Introduction to Electrical and Electronics Engineering. Module 2, module 2.

Electromagnetic induction: Faraday's laws and Lenz's law

Syllabus: Electromagnetic Induction

• Faraday's laws, Lenz's law
• Statically induced and dynamically induced emf
• Self-inductance and mutual inductance, coefficient of coupling (numerical problems not needed)

Concept of Electromagnetic Induction

Electromagnetic Induction

Electromagnetic induction is the phenomenon in which an electromotive force (emf) is induced in a conductor or a coil whenever the magnetic flux linking with it changes. This phenomenon was discovered by Michael Faraday in 1831 and forms the operating principle of generators, transformers, and induction motors.

Magnetic Flux Linkage ($\psi$)

If a coil has $N$ turns and a magnetic flux $\phi$ (in webers, Wb) passes through each turn, the total flux linkage $\psi$ is the product of the number of turns and the flux linking each turn:

Faraday's Laws of Electromagnetic Induction

Faraday summarized the principles of electromagnetic induction into two fundamental laws:

Faraday's First Law

Whenever the magnetic flux linking a closed circuit or conductor changes, an electromotive force (emf) is induced in the conductor. If the conductor forms part of a closed circuit, an induced current flows through it for as long as the change in flux continues.

Faraday's Second Law

The magnitude of the induced emf in a coil is directly proportional to the rate of change of magnetic flux linkages with the coil.

If a coil has $N$ turns and the magnetic flux linking it changes from $\phi_1$ to $\phi_2$ in a small time interval $dt$, the magnitude of the induced emf $e$ in volts is given in differential form by:

If the flux changes uniformly from $\phi_1$ to $\phi_2$ in time $t$ seconds:

Lenz's Law

Lenz's Law

Proposed by Heinrich Lenz in 1834, Lenz's law states that the direction of an electromagnetically induced emf (and hence the induced current in a closed circuit) is always such that it opposes the very cause that produces it—namely, the change in magnetic flux.

Incorporating Lenz's law into Faraday's second law gives the complete mathematical expression for induced emf:

Here, the negative sign indicates that the induced emf sets up a current whose magnetic field opposes the change in the original magnetic flux $\phi$.

Lenz's Law and Conservation of Energy

Lenz's law is a direct consequence of the law of conservation of energy:

1. Suppose the North pole of a permanent magnet is moved toward a closed coil. The magnetic flux linking the coil increases.
2. By Lenz's law, the induced current in the coil flows in a direction (counter-clockwise when viewed from the magnet side) that makes the near face of the coil behave like a North pole.
3. The two North poles repel each other. Therefore, external mechanical work must be done against this force of repulsion to push the magnet toward the coil.
4. This mechanical work done by the external agent is converted directly into electrical energy (and eventually dissipated as heat $I^2Rt$ in the coil). If the induced current aided the motion instead of opposing it, the magnet would accelerate on its own without any energy input, violating conservation of energy.

Types of Induced EMF

An emf can be induced in a conductor by changing the flux linkage in two fundamental ways:

                   Induced EMF
                        |
        +---------------+---------------+
        |                               |
Dynamically Induced EMF        Statically Induced EMF
(Conductor moves OR            (Both conductor and magnet
 field moves physically)        are stationary; flux magnitude
                                varies with time)
                                        |
                        +---------------+---------------+
                        |                               |
               Self-Induced EMF               Mutually Induced EMF
               (In the same coil)             (In a neighboring coil)

Dynamically Induced EMF

Dynamically Induced EMF

When the magnetic field is stationary (constant in time) and the conductor moves across the magnetic flux lines—or when the conductor is stationary and the magnetic field system moves physically past it—the emf induced in the conductor is called dynamically induced emf (or motional emf). Examples include DC generators and AC synchronous generators (alternators).

Expression for Dynamically Induced EMF:

1. Consider a straight conductor of active length $l$ meters moving with a uniform velocity $v$ m/s in a uniform magnetic field of flux density $B$ tesla ($\text{Wb/m}^2$).
2. Let the conductor move in a direction making an angle $\theta$ with the direction of the magnetic field lines.
3. In a small time interval $dt$, the conductor moves a distance $dx = v \, dt$. The component of displacement perpendicular to the magnetic field is $dx_{\perp} = v \, dt \sin\theta$.
4. The area swept by the conductor perpendicular to the flux lines in time $dt$ is: $dA = l \cdot v \sin\theta \, dt$
5. Therefore, the magnetic flux cut by the conductor in time $dt$ is: $d\phi = B \cdot dA = B l v \sin\theta \, dt$
6. By Faraday's law, the magnitude of the induced emf in the single conductor is: $e = \frac{d\phi}{dt} = B l v \sin\theta \quad \text{(volts)}$

• Case 1 (Conductor moves parallel to the field, $\theta = 0^\circ$): $\sin 0^\circ = 0 \implies e = 0\text{ V}$. No flux lines are cut.
• Case 2 (Conductor moves perpendicular to the field, $\theta = 90^\circ$): $\sin 90^\circ = 1 \implies e = Blv\text{ V}$ (maximum induced emf).

Fleming's Right-Hand Rule

The direction of dynamically induced emf is found using Fleming's Right-Hand Rule: Hold the thumb, forefinger, and middle finger of the right hand mutually perpendicular to one another. If the Forefinger points in the direction of the magnetic Field (North to South), and the Thumb points in the direction of Motion of the conductor (relative to the field), then the Middle finger points in the direction of the Induced EMF (or current).

