\n    \u0394v - L \frac{di}{dt} = 0\n    L \frac{di}{dt} = \u0394V_{max} \sin(\omega t)\n- Derivation of Instantaneous Inductor Current:\n - Integrating the equation derived from Kirchhoff's loop rule gives:\n    di = \frac{\u0394V_{max}}{L} \sin(\omega t)\,dt\n    i_L = \int \frac{\u0394V_{max}}{L} \sin(\omega t)\,dt = -\frac{\u0394V_{max}}{\omega L} \cos(\omega t)\n - Using trigonometric identity -\cos(\theta) = \sin\left(\theta - \frac{\pi}{2}\right), the current is:\n    i_L = \frac{\u0394V_{max}}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right) = I_{max} \sin\left(\omega t - \frac{\pi}{2}\right)\n - The instantaneous current i_L$and instantaneous voltage$\u0394v_L$are out of phase by$\frac{\pi}{2}\,rad = 90^\circ\n- Inductive Reactance (X_L):\n - The factor \omega L$has units of ohms ($\Omega) and opposes current in the same way as resistance.\n - Inductive reactance X_L is defined as:\n    X_L \equiv \omega L\n - Maximum current expressed in terms of inductive reactance:\n    I_{max} = \frac{\u0394V_{max}}{X_L}\n - RMS current expressed in terms of inductive reactance:\n    I_{rms} = \frac{\u0394V_{rms}}{X_L}\n- Frequency Dependence and Physical Mechanism:\n - As frequency f$increases,$\omega$increases, causing inductive reactance$X_L to increase.\n - This behavior aligns with Faraday's Law of Induction: a larger rate of change of current in the inductor creates a larger back electromotive force (emf), giving an increase in reactance and a decrease in current.\n- Phase Relationship in Inductors:\n - For a sinusoidal applied voltage, current in an inductor always lags behind the voltage across the inductor by 90^\circ$($\frac{\pi}{2}).\n - The current is at a maximum when the voltage across the inductor is zero (when the current is momentarily not changing, \frac{di}{dt} = 0).\n- Instantaneous Voltage Formulation across Inductor:\n - The instantaneous voltage across the inductor is:\n    \u0394v_L = -L \frac{di}{dt} = -\u0394V_{max} \sin(\omega t) = -I_{max} X_L \sin(\omega t)\n- Inductor Phasor Representation:\n - Voltage and current phasors are at 90^\circ relative to each other.\n - The current phasor I_{max}$lags behind the voltage phasor$\u0394V_L$by$90^\circ.\n\n# Capacitors in AC Circuits\n\n- Circuit Configuration and Governing Equations:\n - An AC circuit consisting of an AC source connected across a capacitor of capacitance C\n - Applying Kirchhoff's loop rule:\n    \u0394v + \u0394v_C = 0\n    \u0394v = \u0394v_C = \u0394V_{max} \sin(\omega t)\n - Instantaneous charge q on the capacitor:\n    q = C \u0394V_{max} \sin(\omega t)\n- Derivation of Instantaneous Capacitor Current:\n - Taking the time derivative of charge q:\n    i_C = \frac{dq}{dt} = \omega C \u0394V_{max} \cos(\omega t)\n - Rewriting using the identity \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right):\n    i_C = \omega C \u0394V_{max} \sin\left(\omega t + \frac{\pi}{2}\right)\n - Current is \frac{\pi}{2}\,rad = 90^\circ out of phase with the voltage across the capacitor.\n- Capacitive Reactance (X_C):\n - Maximum current occurs when \cos(\omega t) = 1:\n    I_{max} = \omega C \u0394V_{max} = \frac{\u0394V_{max}}{1/(\omega C)}\n - The impeding effect of a capacitor on current in an AC circuit is called capacitive reactance X_C:\n    X_C \equiv \frac{1}{\omega C}\n - Maximum current expressed in terms of capacitive reactance:\n    I_{max} = \frac{\u0394V_{max}}{X_C}\n- Frequency Dependence and Physical Mechanism:\n - As source frequency increases, capacitive reactance X_C$decreases, and maximum current$I_{max} increases.\n - As frequency approaches zero (\omega \rightarrow 0$),$X_C$approaches infinity ($\infty$), and current approaches zero ($I \rightarrow 0).\n - At zero frequency (DC voltage), a capacitor acts as an open circuit.\n- Phase Relationship in Capacitors:\n - Current reaches its maximum value one quarter of a cycle sooner than the voltage reaches its maximum value.\n - Current in a capacitor always leads the voltage across the capacitor by 90^\circ$($\frac{\pi}{2}).