Comprehensive Study Guide on Alternating Current Theory and Circuits

Foundations of Alternating Current and Historical Development

• Direct Current (DC) versus Alternating Current (AC):

  • Direct current sources produce voltages and currents that do not vary in direction over time.

  • Alternating voltage refers to a potential difference that oscillates with time, typically following a simple harmonic sine function.

  • Alternating current is the electric current driven by an alternating voltage through a closed circuit loop.

  • In common technical usage, the phrases ac voltage and ac current are widely adopted despite being literal redundancies (i.e., alternating current voltage and alternating current current).

  • Voltage designates the electrical potential difference established between two points.

• Technological Advantages of AC Systems:

  • AC voltages can be stepped up or stepped down easily and with high energy efficiency by using mutual-induction transformers.

  • Stepping up voltage enables the economical transmission of electric power over long distances with minimal Joule heat dissipation.

  • Modern consumer appliances, office equipment, industrial motors, and telecommunication devices depend primarily on alternating current grids.

• Pioneers of Alternating Current Technology:

  • Nicola Tesla (1856–1943):

  • Serbian-American scientist, inventor, and engineer who formulated the concept of the rotating magnetic field, which serves as the foundational operating principle for nearly all AC machinery.

  • Invented the polyphase system of AC power generation and distribution, the AC induction motor, and the high-frequency induction coil known as the Tesla coil (widely utilized in early radio, television receivers, and electronics).

  • The SI unit of magnetic field strength (the tesla, $\text{T}$) is named in his honor.

• George Westinghouse (1846–1914):

  • American industrialist and engineer who championed alternating current technology over direct current, entering into direct technological and commercial conflict with Thomas Alva Edison.

  • Founded the Westinghouse Electric Company, enlisted Nicola Tesla's patents and engineering expertise, and pioneered the commercial implementation of high-voltage transmission lines and municipal AC electric lighting grids.

AC Voltage Applied to a Pure Resistor

• Circuit Formulation and Governing Equations:

  • Consider a pure resistor of resistance $R$ connected across an alternating voltage source denoted by $\varepsilon$.

  • The sinusoidally alternating potential difference across the source terminals is expressed as:     $v = v_m \sin(\omega t)$     where $v_m$ represents the voltage amplitude (peak voltage) and $\omega$ denotes the angular frequency of oscillation.

• Applying Kirchhoff's loop rule to the circuit:     $\sum \varepsilon(t) = 0$     $v_m \sin(\omega t) - i R = 0$     $i = \frac{v_m}{R}\sin(\omega t)$

• Because the resistance $R$ is a time-independent scalar, the current is expressed as:     $i = i_m \sin(\omega t)$     where the peak current amplitude $i_m$ is governed by Ohm's law:     $i_m = \frac{v_m}{R}$

  • Phase Relationship:

• The instantaneous voltage $v$ and current $i$ are in phase; both reach zero, positive maximum, zero, and negative minimum at the identical instants in time.

• Average Current and Instantaneous Power:

  • Because current oscillates symmetrically between positive and negative half-cycles, the algebraic sum of the instantaneous current over a complete cycle period $T$ evaluates to zero:     $\langle i \rangle = \frac{1}{T}\int_0^T i\,dt = 0$

  • Although the average current is zero, electrical energy is dissipated continuously as heat through the mechanism of Joule heating, which depends on the square of current ($i^2 R$):     $p = i^2 R = i_m^2 R \sin^2(\omega t)$

• Derivation of Average Power over a Full Cycle:

  • The time average of an arbitrary time-dependent function $F(t)$ evaluated over a complete period $T = \frac{2\pi}{\omega}$ is defined by:     $\langle F(t) \rangle = \frac{1}{T}\int_0^T F(t)\,dt$

  • The average power dissipated in the resistor is given by:     $\bar{p} = \langle p \rangle = i_m^2 R \langle \sin^2(\omega t) \rangle$

  • Employing the trigonometric half-angle identity:     $\sin^2(\omega t) = \frac{1}{2}(1 - \cos(2\omega t))$     $\langle \sin^2(\omega t) \rangle = \frac{1}{2}\left(1 - \langle \cos(2\omega t) \rangle\right)$

  • Evaluating the integral of $\cos(2\omega t)$ over a period $T$:     $\langle \cos(2\omega t) \rangle = \frac{1}{T}\int_0^T \cos(2\omega t)\,dt = \frac{1}{T}\left[\frac{\sin(2\omega t)}{2\omega}\right]_0^T = \frac{1}{2\omega T}\left[\sin(2\omega T) - \sin(0)\right] = 0$

  • Therefore:     $\langle \sin^2(\omega t) \rangle = \frac{1}{2}$     $\bar{p} = \frac{1}{2}i_m^2 R$

• Root Mean Square (RMS) Values and Equivalent DC Power:

  • To format the expression for AC power into the standard form established for direct current ($P = I^2 R$), root mean square (rms) or effective current is defined as:     $I = I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \frac{i_m}{\sqrt{2}} \approx 0.707\,i_m$

• Similarly, the effective or rms voltage is defined as:     $V = V_{\text{rms}} = \frac{v_m}{\sqrt{2}} \approx 0.707\,v_m$

• In terms of rms parameters, the average power equation and Ohm's law become identical to their DC counterparts:     $P = I^2 R$     $V = I R$     $P = \frac{V^2}{R} = V I$

• RMS current represents the exact magnitude of steady direct current that produces the equivalent rate of Joule heating under identical resistance.