Statically Induced EMF

Statically Induced EMF

When both the conductor (coil) and the magnetic field system are completely stationary, and the magnetic flux linking the coil is changed purely by varying the magnitude of the electric current producing the flux, the induced emf is called statically induced emf (or transformer emf). An example is the electrical transformer.

Statically induced emf is further classified into two types:

1. Self-induced emf: Induced in a coil due to the change of flux produced by its own current.
2. Mutually induced emf: Induced in a coil due to the change of flux produced by a current flowing in a neighboring magnetically coupled coil.

Comparison: Dynamically vs. Statically Induced EMF

Summary

• Faraday's First Law: A changing magnetic flux linking a conductor induces an emf in it.
• Faraday's Second Law & Lenz's Law: $e = -N \frac{d\phi}{dt}$, where the negative sign shows opposition to the change in flux.
• Dynamically induced emf: $e = Blv\sin\theta$, produced by relative motion between conductor and field.
• Statically induced emf: Produced by time-varying current in a stationary coil system.

Self-inductance, mutual inductance and coefficient of coupling

Self-Inductance ($L$)

Self-Induced EMF

When a time-varying current $i$ flows through a coil of $N$ turns, it sets up a time-varying magnetic flux $\phi$ through the coil itself. By Faraday's law, this changing flux induces an emf in the very same coil that opposes the change in current. This emf is called self-induced emf.

Self-Inductance (Coefficient of Self-Induction)

Self-inductance ($L$) is the property of a coil by virtue of which it opposes any change in the magnitude or direction of electric current flowing through it by inducing an emf in itself. The SI unit of inductance is the henry (H).

Expressions for Self-Inductance

Assuming the magnetic core has constant permeability (no magnetic saturation), the flux $\phi$ is directly proportional to the current $i$:

1. In terms of flux linkage per unit current: $\psi = N\phi \propto i \implies N\phi = L i$ $L = \frac{N\phi}{i} \quad \text{(henrys, H)}$ Definition: A coil has a self-inductance of 1 henry if a current of 1 ampere flowing through it produces a flux linkage of 1 weber-turn.

2. In terms of self-induced emf: By Faraday's law, the magnitude of self-induced emf is: $e_L = N \frac{d\phi}{dt} = \frac{d}{dt}(N\phi) = \frac{d}{dt}(Li)$ Since $L$ is constant for an air core or unsaturated magnetic core: $e_L = L \frac{di}{dt} \quad \text{(volts)}$ (Taking opposition into account, $e_L = -L \frac{di}{dt}$.) Definition: A coil has a self-inductance of 1 henry if a current changing at the rate of 1 ampere per second induces an emf of 1 volt in the coil.

3. In terms of physical dimensions and magnetic reluctance: From magnetic circuit theory, flux $\phi = \frac{\text{MMF}}{\text{Reluctance}} = \frac{Ni}{\mathcal{S}}$, where reluctance $\mathcal{S} = \frac{l}{\mu_0 \mu_r A}$. Substituting $\phi$ into $L = \frac{N\phi}{i}$: $L = \frac{N}{i} \left(\frac{Ni}{\mathcal{S}}\right) = \frac{N^2}{\mathcal{S}}$ $L = \frac{\mu_0 \mu_r A N^2}{l} \quad \text{(henrys, H)}$ where $N$ is the number of turns, $A$ is the cross-sectional area of the core ($\text{m}^2$), $l$ is the mean length of the magnetic path ($\text{m}$), $\mu_0 = 4\pi \times 10^{-7}\text{ H/m}$, and $\mu_r$ is the relative permeability of the core.

Mutual Inductance ($M$)

Mutually Induced EMF

Consider two coils, Coil 1 (having $N_1$ turns) and Coil 2 (having $N_2$ turns), placed close to each other so that a portion of the magnetic flux produced by one coil links with the other:

   i1 -->  +--------+           +--------+
           | Coil 1 |  Phi_12   | Coil 2 |  --> Induced emf e_M2
           | (N1)   | ========> | (N2)   |
           +--------+           +--------+

When a time-varying current $i_1$ flows in Coil 1, it produces a total flux $\phi_1$. A part of this flux, denoted $\phi_{12}$, links with Coil 2. As $i_1$ changes, $\phi_{12}$ changes, inducing an emf $e_{M2}$ in Coil 2. This is called mutually induced emf.

Mutual Inductance (Coefficient of Mutual Induction)

Mutual inductance ($M$) between two coils is defined as the property by which a change of current in one coil induces an emf in the neighboring coil. Its SI unit is also the henry (H).

Expressions for Mutual Inductance

1. In terms of mutual flux linkage per unit current: The flux linkage in Coil 2 due to current $i_1$ in Coil 1 is $N_2 \phi_{12}$, which is proportional to $i_1$: $N_2 \phi_{12} = M i_1 \implies M = \frac{N_2 \phi_{12}}{i_1} \quad \text{(H)}$ Similarly, if current $i_2$ flows in Coil 2 and produces flux $\phi_{21}$ linking Coil 1: $M = \frac{N_1 \phi_{21}}{i_2} \quad \text{(H)}$ For a linear (unsaturated) magnetic medium, $M_{12} = M_{21} = M$.

2. In terms of mutually induced emf: By Faraday's second law, the magnitude of mutually induced emf in Coil 2 when current $i_1$ changes in Coil 1 is: $e_{M2} = N_2 \frac{d\phi_{12}}{dt} = \frac{d}{dt}(N_2 \phi_{12}) = \frac{d}{dt}(M i_1)$ $e_{M2} = M \frac{di_1}{dt} \quad \text{(volts)}$ Definition: Two coils have a mutual inductance of 1 henry if a current changing at the rate of 1 ampere per second in one coil induces an emf of 1 volt in the other coil.