\n- Instantaneous Voltage Formulation across Capacitor:\n - Instantaneous voltage across the capacitor:\n    \u0394v_C = \u0394V_{max} \sin(\omega t) = I_{max} X_C \sin(\omega t)\n- Capacitor Phasor Representation:\n - Phasor diagram shows that for sinusoidally applied voltage, current always leads voltage across a capacitor by 90^\circ\n\n# RLC Series Circuits\n\n- Circuit Structure:\n - A series combination of a resistor R$, an inductor$L$, and a capacitor$C connected across an AC power source.\n - Voltage and current in the circuit vary sinusoidally with time.\n- Governing Current and Voltage Equations:\n - Instantaneous applied voltage:\n    \u0394v = \u0394V_{max} \sin(\omega t)\n - Instantaneous circuit current:\n    i = I_{max} \sin(\omega t - \phi)\n - \phi is the phase angle between current and applied voltage.\n - Since elements are connected in series, current at all points in the circuit has identical amplitude and phase.\n- Individual Component Phase Relationships:\n - Resistor: Instantaneous voltage \u0394v_R$is in phase with current$i\n    \u0394v_R = I_{max} R \sin(\omega t) = \u0394V_R \sin(\omega t)\n - Inductor: Instantaneous voltage \u0394v_L$leads current$i$by$90^\circ\n    \u0394v_L = I_{max} X_L \sin\left(\omega t + \frac{\pi}{2}\right) = \u0394V_L \cos(\omega t)\n - Capacitor: Instantaneous voltage \u0394v_C$lags current$i$by$90^\circ\n    \u0394v_C = I_{max} X_C \sin\left(\omega t - \frac{\pi}{2}\right) = -\u0394V_C \cos(\omega t)\n- Maximum Element Voltages:\n - Maximum voltage across resistor: \u0394V_R = I_{max} R\n - Maximum voltage across inductor: \u0394V_L = I_{max} X_L\n - Maximum voltage across capacitor: \u0394V_C = I_{max} X_C\n- Non-Algebraic Voltage Addition:\n - The sum of instantaneous voltages across individual elements equals source voltage.\n - Maximum component voltages cannot be added directly algebraically (\u0394V_{max} \neq \u0394V_R + \u0394V_L + \u0394V_C) because of phase differences.\n\n# Detailed Analysis of Series RLC Parameters and Phasors\n\n- Phasor Vector Addition Method:\n - Phasors for \u0394V_L$and$\u0394V_C lie along a single line in opposite directions, so they subtract directly.\n - Their resultant (\u0394V_L - \u0394V_C)$is perpendicular to$\u0394V_R\n - Applying Pythagorean theorem to the phasor triangle:\n    \u0394V_{max} = \sqrt{(\u0394V_R)^2 + (\u0394V_L - \u0394V_C)^2}\n    \u0394V_{max} = \sqrt{(I_{max} R)^2 + (I_{max} X_L - I_{max} X_C)^2}\n    \u0394V_{max} = I_{max} \sqrt{R^2 + (X_L - X_C)^2}\n- Impedance (Z):\n - Current in an RLC circuit is given by:\n    I_{max} = \frac{\u0394V_{max}}{Z}\n - Z is called the impedance of the circuit and plays the role of resistance in AC circuits:\n    Z \equiv \sqrt{R^2 + (X_L - X_C)^2}\n - Impedance has units of ohms (\Omega).\n- Phase Angle (\phi):\n - Trigonometric right triangle in phasor diagram yields phase angle \phi:\n    \tan(\phi) = \frac{X_L - X_C}{R}\n    \phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\n - Phase angle can be positive or negative, determining circuit behavior:\n - If X_L > X_C$:$\phi > 0, circuit is predominantly inductive, voltage leads current.\n - If X_L < X_C$:$\phi < 0, circuit is predominantly capacitive, current leads voltage.\n - If X_L = X_C$:$\phi = 0, circuit is in resonance and behaves purely resistively.\n\n# Practice Problems and Worked Examples\n\n- Example 33.3: Purely Capacitive AC Circuit\n - Problem Statement: An 8.00\,\mu F$capacitor is connected to the terminals of a$60.0\,Hz$AC source whose rms voltage is$150\,V. Find the capacitive reactance and the rms current in the circuit.\n - Solution:\n - Calculate capacitive reactance X_C:\n      X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2\pi (60.0\,Hz)(8.00 \times 10^{-6}\,F)} = 332\,\Omega\n - Calculate rms current I_{rms}:\n      I_{rms} = \frac{\u0394V_{rms}}{X_C} = \frac{150\,V}{332\,\Omega} = 0.452\,A\n\n- Example 33.4: Analyzing a Series RLC Circuit\n - Problem Statement: A series RLC circuit has R = 425\,\Omega$,$L = 1.25\,H$, and$C = 3.50\,\mu F$. It is connected to an AC source with$f = 60.0\,Hz$and$\u0394V_{max} = 150\,V\n - (A) Determine inductive reactance, capacitive reactance, and impedance of the circuit:\n - Angular frequency \omega:\n      \omega = 2\pi f = 2\pi (60.0\,Hz) = 377\,s^{-1}\n - Inductive reactance X_L:\n      X_L = \omega L = (377\,s^{-1})(1.25\,H) = 471\,\Omega\n - Capacitive reactance X_C:\n      X_C = \frac{1}{\omega C} = \frac{1}{(377\,s^{-1})(3.50 \times 10^{-6}\,F)} = 758\,\Omega\n - Impedance Z:\n      Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(425\,\Omega)^2 + (471\,\Omega - 758\,\Omega)^2} = \sqrt{(425\,\Omega)^2 + (-287\,\Omega)^2} = 513\,\Omega\n - (B) Find maximum current in the circuit:\n    I_{max} = \frac{\u0394V_{max}}{Z} = \frac{150\,V}{513\,\Omega} = 0.293\,A\n - (C) Find phase angle between current and voltage:\n    \phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right) = \tan^{-1}\left(\frac{471\,\Omega - 758\,\Omega}{425\,\Omega}\right) = -34.0^\circ\n - (D) Find maximum voltage across each element:\n    \u0394V_R = I_{max} R = (0.293\,A)(425\,\Omega) = 124\,V\n    \u0394V_L = I_{max} X_L = (0.293\,A)(471\,\Omega) = 138\,V\n    \u0394V_C = I_{max} X_C = (0.293\,A)(758\,\Omega) = 222\,V\n - (E) Replacement value of L$such that current leads applied voltage by$30.0^\circ$rather than$34.0^\circ (all other values constant):\n - Phase angle constraint: \phi = -30.0^\circ\n - Solve for X_L:\n      X_L = X_C + R \tan(\phi)\n      \omega L = \frac{1}{\omega C} + R \tan(\phi)\n      L = \frac{1}{\omega} \left(\frac{1}{\omega C} + R \tan(\phi)\right)\n - Substitute known values:\n      L = \frac{1}{377\,s^{-1}} \left(\frac{1}{(377\,s^{-1})(3.50 \times 10^{-6}\,F)} + (425\,\Omega) \tan(-30.0^\circ)\right)\n      L = \frac{1}{377\,s^{-1}} \left(758\,\Omega + (425\,\Omega)(-0.5774)\right) = 1.36\,H\n\n- End-of-Chapter Practice Problem 1:\n - Statement: An inductor is connected to a 20.0\,Hz$power supply that produces a$50.0\,V$rms voltage. What inductance is needed to keep the instantaneous current in the circuit below$80.0\,mA?\n\n- End-of-Chapter Practice Problem 2:\n - Statement: In a purely inductive AC circuit, \u0394V_{max} = 100\,V\n - (a) The maximum current is 7.50\,A$at$50.0\,Hz$. Calculate inductance$L\n - (b) What If? At what angular frequency \omega$is maximum current$2.50\,A?\n\n- End-of-Chapter Practice Problem 3:\n - Statement: An inductor has a 54\,\Omega$reactance at$60.0\,Hz$. What is the maximum current if this inductor is connected to a$50.0\,Hz$source that produces a$100\,V rms voltage?\n\n- End-of-Chapter Practice Problem 4:\n - Statement:\n - (a) For what frequencies does a 22.0\,\mu F$capacitor have a reactance below$175\,\Omega?\n - (b) What If? Over this same frequency range, what is the reactance of a 44.0\,\mu F capacitor?\n\n- End-of-Chapter Practice Problem 5:\n - Statement: What is the maximum current in a 2.20\,\mu F capacitor when connected across:\n - (a) Electrical outlet having \u0394V_{rms} = 120\,V$,$f = 60.0\,Hz\n - (b) What If? Electrical outlet having \u0394V_{rms} = 240\,V$,$f = 50.0\,Hz?\n\n- End-of-Chapter Practice Problem 6:\n - Statement: A 1.00\,mF$capacitor is connected to a standard electrical outlet ($\u0394V_{rms} = 120\,V$;$f = 60.0\,Hz$). Determine current in capacitor at$t = \frac{1}{180}\,s$, assuming at$t = 0 stored energy in capacitor is zero.\n\n- End-of-Chapter Practice Problem 7:\n - Statement: An inductor (L = 400\,mH$), capacitor ($C = 4.43\,\mu F$), and resistor ($R = 500\,\Omega$) are connected in series. A$50.0\,Hz$AC source produces peak current of$250\,mA in circuit.\n - (a) Calculate required peak voltage \u0394V_{max}\n - (b) Determine phase angle by which current leads or lags applied voltage.\n\n- End-of-Chapter Practice Problem 8:\n - Statement: For an inductive circuit, \u0394V_{max} = 80\,V$,$\omega = 65\pi\,rad/s$, and$L = 70\,mH$. Calculate current in inductor at$t = 15.5\,ms\n\n- End-of-Chapter Practice Problem 9:\n - Statement: Series AC circuit contains: R = 150\,\Omega$,$L = 250\,mH$,$C = 2.00\,\mu F$, and source with$\u0394V_{max} = 210\,V$operating at$50.0\,Hz. Calculate:\n - (a) Inductive reactance\n - (b) Capacitive reactance\n - (c) Impedance\n - (d) Maximum current\n - (e) Phase angle between current and source voltage\n\n- End-of-Chapter Practice Problem 10:\n - Statement: Sinusoidal voltage \u0394v(t) = (40.0\,V) \sin(100t)$is applied to series RLC circuit with$L = 160\,mH$,$C = 99.0\,\mu F$, and$R = 68.0\,\Omega\n - (a) What is impedance of circuit?\n - (b) What is maximum current?\n - (c) Determine numerical values for I_{max}$,$\omega$, and$\phi$$