• Standard domestic line ratings reflect rms magnitudes; for instance, a $220\,V$ domestic AC supply exhibits a peak amplitude of:     $v_m = \sqrt{2} V = 1.414 \times 220\,V \approx 311\,V$

  • Example 7.1 (Light Bulb Calculations):

• Problem: A light bulb is rated at $100\,W$ for a $220\,V$ supply. Calculate (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb.

• Solution: Given $P = 100\,W$ and $V = 220\,V$:

  • (a) Resistance:       $R = \frac{V^2}{P} = \frac{(220\,V)^2}{100\,W} = 484\,\Omega$

  • (b) Peak voltage:       $v_m = \sqrt{2} V = \sqrt{2}(220\,V) = 311\,V$

  • (c) RMS current:       $I = \frac{P}{V} = \frac{100\,W}{220\,V} = 0.454\,A$

Phasor Representation of Alternating Current and Voltage

• Definition and Mechanics of Phasors:

  • A phasor is a directed line segment (vector) that rotates counterclockwise about the origin with a constant angular velocity $\omega$.

  • The length of the phasor represents the amplitude or peak value of the oscillating scalar ($v_m$ for voltage and $i_m$ for current).

  • The vertical projection of the phasor onto the vertical axis at any instant $t$ yields the instantaneous physical value of the oscillating quantity ($v = v_m \sin(\omega t)$ and $i = i_m \sin(\omega t)$).

• Scalar Reality versus Vector Analogy:

  • Voltage and current in alternating circuits are physical scalars, not vectors.

  • However, sinusoidally varying scalars combine mathematically in the exact same manner as spatial projections of rotating vectors.

  • The phasor construction provides a visual geometric method for computing trigonometric additions, phase angles, and amplitudes according to standard vector algebra rules.

• Phasor Configuration for a Pure Resistor:

  • Because the phase difference between voltage and current in an ideal resistor is zero, the voltage phasor $\mathbf{V}$ and current phasor $\mathbf{I}$ are collinear (parallel) at all times.

AC Voltage Applied to an Inductor

• Purely Inductive Circuit Dynamics:

  • Consider a circuit consisting of a pure inductor of self-inductance $L$ and zero internal resistance connected across an AC voltage source:     $v = v_m \sin(\omega t)$

• Applying Kirchhoff's loop rule:     $v - L\frac{di}{dt} = 0$     $\frac{di}{dt} = \frac{v}{L} = \frac{v_m}{L}\sin(\omega t)$

• The term $-L\frac{di}{dt}$ represents the self-induced Faraday back-emf opposing the change in current, adhering to Lenz's law.

• Integrating with respect to time to extract the current:     $\int \frac{di}{dt}\,dt = \frac{v_m}{L}\int \sin(\omega t)\,dt$     $i(t) = -\frac{v_m}{\omega L}\cos(\omega t) + \text{constant}$

• The integration constant possesses dimensions of current and is time-independent. Because the source oscillates symmetrically about zero, the sustained current possesses no DC offset; hence, the constant equals zero.

  • Phase Lag and Inductive Reactance:

• Applying the trigonometric relation $-\cos(\omega t) = \sin\left(\omega t - \frac{\pi}{2}\right)$:     $i = i_m \sin\left(\omega t - \frac{\pi}{2}\right)$     where the current amplitude is:     $i_m = \frac{v_m}{\omega L} = \frac{v_m}{X_L}$

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

• Inductive reactance limits the flow of AC current analogously to electrical resistance in a resistive circuit.

• The dimension of inductive reactance is $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-2}]$, and its SI unit is the ohm ($\Omega$).

• Inductive reactance is directly proportional to both self-inductance $L$ and source frequency $\nu$ ($X_L = 2\pi \nu L$).

  • In a purely inductive circuit, the current lags the applied voltage by an angle of $\frac{\pi}{2}$ radians ($90^\circ$), or one-quarter of a full oscillation period ($\frac{T}{4} = \frac{\pi}{2\omega}$).

• Power Dissipation in an Inductor:

  • The instantaneous power delivered to an ideal inductor is:     $p_L = i v = \left[i_m \sin\left(\omega t - \frac{\pi}{2}\right)\right] \left[v_m \sin(\omega t)\right] = -i_m v_m \cos(\omega t) \sin(\omega t) = -\frac{i_m v_m}{2}\sin(2\omega t)$

  • Averaging over a complete oscillation cycle:     $P_L = -\frac{i_m v_m}{2}\langle \sin(2\omega t) \rangle = 0$

  • Over a full cycle, the net average power absorbed by a pure inductor is zero because energy stored in the magnetic field during current buildup is returned completely to the source during current decay.