3. In terms of magnetic circuit parameters: If both coils are wound on the same core of reluctance $\mathcal{S} = \frac{l}{\mu_0 \mu_r A}$ and all the flux of Coil 1 links Coil 2 ($\phi_{12} = \phi_1 = \frac{N_1 i_1}{\mathcal{S}}$): $M = \frac{N_2}{i_1}\left(\frac{N_1 i_1}{\mathcal{S}}\right) = \frac{N_1 N_2}{\mathcal{S}} = \frac{\mu_0 \mu_r A N_1 N_2}{l} \quad \text{(H)}$

Coefficient of Coupling ($k$)

Coefficient of Coupling

The coefficient of coupling ($k$) between two magnetically coupled coils is defined as the fraction of the total magnetic flux produced by one coil that actually links with the second coil. It is a dimensionless number between $0$ and $1$ ($0 \le k \le 1$).

Derivation of $M = k\sqrt{L_1 L_2}$:

1. Let current $i_1$ in Coil 1 ($N_1$ turns) produce total flux $\phi_1$. The self-inductance of Coil 1 is: $L_1 = \frac{N_1 \phi_1}{i_1} \quad \text{--- (Equation 1)}$
2. If $k$ is the coefficient of coupling, the flux linking Coil 2 is $\phi_{12} = k \phi_1$. Therefore, the mutual inductance is: $M = \frac{N_2 \phi_{12}}{i_1} = \frac{N_2 (k \phi_1)}{i_1} \quad \text{--- (Equation 2)}$
3. Similarly, let current $i_2$ in Coil 2 ($N_2$ turns) produce total flux $\phi_2$. The self-inductance of Coil 2 is: $L_2 = \frac{N_2 \phi_2}{i_2} \quad \text{--- (Equation 3)}$
4. A fraction $k$ of $\phi_2$ links with Coil 1, so $\phi_{21} = k \phi_2$. Expressing $M$ from Coil 2 to Coil 1: $M = \frac{N_1 \phi_{21}}{i_2} = \frac{N_1 (k \phi_2)}{i_2} \quad \text{--- (Equation 4)}$
5. Multiplying Equation 2 and Equation 4 together: $M \times M = \left(\frac{k N_2 \phi_1}{i_1}\right) \left(\frac{k N_1 \phi_2}{i_2}\right) = k^2 \left(\frac{N_1 \phi_1}{i_1}\right) \left(\frac{N_2 \phi_2}{i_2}\right)$
6. Substituting $L_1$ and $L_2$ from Equations 1 and 3: $M^2 = k^2 L_1 L_2$ $M = k \sqrt{L_1 L_2} \quad \Longleftrightarrow \quad k = \frac{M}{\sqrt{L_1 L_2}}$

Special Cases of Coupling:

• Tightly coupled coils ($k \approx 1$): Nearly 100% of the flux of one coil links the other (e.g., transformer windings wound on a common iron core, where $k \approx 0.98\text{ to }0.99$). For ideal coupling ($k = 1$), $M_{\max} = \sqrt{L_1 L_2}$.
• Loosely coupled coils ($k < 0.5$): Only a small fraction of flux links the second coil (e.g., air-core radio-frequency coils separated in space).
• Magnetically isolated coils ($k = 0$): Coils are placed far apart or perpendicular to each other so that $\phi_{12} = 0$ and $M = 0$.

Comparison: Self-Inductance vs. Mutual Inductance

Summary

• Self-inductance: $L = \frac{N\phi}{i} = \frac{N^2}{\mathcal{S}}$, and self-induced emf $e_L = L\frac{di}{dt}$.
• Mutual inductance: $M = \frac{N_2 \phi_{12}}{i_1}$, and mutually induced emf $e_{M2} = M\frac{di_1}{dt}$.
• Coefficient of coupling: $k = \frac{\phi_{12}}{\phi_1} = \frac{M}{\sqrt{L_1 L_2}}$, with $0 \le k \le 1$.

Alternating current fundamentals: Generation and waveform representation

Syllabus: Alternating Current Fundamentals

• Generation of alternating voltages
• Representation of sinusoidal waveforms: frequency, period, average value, RMS value and form factor - numerical problems

Alternating Quantity vs. Direct Quantity

• Direct Current (DC): An electric current or voltage whose magnitude and direction (polarity) remain constant with time (e.g., output of a storage battery).
• Alternating Current (AC): An electric current or voltage whose magnitude changes continuously with time and whose direction reverses periodically at regular intervals. The most standard and widely used alternating waveform is the sinusoidal waveform (sine wave).

Generation of Alternating Voltage

An alternating voltage is generated based on Faraday's law of electromagnetic induction either by rotating a rectangular coil with constant angular velocity inside a uniform stationary magnetic field, or by rotating a magnetic field inside a stationary set of coils (as in large commercial alternators).

      N-Pole                     S-Pole
     +------+       (b)         +------+
     |      |        o          |      |
     |      |       /           |      |
     |  B   |  --->  O  omega   |  B   |
     |      |       /           |      |
     |      |      o            |      |
     +------+      (a)          +------+
            Axis of coil rotation

Derivation of the Sinusoidal EMF Equation:

1. Consider a rectangular coil of $N$ turns and cross-sectional area $A\text{ m}^2$ placed in a uniform magnetic field of flux density $B$ tesla ($\text{Wb/m}^2$).
2. Let the coil rotate counter-clockwise at a uniform angular velocity of $\omega$ radians per second ($\text{rad/s}$).
3. At time $t = 0$, assume the plane of the coil is perpendicular to the magnetic field lines (i.e., the normal to the coil plane is parallel to $\vec{B}$, so angle $\theta = 0$). At this instant, the magnetic flux passing through the coil is maximum: $\Phi_m = B \cdot A \quad \text{(webers)}$
4. After time $t$ seconds, the coil has rotated through an angle $\theta = \omega t$ radians. The component of magnetic flux perpendicular to the plane of the coil at instant $t$ is: $\phi(t) = B A \cos\theta = \Phi_m \cos(\omega t)$
5. By Faraday's law of electromagnetic induction, the instantaneous emf $e$ induced in the coil of $N$ turns is: $e = -N \frac{d\phi}{dt} = -N \frac{d}{dt} \left(\Phi_m \cos\omega t\right)$ $e = -N \Phi_m (-\omega \sin\omega t) = \omega N \Phi_m \sin(\omega t)$
6. The induced emf reaches its maximum (peak) value $E_m$ when $\sin(\omega t) = 1$, i.e., when $\theta = \omega t = 90^\circ$ (at this position, the plane of the coil is parallel to the magnetic field lines and the coil sides cut the flux lines at maximum rate): $E_m = \omega N \Phi_m = 2\pi f N B A \quad \text{(volts)}$
7. Substituting $E_m$ gives the standard equation for instantaneous alternating emf: $e(t) = E_m \sin(\omega t) = E_m \sin\theta \quad \text{(volts)}$

Key Terminology of Sinusoidal Waveforms

A general sinusoidal voltage or current is written as:

1. Instantaneous Value ($v$ or $i$): The value of an alternating quantity at any particular instant of time $t$. It is denoted by lowercase letters ($e, v, i$).
2. Maximum Value / Peak Value / Amplitude ($V_m$ or $I_m$): The highest positive or negative value attained by an alternating quantity during one half-cycle. The peak-to-peak value is $V_{p-p} = 2V_m$.
3. Cycle: One complete set of positive and negative values of an alternating quantity before it repeats itself. One complete cycle corresponds to $360^\circ$ or $2\pi$ electrical radians.
4. Time Period ($T$): The time taken in seconds by an alternating quantity to complete one full cycle.
5. Frequency ($f$): The number of cycles completed per second. Its SI unit is hertz (Hz) (or cycles/second). $f = \frac{1}{T} \quad \text{and} \quad T = \frac{1}{f}$ (In India, the standard AC power supply frequency is $f = 50\text{ Hz}$, giving $T = \frac{1}{50} = 0.02\text{ s} = 20\text{ ms}$.)
6. Angular Frequency ($\omega$): The rate of change of angle in radians per second. Since one cycle equals $2\pi$ radians in time $T$: $\omega = \frac{2\pi}{T} = 2\pi f \quad \text{(rad/s)}$
7. Electrical Angle vs. Mechanical Angle: In an AC machine having $P$ magnetic poles, one complete mechanical revolution ($360^\circ$ mechanical) produces $P/2$ complete electrical cycles. Therefore: $\theta_{\text{electrical}} = \frac{P}{2} \theta_{\text{mechanical}} \quad \text{and} \quad f = \frac{PN}{120}$ where $N$ is the rotational speed in revolutions per minute (rpm).

Worked Example 1

Problem: A rectangular coil of $200$ turns having dimensions $30\text{ cm} \times 20\text{ cm}$ is rotated at a constant speed of $1500\text{ rpm}$ in a uniform magnetic field of flux density $0.1\text{ T}$. Calculate: (a) Frequency and time period of the generated voltage, (b) Maximum value of induced emf, (c) Equation of the instantaneous emf, and (d) Instantaneous emf at $t = 2\text{ ms}$ from the start of the cycle.

Solution:

1. Identify given data:

   • Number of turns, $N = 200$
   • Area of coil, $A = 0.30\text{ m} \times 0.20\text{ m} = 0.06\text{ m}^2$
   • Speed, $n = 1500\text{ rpm}$ (for a basic 2-pole field, $1\text{ rev} = 1\text{ cycle}$)
   • Flux density, $B = 0.1\text{ T}$

2. Calculate frequency ($f$) and time period ($T$): $f = \frac{n}{60} = \frac{1500}{60} = 25\text{ Hz}$ $T = \frac{1}{f} = \frac{1}{25} = 0.04\text{ s} = 40\text{ ms}$ Angular frequency: $\omega = 2\pi f = 2\pi \times 25 = 50\pi \approx 157.08\text{ rad/s}$

3. Calculate maximum induced emf ($E_m$): $E_m = \omega N B A = 157.08 \times 200 \times 0.1 \times 0.06 = 188.5\text{ V}$

4. Write the equation of instantaneous emf: $e(t) = E_m \sin(\omega t) = 188.5 \sin(157.08 t)\text{ V}$

5. Calculate instantaneous emf at $t = 2\text{ ms} = 0.002\text{ s}$: $\omega t = 157.08 \times 0.002 = 0.31416\text{ rad} = 18^\circ$ $e(0.002) = 188.5 \sin(18^\circ) = 188.5 \times 0.3090 = 58.25\text{ V}$

Answer: (a) $f = 25\text{ Hz}, T = 0.04\text{ s}$; (b) $E_m = 188.5\text{ V}$; (c) $e(t) = 188.5\sin(157.08t)\text{ V}$; (d) $e = 58.25\text{ V}$.