• Example 7.2 (Inductive Circuit Calculations):

  • Problem: A pure inductor of $25.0\,\text{mH}$ is connected to a $220\,V$, $50\,\text{Hz}$ source. Determine the inductive reactance and the rms current in the circuit.

  • Solution:

  • Inductance $L = 25.0 \times 10^{-3}\,\text{H}$, $V = 220\,V$, $\nu = 50\,\text{Hz}$.

• Inductive reactance:   $X_L = 2\pi \nu L = 2 \times 3.14 \times 50\,\text{s}^{-1} \times (25.0 \times 10^{-3}\,\text{H}) = 7.85\,\Omega$

• RMS current:   $I = \frac{V}{X_L} = \frac{220\,V}{7.85\,\Omega} \approx 28\,A$

AC Voltage Applied to a Capacitor

• Capacitive Circuit Behavior across DC and AC:

  • In a direct current circuit, a capacitor allows current to flow only transiently while accumulating charge on its plates. Once fully charged to the source potential, the opposing electric field halts current entirely ($I = 0$).

  • In an alternating current circuit, the periodic reversal of polarity causes continuous charging and discharging of the plates, sustaining an oscillating current throughout the network.

• Current Derivation and Capacitive Reactance:

  • The instantaneous voltage across a capacitor of capacitance $C$ bearing instantaneous plate charge $q$ is:     $v = \frac{q}{C}$

  • Equating to the source voltage $v = v_m \sin(\omega t)$:     $q = C v_m \sin(\omega t)$

  • Differentiating with respect to time yields the instantaneous current:     $i = \frac{dq}{dt} = \frac{d}{dt}\left[C v_m \sin(\omega t)\right] = \omega C v_m \cos(\omega t)$

  • Utilizing the identity $\cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right)$, the expression becomes:     $i = i_m \sin\left(\omega t + \frac{\pi}{2}\right)$     where the current amplitude is:     $i_m = \omega C v_m = \frac{v_m}{\left(\frac{1}{\omega C}\right)}$

  • The quantity $X_C = \frac{1}{\omega C}$ represents the capacitive reactance.

  • Capacitive reactance limits the amplitude of alternating current inversely proportional to frequency and capacitance.

  • The dimension of capacitive reactance is identical to that of resistance, $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-2}]$, measured in ohms ($\Omega$).

  • The current leads the applied voltage by a phase difference of $\frac{\pi}{2}$ radians ($90^\circ$), reaching its peak value one-quarter period ahead of the voltage.

• Power Dissipation in a Capacitor:

  • The instantaneous power supplied to the capacitor is:     $p_C = i v = \left[i_m \cos(\omega t)\right] \left[v_m \sin(\omega t)\right] = \frac{i_m v_m}{2}\sin(2\omega t)$

  • Evaluating over a full cycle:     $P_C = \frac{i_m v_m}{2}\langle \sin(2\omega t) \rangle = 0$

  • The net power consumed over a complete cycle by an ideal capacitor is zero.

• Worked Examples for Capacitive and Inductive Systems:

  • Example 7.3 (Lamp and Capacitor in Series):

  • Scenario: A lamp is connected in series with a capacitor across DC and AC sources. Predict the illumination, and examine the effect of reducing capacitance $C$.

  • Observations: Connected to DC, the capacitor charges rapidly, after which current drops to zero; the lamp does not glow. Reducing $C$ produces no change under DC. Connected to AC, the capacitor presents finite reactance $X_C = \frac{1}{\omega C}$, sustaining current so the lamp glows. Decreasing $C$ elevates $X_C$, lowering the current and causing the lamp to glow less brightly.

  • Example 7.4 (Capacitor Calculations and Frequency Variation):

  • Problem: A $15.0\,\mu\text{F}$ capacitor is connected to a $220\,V$, $50\,\text{Hz}$ source. Determine $X_C$, rms current, and peak current. State what occurs if frequency is doubled.

  • Solution:       $X_C = \frac{1}{2\pi \nu C} = \frac{1}{2 \times 3.14 \times 50\,\text{s}^{-1} \times (15.0 \times 10^{-6}\,\text{F})} = 212\,\Omega$       $I = \frac{V}{X_C} = \frac{220\,V}{212\,\Omega} = 1.04\,A$       $i_m = \sqrt{2} I = 1.414 \times 1.04\,A = 1.47\,A$       The current alternates between $+1.47\,A$ and $-1.47\,A$, leading the voltage by $\frac{\pi}{2}$.       Doubling the frequency halves the reactance ($X_C \propto \nu^{-1}$), thereby doubling both the rms and peak currents.

• Example 7.5 (Inductor with Core Insertion):

  • Scenario: A light bulb and an open-coil inductor are connected in series with an AC source through a switch. An iron rod is inserted into the interior of the coil. Determine whether the bulb's glow increases, decreases, or remains unchanged.

• Solution: Inserting an iron rod magnetizes the ferromagnetic core, substantially increasing the internal magnetic flux density. This elevates the coil's self-inductance $L$, driving up the inductive reactance $X_L = \omega L$. Consequently, a larger portion of the supply voltage appears across the inductor, leaving a smaller potential difference across the light bulb. The lamp brightness decreases.