Summary

• Instantaneous sinusoidal emf: $e(t) = E_m \sin(\omega t)$, where $E_m = 2\pi f N B A$.
• Frequency $f = 1/T$ and angular frequency $\omega = 2\pi f\text{ rad/s}$.
• For a $P$-pole machine rotating at $N\text{ rpm}$, $f = \frac{PN}{120}\text{ Hz}$.

Average value, RMS value, form factor and peak factor

Average Value ($I_{\text{avg}}$ or $V_{\text{avg}}$)

Definition of Average Value

The average value of an alternating current is defined as that steady (DC) current which transfers across any circuit the same amount of electric charge in a given time interval as is transferred by the alternating current in the same time interval.

For a symmetrical alternating waveform (like a pure sine wave), the positive half-cycle is identical in area to the negative half-cycle. Hence, the average value over one complete cycle is always zero. Therefore, for symmetrical waveforms, the average value is always defined over a half-cycle (from $\theta = 0$ to $\theta = \pi$).

In general, for a waveform $i(\theta)$ over an interval of base length $B$:

Derivation of Average Value for a Sinusoidal Waveform:

1. Consider a sinusoidal current $i = I_m \sin\theta$.
2. Integrating over one positive half-cycle ($\theta = 0$ to $\theta = \pi$, so base $B = \pi$): $I_{\text{avg}} = \frac{1}{\pi} \int_{0}^{\pi} I_m \sin\theta \, d\theta = \frac{I_m}{\pi} \Big[ -\cos\theta \Big]_{0}^{\pi}$
3. Substituting the limits ($\cos\pi = -1$ and $\cos 0 = 1$): $I_{\text{avg}} = \frac{I_m}{\pi} \left[ -(-1) - (-1) \right] = \frac{2 I_m}{\pi} \approx 0.637 I_m$ Similarly, for a sinusoidal voltage:

Root Mean Square (RMS) or Effective Value ($I_{\text{rms}}$ or $V_{\text{rms}}$)

Definition of RMS (Effective) Value

The RMS (Root Mean Square) or effective value of an alternating current is defined as that steady (DC) current which, when flowing through a given resistance for a given time, produces the exact same amount of heat energy as is produced by the alternating current when flowing through the same resistance for the same time.

(Note: All standard AC voltmeters and ammeters are calibrated to read RMS values, and domestic supply ratings like "$230\text{ V}$ AC" always specify the RMS value unless stated otherwise.)

Mathematically, for a periodic waveform $i(\theta)$ of period $T$ (or $2\pi$ radians):

Derivation of RMS Value for a Sinusoidal Waveform:

1. Consider a sinusoidal current $i = I_m \sin\theta$. Because $[i(\theta)]^2 = I_m^2 \sin^2\theta$ is identical over both positive and negative half-cycles, the mean of the square over a half-cycle ($0$ to $\pi$) is the same as over a full cycle ($0$ to $2\pi$).
2. Taking the integral over $0$ to $\pi$: $I_{\text{rms}}^2 = \frac{1}{\pi} \int_{0}^{\pi} I_m^2 \sin^2\theta \, d\theta$
3. Using the trigonometric identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$: $I_{\text{rms}}^2 = \frac{I_m^2}{2\pi} \int_{0}^{\pi} (1 - \cos 2\theta) \, d\theta = \frac{I_m^2}{2\pi} \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\pi}$
4. Since $\sin(2\pi) = 0$ and $\sin(0) = 0$: $I_{\text{rms}}^2 = \frac{I_m^2}{2\pi} (\pi - 0) = \frac{I_m^2}{2}$
5. Taking the square root on both sides: $I_{\text{rms}} = \frac{I_m}{\sqrt{2}} \approx 0.707 I_m$ Similarly, for a sinusoidal voltage:

Form Factor ($K_f$) and Peak (Crest) Factor ($K_p$)

Form Factor ($K_f$)

The form factor of an alternating waveform is defined as the ratio of its RMS value to its average value:

For a pure sinusoidal waveform:

Peak Factor / Crest Factor / Amplitude Factor ($K_p$)

The peak factor (or crest factor) of an alternating waveform is defined as the ratio of its maximum (peak) value to its RMS value:

For a pure sinusoidal waveform:

Summary Table for Standard Waveforms

Worked Example 1

Problem: An alternating voltage is given by $v(t) = 325.27 \sin(314 t)\text{ V}$. Find: (a) Maximum voltage, (b) RMS voltage, (c) Average voltage, (d) Frequency, (e) Time period, and (f) Form factor.

Solution:

1. Compare $v(t) = 325.27 \sin(314 t)$ with $v(t) = V_m \sin(\omega t)$:
   • (a) Maximum voltage: $V_m = 325.27\text{ V}$
   • Angular frequency: $\omega = 314\text{ rad/s}$
2. (b) RMS voltage: $V_{\text{rms}} = \frac{V_m}{\sqrt{2}} = \frac{325.27}{1.4142} = 230\text{ V}$
3. (c) Average voltage over a half-cycle: $V_{\text{avg}} = \frac{2 V_m}{\pi} = 0.637 \times 325.27 = 207.2\text{ V}$
4. (d) Frequency: $f = \frac{\omega}{2\pi} = \frac{314}{2 \times 3.1416} = 50\text{ Hz}$
5. (e) Time period: $T = \frac{1}{f} = \frac{1}{50} = 0.02\text{ s} = 20\text{ ms}$
6. (f) Form factor: $K_f = \frac{V_{\text{rms}}}{V_{\text{avg}}} = \frac{230}{207.2} = 1.11$

Answer: $V_m = 325.27\text{ V}$, $V_{\text{rms}} = 230\text{ V}$, $V_{\text{avg}} = 207.2\text{ V}$, $f = 50\text{ Hz}$, $T = 0.02\text{ s}$, $K_f = 1.11$.

Worked Example 2

Problem: Find the average value, RMS value, and form factor of a periodic voltage waveform that increases linearly from $0\text{ V}$ to $100\text{ V}$ in $0.01\text{ s}$ and then drops instantly to $0\text{ V}$, repeating with a time period $T = 0.01\text{ s}$ (a sawtooth waveform).

Solution:

1. Write the equation of the waveform for one period ($0 \le t \le T$): With $V_m = 100\text{ V}$ and $T = 0.01\text{ s}$, the straight line through the origin is: $v(t) = \frac{V_m}{T} t$
2. Calculate the average value ($V_{\text{avg}}$): $V_{\text{avg}} = \frac{1}{T} \int_{0}^{T} \left(\frac{V_m}{T} t\right) dt = \frac{V_m}{T^2} \left[\frac{t^2}{2}\right]_{0}^{T} = \frac{V_m}{2} = \frac{100}{2} = 50\text{ V}$
3. Calculate the RMS value ($V_{\text{rms}}$): $V_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} \left(\frac{V_m}{T} t\right)^2 dt} = \sqrt{\frac{V_m^2}{T^3} \left[\frac{t^3}{3}\right]_{0}^{T}} = \frac{V_m}{\sqrt{3}} = \frac{100}{1.732} = 57.74\text{ V}$
4. Calculate the form factor ($K_f$): $K_f = \frac{V_{\text{rms}}}{V_{\text{avg}}} = \frac{57.74}{50} = 1.155$

Answer: $V_{\text{avg}} = 50\text{ V}$, $V_{\text{rms}} = 57.74\text{ V}$, $K_f = 1.155$.

Summary

• For a sine wave: $I_{\text{avg}} = \frac{2I_m}{\pi} = 0.637 I_m$ and $I_{\text{rms}} = \frac{I_m}{\sqrt{2}} = 0.707 I_m$.
• Form factor $K_f = \frac{\text{RMS}}{\text{Average}} = 1.11$ (for sine wave).
• Peak factor $K_p = \frac{\text{Peak}}{\text{RMS}} = 1.414$ (for sine wave).

Phasor representation of sinusoidal quantities

Syllabus: AC Circuits – Phasor Representation

• Phasor representation of sinusoidal quantities
• Trigonometric, Rectangular, Polar and complex forms

Concept of a Phasor

Phasor

A phasor is a directed line segment (rotating vector) rotating counter-clockwise in the complex plane with a constant angular velocity $\omega$ equal to the angular frequency of the sinusoidal quantity it represents.

• The length of the phasor represents either the maximum value ($V_m$) or, by standard engineering convention, the RMS value ($V = V_m/\sqrt{2}$) of the sinusoid.
• The angle ($\phi$) that the phasor makes with the positive real (reference) axis at $t = 0$ represents the phase angle of the sinusoid.
• The projection of a rotating phasor of length $V_m$ on the vertical (imaginary) axis at any instant $t$ gives the instantaneous value $v(t) = V_m \sin(\omega t + \phi)$.

Phase and Phase Difference

Phase

The phase of an alternating quantity at any instant is the fraction of a time period (or the angle in radians/degrees) that has elapsed since the quantity last passed through its zero reference position in the positive direction. In $v(t) = V_m \sin(\omega t + \phi_1)$, the argument $(\omega t + \phi_1)$ is the phase at time $t$, and $\phi_1$ is the initial phase angle at $t = 0$.

Phase Difference

When two alternating quantities of the same frequency $\omega$ are given by:

the difference between their phase angles, $\phi = \phi_1 - \phi_2$, is called the phase difference, which remains constant with time:

1. In Phase ($\phi_1 - \phi_2 = 0^\circ$): Both quantities pass through their zero and peak values of the same polarity at the exact same instant. Their phasors are collinear and point in the same direction.
2. Leading ($\phi_1 > \phi_2$): Voltage $v(t)$ reaches its positive zero and positive peak earlier in time by angle $\phi = \phi_1 - \phi_2$ than current $i(t)$. We say $v$ leads $i$ by angle $\phi$. In a phasor diagram (where rotation is counter-clockwise), the phasor $\mathbf{V}$ lies ahead of $\mathbf{I}$ in the counter-clockwise direction by angle $\phi$.
3. Lagging ($\phi_1 < \phi_2$): Current $i(t)$ reaches its peak behind voltage $v(t)$ by angle $\phi$. We say $i$ lags behind $v$ by angle $\phi$ (located clockwise relative to $\mathbf{V}$).

      Imaginary (+j)
          ^
          |      / V (Leads I by phi)
          |     /
          |    /  phi1
          |   /    \  phi2
   -------+--o----------------> Real (Reference Axis)
          |   \
          |    \ I (Lags V)

Mathematical Forms of a Phasor

Using the RMS value $V = V_m/\sqrt{2}$ and phase angle $\phi$, a sinusoidal quantity $v(t) = \sqrt{2} V \sin(\omega t + \phi)$ is represented as a complex phasor $\mathbf{V}$ in four equivalent mathematical forms:

1. Polar Form: Specifies the RMS magnitude $V$ and the phase angle $\phi$ (usually in degrees): $\mathbf{V} = V \angle \phi$