AC Voltage Applied to a Series LCR Circuit

• Circuit Topology and Governing Differential Equation:

  • A resistor $R$, inductor $L$, and capacitor $C$ are connected in series across an alternating voltage source $v = v_m \sin(\omega t)$.

• Applying Kirchhoff's loop rule:     $L\frac{di}{dt} + i R + \frac{q}{C} = v$     where $i = \frac{dq}{dt}$.

  • Phasor-Diagram Solution:

• In a series configuration, identical current passes through every element at every instant:     $i = i_m \sin(\omega t + \phi)$     where $\phi$ represents the phase angle between total voltage and circuit current.

• Let $\mathbf{I}$ be the reference current phasor. The constituent voltage phasors are:

  • $\mathbf{V}_R$ is parallel to $\mathbf{I}$, with amplitude $v_{Rm} = i_m R$.

  • $\mathbf{V}_L$ leads $\mathbf{I}$ by $\frac{\pi}{2}$, with amplitude $v_{Lm} = i_m X_L$.

  • $\mathbf{V}_C$ lags $\mathbf{I}$ by $\frac{\pi}{2}$, with amplitude $v_{Cm} = i_m X_C$.

• The vector loop relation equates the source voltage phasor to the vector sum of components:     $\mathbf{V} = \mathbf{V}_R + \mathbf{V}_L + \mathbf{V}_C$

• Because $\mathbf{V}_C$ and $\mathbf{V}_L$ are collinear and directed oppositely, their combination forms a single phasor of magnitude $|v_{Cm} - v_{Lm}|$.

• Applying the Pythagorean theorem to the right triangle formed by $\mathbf{V}_R$, $(\mathbf{V}_C + \mathbf{V}_L)$, and hypotenuse $\mathbf{V}$:     $v_m^2 = v_{Rm}^2 + (v_{Cm} - v_{Lm})^2$     $v_m^2 = (i_m R)^2 + (i_m X_C - i_m X_L)^2 = i_m^2 \left[R^2 + (X_C - X_L)^2\right]$     $i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}$

  • Impedance and the Impedance Triangle:

• The overall opposition of an LCR circuit to alternating current is called impedance ($Z$):     $Z = \sqrt{R^2 + (X_C - X_L)^2}$     $i_m = \frac{v_m}{Z}$

• The phase angle $\phi$ satisfies:     $\tan(\phi) = \frac{v_{Cm} - v_{Lm}}{v_{Rm}} = \frac{X_C - X_L}{R}$

• Operating Regimes:

  • When $X_C > X_L$, $\tan(\phi) > 0$: the circuit is predominantly capacitive, and current leads source voltage.

  • When $X_C < X_L$, $\tan(\phi) < 0$: the circuit is predominantly inductive, and current lags source voltage.

  • When $X_C = X_L$, $\phi = 0$: the circuit is purely resistive, operating at electrical resonance.

• Steady-State versus Transient Solutions:

  • The phasor analysis method directly computes the steady-state AC response of the circuit.

  • The complete general mathematical solution includes both the transient solution (which satisfies the homogeneous differential equation with $v = 0$) and the steady-state forced response.

  • Due to circuit damping via resistance, the transient currents decay exponentially to zero over time, leaving only the steady-state sinusoidal response.

Electrical Resonance in Series LCR Circuits

• Resonance Condition:

  • Resonance occurs when a system driven by a periodic source oscillates with maximal amplitude at an excitation frequency matching the system's natural frequency.

  • In a series LCR circuit, current amplitude is expressed by:     $i_m = \frac{v_m}{\sqrt{R^2 + \left(\frac{1}{\omega C} - \omega L\right)^2}}$

  • By varying the driving frequency $\omega$, impedance reaches its minimum possible value ($Z = R$) when the inductive and capacitive reactances cancel exactly:     $X_L = X_C$     $\omega_0 L = \frac{1}{\omega_0 C}$     $\omega_0 = \frac{1}{\sqrt{LC}}$

  • At the resonant angular frequency $\omega_0$, the impedance is purely resistive, and current amplitude peaks at:     $i_m = \frac{v_m}{R}$

• Resonance Characteristics and Resistance Dependence:

  • The sharpness of the current amplitude curve versus driving frequency depends critically on $R$.

  • Lower values of resistance produce sharp, high-amplitude resonance peaks, while larger resistances flatten and broaden the response.

  • Resonance requires the simultaneous presence of both an inductor ($L$) and a capacitor ($C$); circuits containing only $RL$ or $RC$ components cannot exhibit resonance because the opposing $\pi$ phase shift required for reactance cancellation cannot occur.

• Practical Applications:

  • Radio and Television Tuning: Radio antenna receivers capture a multitude of broadcast signals across various frequencies. By adjusting a variable capacitor in the tuning circuit, the resonant frequency is tuned to match the carrier frequency of a desired station, selectively maximizing its current response while rejecting non-resonant frequencies.