2. Exponential (Complex) Form: Based on Euler's identity $e^{j\phi} = \cos\phi + j\sin\phi$, where $j = \sqrt{-1}$ is the imaginary operator in electrical engineering (used instead of $i$ to avoid confusion with current): $\mathbf{V} = V e^{j\phi}$

3. Trigonometric Form: Expanding the exponential form using Euler's identity: $\mathbf{V} = V (\cos\phi + j\sin\phi)$

4. Rectangular (Cartesian) Form: Expresses the phasor in terms of its real (in-phase) component $a$ and imaginary (quadrature) component $b$ in the complex plane: $\mathbf{V} = a + jb$ where: $a = V \cos\phi \quad \text{and} \quad b = V \sin\phi$

Conversions Between Rectangular and Polar Forms

• Polar to Rectangular ($V \angle \phi \longrightarrow a + jb$): $a = V \cos\phi, \qquad b = V \sin\phi$

• Rectangular to Polar ($a + jb \longrightarrow V \angle \phi$): $V = \sqrt{a^2 + b^2}, \qquad \phi = \tan^{-1}\left(\frac{b}{a}\right)$ (Care must be taken with the signs of $a$ and $b$ to place $\phi$ in the correct quadrant: if $a < 0$ and $b > 0$, $\phi = 180^\circ - \tan^{-1}|b/a|$; if $a < 0$ and $b < 0$, $\phi = -180^\circ + \tan^{-1}|b/a|$.)

Properties of the $j$-Operator and Phasor Arithmetic

• Properties of $j$:

  • $j = \sqrt{-1} = 1\angle 90^\circ$ (multiplying a phasor by $j$ rotates it counter-clockwise by $90^\circ$).
  • $j^2 = -1 = 1\angle 180^\circ$
  • $j^3 = -j = 1\angle 270^\circ = 1\angle -90^\circ$
  • $j^4 = +1 = 1\angle 0^\circ$
  • $\frac{1}{j} = -j = 1\angle -90^\circ$ (dividing by $j$ or multiplying by $-j$ rotates a phasor clockwise by $90^\circ$).

• Addition and Subtraction (best done in Rectangular Form): If $\mathbf{V}_1 = a_1 + jb_1$ and $\mathbf{V}_2 = a_2 + jb_2$: $\mathbf{V}_1 \pm \mathbf{V}_2 = (a_1 \pm a_2) + j(b_1 \pm b_2)$

• Multiplication and Division (best done in Polar Form): If $\mathbf{V}_1 = V_1 \angle \phi_1$ and $\mathbf{V}_2 = V_2 \angle \phi_2$: $\mathbf{V}_1 \cdot \mathbf{V}_2 = (V_1 V_2) \angle (\phi_1 + \phi_2)$ $\frac{\mathbf{V}_1}{\mathbf{V}_2} = \left(\frac{V_1}{V_2}\right) \angle (\phi_1 - \phi_2)$

Worked Example 1

Problem: Express the sinusoidal voltage $v(t) = 141.42 \sin(314t - 30^\circ)\text{ V}$ as an RMS phasor in Polar, Exponential, Trigonometric, and Rectangular forms.

Solution:

1. Find the RMS magnitude and phase angle:
   • Peak voltage $V_m = 141.42\text{ V}$
   • RMS voltage $V = \frac{141.42}{\sqrt{2}} = 100\text{ V}$
   • Phase angle $\phi = -30^\circ = -\frac{\pi}{6}\text{ rad}$
2. Write the four forms:
   • Polar form: $\mathbf{V} = 100 \angle -30^\circ\text{ V}$
   • Exponential form: $\mathbf{V} = 100 e^{-j30^\circ}\text{ V} = 100 e^{-j\pi/6}\text{ V}$
   • Trigonometric form: $\mathbf{V} = 100 \left[\cos(-30^\circ) + j\sin(-30^\circ)\right] = 100 (\cos 30^\circ - j\sin 30^\circ)\text{ V}$
   • Rectangular form: $a = 100 \cos(-30^\circ) = 100 \times 0.866 = 86.6\text{ V}$ $b = 100 \sin(-30^\circ) = 100 \times (-0.5) = -50\text{ V}$ $\mathbf{V} = (86.6 - j50)\text{ V}$

Answer: Polar: $100\angle -30^\circ\text{ V}$; Exponential: $100e^{-j30^\circ}\text{ V}$; Trigonometric: $100(\cos 30^\circ - j\sin 30^\circ)\text{ V}$; Rectangular: $(86.6 - j50)\text{ V}$.

Worked Example 2

Problem: Two alternating currents entering a node are given by $i_1(t) = 10\sqrt{2}\sin(\omega t + 30^\circ)\text{ A}$ and $i_2(t) = 20\sqrt{2}\sin(\omega t - 45^\circ)\text{ A}$. Find the total current $i(t) = i_1(t) + i_2(t)$ in time-domain sinusoidal form.

Solution:

1. Convert $i_1(t)$ and $i_2(t)$ to RMS Polar and Rectangular forms:
   • $\mathbf{I}_1 = 10\angle 30^\circ = 10(\cos 30^\circ + j\sin 30^\circ) = 8.66 + j5.0\text{ A}$
   • $\mathbf{I}_2 = 20\angle -45^\circ = 20(\cos 45^\circ - j\sin 45^\circ) = 14.14 - j14.14\text{ A}$
2. Add the two phasors in rectangular form: $\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 = (8.66 + 14.14) + j(5.0 - 14.14) = 22.80 - j9.14\text{ A}$
3. Convert $\mathbf{I}$ back to polar form: $I = \sqrt{(22.80)^2 + (-9.14)^2} = \sqrt{519.84 + 83.54} = \sqrt{603.38} = 24.56\text{ A}$ $\phi = \tan^{-1}\left(\frac{-9.14}{22.80}\right) = -21.84^\circ$ $\mathbf{I} = 24.56 \angle -21.84^\circ\text{ A}$
4. Write the time-domain expression ($I_m = \sqrt{2} I$): $i(t) = 24.56\sqrt{2} \sin(\omega t - 21.84^\circ) = 34.73 \sin(\omega t - 21.84^\circ)\text{ A}$

Answer: $i(t) = 34.73 \sin(\omega t - 21.84^\circ)\text{ A}$.

Summary

• A sinusoid $v(t) = \sqrt{2}V\sin(\omega t + \phi)$ is represented as an RMS phasor $\mathbf{V} = V\angle\phi = V(\cos\phi + j\sin\phi) = a + jb$.
• Use rectangular form for addition/subtraction and polar form for multiplication/division.
• Multiplying by $j$ advances the phase by $+90^\circ$; multiplying by $-j$ (or dividing by $j$) retards the phase by $-90^\circ$.

Analysis of simple AC circuits: Purely resistive, inductive and capacitive circuits

Syllabus: Analysis of Simple AC Circuits

• Purely resistive, inductive & capacitive circuits
• Inductive and capacitive reactance, concept of impedance - numerical problems

1. AC Circuit with a Pure Resistor ($R$)

Consider an AC voltage source $v(t) = V_m \sin(\omega t)$ connected across a pure resistor of resistance $R$ ohms.

         i(t) --->
      +-----/\/\/\-----+
      |       R        |
     (~) v(t)          |
      |                |
      +----------------+

1. Current Equation: By Ohm's law, the instantaneous current is: $i(t) = \frac{v(t)}{R} = \frac{V_m \sin(\omega t)}{R} = I_m \sin(\omega t)$ where peak current $I_m = \frac{V_m}{R}$ and RMS current $I = \frac{V}{R}$.
2. Phase Relationship: Comparing $v(t) = V_m \sin(\omega t)$ and $i(t) = I_m \sin(\omega t)$, the phase angle between voltage and current in a pure resistor is zero ($\phi = 0^\circ$). Voltage and current are in phase.
3. Phasor Relation and Complex Impedance: If $\mathbf{V} = V\angle 0^\circ$, then $\mathbf{I} = I\angle 0^\circ$. $\mathbf{Z}_R = \frac{\mathbf{V}}{\mathbf{I}} = R\angle 0^\circ = R + j0 \quad (\Omega)$
4. Instantaneous and Average Power: $p(t) = v(t) \cdot i(t) = (V_m \sin\omega t)(I_m \sin\omega t) = V_m I_m \sin^2(\omega t)$ Using $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$ and $V_m I_m = 2VI$: $p(t) = V I (1 - \cos 2\omega t) = V I - V I \cos(2\omega t)$ Since the average of $\cos(2\omega t)$ over a full cycle is zero, the average power consumed over a cycle is: $P = V I = I^2 R = \frac{V^2}{R} \quad \text{(watts, W)}$

2. AC Circuit with a Pure Inductor ($L$)

Consider an AC voltage source $v(t) = V_m \sin(\omega t)$ connected across a pure (lossless, $R = 0$) inductor of inductance $L$ henrys.

1. Current Equation: By KVL and Faraday's law, the applied voltage equals the self-induced voltage drop across the inductor: $v(t) = L \frac{di}{dt} = V_m \sin(\omega t)$ Integrating both sides with respect to $t$: $i(t) = \frac{V_m}{L} \int \sin(\omega t) \, dt = -\frac{V_m}{\omega L} \cos(\omega t)$ Using $-\cos(\omega t) = \sin(\omega t - 90^\circ)$: $i(t) = I_m \sin(\omega t - 90^\circ) = I_m \sin\left(\omega t - \frac{\pi}{2}\right)$ where the peak current is $I_m = \frac{V_m}{\omega L} = \frac{V_m}{X_L}$.
2. Inductive Reactance ($X_L$): The opposition offered by an inductor to the flow of alternating current is called inductive reactance ($X_L$), measured in ohms ($\Omega$): $X_L = \omega L = 2\pi f L \quad (\Omega)$
   • $X_L$ is directly proportional to frequency $f$. For DC ($f = 0$), $X_L = 0\ \Omega$, so a pure inductor acts as a short circuit to steady DC.
3. Phase Relationship: In a pure inductor, the current lags behind the applied voltage by $90^\circ$ ($\pi/2\text{ rad}$), or equivalently, voltage leads current by $90^\circ$.
4. Phasor Relation and Complex Impedance: If $\mathbf{V} = V\angle 0^\circ$, then $\mathbf{I} = I\angle -90^\circ$. $\mathbf{Z}_L = \frac{\mathbf{V}}{\mathbf{I}} = \frac{V\angle 0^\circ}{I\angle -90^\circ} = X_L \angle 90^\circ = j X_L = j\omega L \quad (\Omega)$
5. Instantaneous and Average Power: $p(t) = v(t) \cdot i(t) = (V_m \sin\omega t)\left(-I_m \cos\omega t\right) = -\frac{V_m I_m}{2} \sin(2\omega t) = -V I \sin(2\omega t)$ The instantaneous power is a double-frequency sine wave ($2f$) that alternates equally between positive (energy absorbed from source to build magnetic field, $\frac{1}{2}Li^2$) and negative (energy returned completely to the source when magnetic field collapses). Therefore, the average power consumed over a complete cycle is zero: $P = 0\text{ W}$