  • Airport Walk-Through Metal Detectors: A detection coil possessing a high turn count is coupled to a capacitor tuned precisely to resonance. When an individual carrying metallic items steps inside, eddy currents and ferromagnetic induction alter the effective inductance $L$ of the loop coil. This shifts the circuit out of resonance, causing a measurable change in impedance and current that triggers an electronic alarm.

• Example 7.6 (RC Series Circuit and Phasor Voltage Addition):

  • Problem: A resistor of $200\,\Omega$ and a capacitor of $15.0\,\mu\text{F}$ are connected in series to a $220\,V$, $50\,\text{Hz}$ source. (a) Calculate the current in the circuit. (b) Calculate the rms voltages across the resistor and capacitor. Resolve the paradox regarding their algebraic sum.

  • Solution:

  • (a) Reactance and impedance:       $X_C = \frac{1}{2\pi \nu C} = \frac{1}{2 \times 3.14 \times 50\,\text{s}^{-1} \times (15.0 \times 10^{-6}\,\text{F})} = 212.3\,\Omega$       $Z = \sqrt{R^2 + X_C^2} = \sqrt{(200\,\Omega)^2 + (212.3\,\Omega)^2} = \sqrt{40000 + 45071} = 291.67\,\Omega$       $I = \frac{V}{Z} = \frac{220\,V}{291.67\,\Omega} = 0.755\,A$

• (b) Component voltages:   $V_R = I R = 0.755\,A \times 200\,\Omega = 151\,V$   $V_C = I X_C = 0.755\,A \times 212.3\,\Omega = 160.3\,V$

• Paradox Resolution: The direct algebraic sum of voltages evaluates to $151\,V + 160.3\,V = 311.3\,V$, which exceeds the supply voltage of $220\,V$. Because $V_R$ and $V_C$ differ in phase by $90^\circ$, they must be combined vectorially according to the Pythagorean theorem:   $V = \sqrt{V_R^2 + V_C^2} = \sqrt{(151\,V)^2 + (160.3\,V)^2} = \sqrt{22801 + 25696} = 220\,V$   The vector sum agrees with the source voltage.

Power in AC Circuits and the Power Factor

• Derivation of Average Power:

  • Let the applied voltage and resulting current be:     $v = v_m \sin(\omega t)$     $i = i_m \sin(\omega t + \phi)$

  • The instantaneous power is:     $p = v i = v_m i_m \sin(\omega t) \sin(\omega t + \phi)$

  • Applying the product-to-sum identity $\sin(A)\sin(B) = \frac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]$:     $p = \frac{v_m i_m}{2}\left[\cos(\phi) - \cos(2\omega t + \phi)\right]$

  • Taking the time average over a complete cycle, the time-dependent term $\langle \cos(2\omega t + \phi) \rangle$ vanishes:     $P = \frac{v_m i_m}{2}\cos(\phi) = \left(\frac{v_m}{\sqrt{2}}\right)\left(\frac{i_m}{\sqrt{2}}\right)\cos(\phi) = V I \cos(\phi)$

  • Since $V = I Z$, power is also expressed as:     $P = I^2 Z \cos(\phi)$

  • The factor $\cos(\phi)$ is termed the power factor, where:     $\cos(\phi) = \frac{R}{Z}$

• Analysis of Circuit Cases:

  • Purely Resistive Circuit: $\phi = 0$, $\cos(\phi) = 1$. Power dissipation is maximized:     $P = V I = I^2 R$

  • Purely Inductive or Capacitive Circuit: Current and voltage differ in phase by $\phi = \pm \frac{\pi}{2}$, yielding $\cos(\phi) = 0$. Average power dissipated over a full cycle is zero despite the presence of physical current flow. This current is termed wattless current.

  • General Series LCR Circuit: Power is dissipated strictly within the resistive element $R$:     $P = I^2 Z \left(\frac{R}{Z}\right) = I^2 R$

  • LCR Circuit at Resonance: $X_C = X_L$, implying $\phi = 0$ and $\cos(\phi) = 1$. Power dissipated peaks at:     $P = I^2 R = \frac{V^2}{R}$

• Power Factor Correction and Engineering Implications:

  • Transmission Line Losses: Power delivered by a supply line is $P = V I \cos(\phi)$. Supplying a designated power level $P$ at fixed line voltage $V$ with a low power factor ($\cos(\phi) \ll 1$) requires drawing an elevated current $I = \frac{P}{V \cos(\phi)}$. This increases Joule heating losses ($I^2 R_{\text{line}}$) along the supply lines.

  • Resolution into Components: Total current $\mathbf{I}$ can be resolved into two orthogonal components relative to voltage $\mathbf{V}$:

  • In-phase active component: $I_p = I \cos(\phi)$ (power-dissipating component).

  • Quadrature component: $I_q = I \sin(\phi)$ (wattless component).

  • Capacitive Correction: In inductive industrial loads where current lags voltage, the lagging wattless current $I_q$ is compensated by connecting a capacitor of calibrated capacitance in parallel. The capacitor draws an equal and opposite leading wattless current ($I_q'$), canceling the reactive current and driving the total circuit power factor $\cos(\phi)$ toward unity.

• Comprehensive Numerical Examples:

  • Example 7.8 (Series LCR Circuit Analysis):

  • Problem: A sinusoidal source with peak amplitude $v_m = 283\,V$ and frequency $\nu = 50\,\text{Hz}$ is applied to a series LCR circuit with $R = 3\,\Omega$, $L = 25.48\,\text{mH}$, and $C = 796\,\mu\text{F}$. Determine (a) circuit impedance, (b) phase difference, (c) power dissipated, and (d) power factor.

• Solution:   $X_L = 2\pi \nu L = 2 \times 3.14 \times 50\,\text{s}^{-1} \times (25.48 \times 10^{-3}\,\text{H}) = 8\,\Omega$       $X_C = \frac{1}{2\pi \nu C} = \frac{1}{2 \times 3.14 \times 50\,\text{s}^{-1} \times (796 \times 10^{-6}\,\text{F})} = 4\,\Omega$       $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(3\,\Omega)^2 + (8\,\Omega - 4\,\Omega)^2} = \sqrt{9 + 16} = 5\,\Omega$       $\tan(\phi) = \frac{X_C - X_L}{R} = \frac{4\,\Omega - 8\,\Omega}{3\,\Omega} = -\frac{4}{3} \approx -1.333$       $\phi = -53.1^\circ$       Because $\phi$ is negative, current lags the source voltage by $53.1^\circ$.       $I = \frac{i_m}{\sqrt{2}} = \frac{v_m}{\sqrt{2} Z} = \frac{283\,V}{1.414 \times 5\,\Omega} = 40\,A$       $P = I^2 R = (40\,A)^2 \times 3\,\Omega = 4800\,W$       $\text{Power factor} = \cos(-53.1^\circ) = 0.6$

• Example 7.9 (Resonant Condition Calculations):

  • Problem: For the parameters in Example 7.8, find (a) the resonant frequency and (b) impedance, current, and power dissipation at resonance.

  • Solution:     $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(25.48 \times 10^{-3}\,\text{H}) \times (796 \times 10^{-6}\,\text{F})}} = 222.1\,\text{rad/s}$     $\nu_0 = \frac{\omega_0}{2\pi} = \frac{222.1}{2 \times 3.14} = 35.4\,\text{Hz}$       At resonance:       $Z = R = 3\,\Omega$       $I = \frac{V}{R} = \frac{283\,V}{\sqrt{2} \times 3\,\Omega} = 66.7\,A$       $P = I^2 R = (66.7\,A)^2 \times 3\,\Omega = 13.35\,\text{kW}$       Resonant power ($13.35\,\text{kW}$) substantially exceeds off-resonance power ($4.8\,\text{kW}$).

Transformers

• Working Principle and Physical Architecture:

  • A transformer converts alternating voltage between higher and lower potentials based on the operating principle of Faraday's mutual induction.

  • It consists of two electrically insulated, high-conductivity coils wound around a shared high-permeability soft-iron laminated core.

  • Primary Coil: The input winding connected to the alternating source, consisting of $N_p$ turns.

  • Secondary Coil: The output winding connected across the external electrical load, consisting of $N_s$ turns.

  • Structural designs encompass core arrangements (windings on separate limbs) and shell arrangements (primary and secondary wound concentrically one over the other).

• Mathematical Derivation for an Ideal Transformer:

  • Let $\Phi$ represent the magnetic flux linked through each individual turn of the core at time $t$.

  • Induced emf across the secondary winding is given by Faraday's law:     $\varepsilon_s = -N_s \frac{d\Phi}{dt}$

  • The alternating flux induces an opposing self-induced back-emf across the primary winding:     $\varepsilon_p = -N_p \frac{d\Phi}{dt}$

  • Under standard ideal operational assumptions:

  1. The primary coil has negligible winding resistance, ensuring primary terminal voltage equals the induced emf: $v_p = \varepsilon_p$.

  2. Flux leakage is negligible, meaning identical magnetic flux $\Phi$ links every single turn of both primary and secondary windings.

  3. The secondary load current remains sufficiently small such that the open-circuit terminal voltage matches induced emf: $v_s = \varepsilon_s$.

  • Taking the ratio of terminal voltages:     $\frac{v_s}{v_p} = \frac{N_s}{N_p}$

  • For a system with zero internal power dissipation ($100\%$ efficiency), input power equals output power:     $i_p v_p = i_s v_s$     $\frac{i_p}{i_s} = \frac{v_s}{v_p} = \frac{N_s}{N_p}$

  • Expressed in terms of effective rms voltages and currents:     $V_s = \left(\frac{N_s}{N_p}\right) V_p$     $I_s = \left(\frac{N_p}{N_s}\right) I_p$

• Classification of Transformers:

  • Step-Up Transformer: $N_s > N_p$, resulting in $V_s > V_p$ and $I_s < I_p$. Output voltage increases while output current decreases proportionally, conserving overall energy.

  • Example: A transformer with $N_p = 100$ turns and $N_s = 200$ turns ($\frac{N_s}{N_p} = 2$) converts a $220\,V$, $10\,A$ input into a $440\,V$, $5.0\,A$ output.

  • Step-Down Transformer: $N_s < N_p$, resulting in $V_s < V_p$ and $I_s > I_p$. Voltage decreases while current capacity increases.

• Mechanisms of Energy Loss in Real Transformers:

  • Flux Leakage: A fraction of the magnetic flux generated by the primary fails to traverse the secondary winding due to air gaps and core geometries. Minimized by winding the primary and secondary coils concentrically over one another.

  • Winding Resistance (Copper Losses): Current flowing through the finite resistance of copper turns dissipates power as heat ($I^2 R$). Minimized in high-current windings by utilizing heavy-gauge, thick cross-section wire.

  • Eddy Current Losses (Iron Losses): The time-varying core flux induces circulating currents within the bulk iron core, causing resistive heating. Minimized by constructing the core from thin, electrically insulated sheets of soft iron (laminated core) oriented parallel to magnetic flux lines to disrupt circulating conductive paths.

  • Hysteresis Loss: Continual cyclical reversal of magnetic dipoles within the ferromagnetic core consumes energy dissipated as heat. Minimized by using core alloys with narrow magnetic hysteresis loops, such as high-grade silicon steel.

• Power Transmission Networks:

  • Electrical power generated at power stations is stepped up to hundreds of kilovolts (reducing line current by orders of magnitude and cutting transmission $I^2 R$ losses) prior to long-distance transport.

  • Near consumer districts, substations employ step-down transformers sequentially to step down the voltages to regional distribution levels, reaching domestic wall sockets at standard potentials ($240\,V$ or $220\,V$).

Systematic Summary of Physical Quantities

• Table of Quantities, Dimensions, and Units:

  • RMS Voltage ($V$):

  • Symbol: $V$

  • Dimensions: $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-1}]$

  • SI Unit: Volt ($\text{V}$

  • Definition: $V = \frac{v_m}{\sqrt{2}}$

  • RMS Current ($I$):

  • Symbol: $I$

  • Dimensions: $[\text{A}]$

  • SI Unit: Ampere ($\text{A}$

  • Definition: $I = \frac{i_m}{\sqrt{2}}$

  • Inductive Reactance ($X_L$):

  • Symbol: $X_L$

  • Dimensions: $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-2}]$

  • SI Unit: Ohm ($\Omega$

  • Definition: $X_L = \omega L$

  • Capacitive Reactance ($X_C$):

  • Symbol: $X_C$

  • Dimensions: $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-2}]$

  • SI Unit: Ohm ($\Omega$

  • Definition: $X_C = \frac{1}{\omega C}$

  • Impedance ($Z$):

  • Symbol: $Z$

  • Dimensions: $[\text{M}\,\text{L}^2\,\text{T}^{-3}\,\text{A}^{-2}]$

  • SI Unit: Ohm ($\Omega$

  • Definition: $Z = \sqrt{R^2 + (X_C - X_L)^2}$

  • Resonant Frequency ($\omega_0$ or $\nu_0$):

  • Symbol: $\omega_0$

  • Dimensions: $[\text{T}^{-1}]$

  • SI Unit: $\text{rad/s}$ or Hertz ($\text{Hz}$

  • Definition: $\omega_0 = \frac{1}{\sqrt{LC}}$

• Quality Factor ($Q$):

  • Symbol: $Q$

  • Dimensions: Dimensionless ($[1]$)

  • Definition: $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$

• Power Factor:

  • Symbol: $\cos(\phi)$

  • Dimensions: Dimensionless ($[1]$)

  • Definition: $\cos(\phi) = \frac{R}{Z}$

Foundational Principles and Insights

• Standard Value Designation: Quoted values of alternating voltage or current refer exclusively to their rms values unless explicitly designated as peak values. A household outlet rated at $240\,V$ possesses an amplitude of $v_m = \sqrt{2} \times 240\,V \approx 340\,V$.

• Power Ratings: The specified power rating of an AC component designates its time-averaged power dissipation over a cycle.

• Positivity of Dissipated Power: Average power consumed within an AC network is always non-negative ($P \ge 0$).

• Definition of the AC Ampere: The AC ampere cannot be defined by magnetic attraction forces between parallel conductors (as with the DC ampere) because alternating current reversals average the magnetic force to zero over a cycle. The AC ampere is defined by thermal equivalence via Joule heating: one ampere of rms AC produces heat in a given resistor at the same rate as one ampere of steady DC under identical conditions.

• Non-Linear Voltage Summation: Voltages across series elements possessing differing phase shifts must be summed vectorially using phasors. In a series RC network, total potential difference is $V = \sqrt{V_R^2 + V_C^2}$, never the simple algebraic sum $V_R + V_C$.

• Zero Loss in Pure Reactive Elements: Pure inductors and capacitors dissipate zero average power over a complete cycle ($P = 0$); resistive elements account for all circuit power losses.

• Resonance Prerequisite: Resonance cannot occur without the simultaneous presence of both inductance and capacitance to provide the necessary canceling $\pi$ phase difference.

• Conservation of Energy in Transformers: Stepping up voltage in a transformer does not violate energy conservation, as output current is scaled down by the reciprocal ratio ($I_s V_s = I_p V_p$).

• Motor-Generator Duality: Generators convert mechanical input into electrical output, whereas motors consume electrical input to produce mechanical output.

Reference Review Exercises

• Exercise 7.1: A $100\,\Omega$ resistor is connected to a $220\,V$, $50\,\text{Hz}$ AC supply.

  • (a) Calculate the rms current:     $I = \frac{V}{R} = \frac{220\,V}{100\,\Omega} = 2.20\,A$

  • (b) Calculate the net power consumed over a full cycle:     $P = V I = 220\,V \times 2.20\,A = 484\,W$

• Exercise 7.2:

  • (a) Peak voltage of an AC supply is $300\,V$. The rms voltage is:     $V = \frac{v_m}{\sqrt{2}} = \frac{300\,V}{1.414} = 212.1\,V$

  • (b) RMS current in an AC circuit is $10\,A$. The peak current is:     $i_m = \sqrt{2} I = 1.414 \times 10\,A = 14.14\,A$

• Exercise 7.3: A $44\,\text{mH}$ inductor is connected to a $220\,V$, $50\,\text{Hz}$ AC supply.

  • Inductive reactance:     $X_L = 2\pi \nu L = 2 \times 3.1416 \times 50\,\text{s}^{-1} \times (44 \times 10^{-3}\,\text{H}) = 13.82\,\Omega$

  • RMS current:     $I = \frac{V}{X_L} = \frac{220\,V}{13.82\,\Omega} = 15.92\,A$

• Exercise 7.4: A $60\,\mu\text{F}$ capacitor is connected to a $110\,V$, $60\,\text{Hz}$ AC supply.

  • Capacitive reactance:     $X_C = \frac{1}{2\pi \nu C} = \frac{1}{2 \times 3.1416 \times 60\,\text{s}^{-1} \times (60 \times 10^{-6}\,\text{F})} = 44.21\,\Omega$

  • RMS current:     $I = \frac{V}{X_C} = \frac{110\,V}{44.21\,\Omega} = 2.49\,A$

• Exercise 7.5: Net power absorbed over a complete cycle in Exercises 7.3 and 7.4 equals zero in both cases ($P = 0$), because the phase difference between current and voltage is $\frac{\pi}{2}$, producing a zero power factor ($\cos(\phi) = 0$).

• Exercise 7.6: A charged $30\,\mu\text{F}$ capacitor is connected to a $27\,\text{mH}$ inductor. The angular frequency of free oscillations is:   $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(27 \times 10^{-3}\,\text{H}) \times (30 \times 10^{-6}\,\text{F})}} = \frac{1}{\sqrt{8.1 \times 10^{-7}}} = \frac{1}{9 \times 10^{-4}} = 1.11 \times 10^3\,\text{rad/s}$

• Exercise 7.7: A series LCR circuit with $R = 20\,\Omega$, $L = 1.5\,\text{H}$, and $C = 35\,\mu\text{F}$ is connected to a variable-frequency $200\,V$ source. At resonance ($\omega = \omega_0$), impedance equals resistance ($Z = R = 20\,\Omega$), and average power transferred is:   $P = \frac{V^2}{R} = \frac{(200\,V)^2}{20\,\Omega} = 2000\,W$

• Exercise 7.8: A series LCR circuit has $L = 5.0\,\text{H}$, $C = 80\,\mu\text{F}$, and $R = 40\,\Omega$ connected to a variable-frequency $230\,V$ source.

  • (a) Resonant angular frequency:     $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0\,\text{H} \times (80 \times 10^{-6}\,\text{F})}} = \frac{1}{\sqrt{4.0 \times 10^{-4}}} = 50\,\text{rad/s}$     $\nu_0 = \frac{50}{2\pi} \approx 7.96\,\text{Hz}$

• (b) Circuit impedance and peak current amplitude at resonance:   $Z = R = 40\,\Omega$   $i_m = \frac{v_m}{R} = \frac{\sqrt{2} \times 230\,V}{40\,\Omega} = \frac{325.27\,V}{40\,\Omega} \approx 8.13\,A$   $I = \frac{230\,V}{40\,\Omega} = 5.75\,A$

• (c) Component potential drops (rms) at resonance:   $V_R = I R = 5.75\,A \times 40\,\Omega = 230\,V$   $V_L = I X_L = I (\omega_0 L) = 5.75\,A \times (50\,\text{s}^{-1} \times 5.0\,\text{H}) = 5.75\,A \times 250\,\Omega = 1437.5\,V$   $V_C = I X_C = I \left(\frac{1}{\omega_0 C}\right) = 5.75\,A \times \left(\frac{1}{50\,\text{s}^{-1} \times 80 \times 10^{-6}\,\text{F}}\right) = 5.75\,A \times 250\,\Omega = 1437.5\,V$   The potential drop across the LC series combination is:   $V_{LC} = |V_L - V_C| = 1437.5\,V - 1437.5\,V = 0\,V$   The entire source voltage drops across the resistor at resonance.