Chapter 7: Alternating Current

Dear Class 12 Student! The electricity that powers your home, charges your phone, and runs entire cities is Alternating Current (AC). This chapter marries the concepts of Electromagnetic Induction (Chapter 6) with trigonometry. The phasor diagrams for LCR circuits, the concept of Resonance, and the working of Transformers are high-priority topics for both CBSE Boards and JEE Main. Let's master the math and physics behind AC!

1. Introduction to Alternating Current

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "A standard sine wave graph representing Alternating Current (AC). X-axis is time 't' or angle '$\omega t$'. Y-axis is Current 'I'. The wave starts at origin, reaches a peak (label as $I_0$ or Peak Value), crosses the x-axis at $T/2$, reaches a negative peak ($-I_0$), and completes one cycle at $T$. Shade the positive half-cycle green and negative half-cycle red. Pure white background #FFFFFF, clean textbook style."

An alternating current/voltage is one whose magnitude changes continuously with time and whose direction reverses periodically.

Equations & Parameters Voltage: $$V = V_0 \sin\omega t$$ Current: $$I = I_0 \sin\omega t$$

Amplitude / Peak Value ($V_0, I_0$): The maximum value reached by the AC in either direction.
Time Period ($T$): Time taken to complete one full cycle.
Frequency ($f$ or $\nu$): Number of cycles completed per second. Standard in India is $50 \text{ Hz}$.
Angular Frequency ($\omega$): $\omega = 2\pi f = \frac{2\pi}{T}$. (Unit: rad/s).

2. Average and RMS Values of AC

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "A single positive half-cycle of a sine wave (from 0 to $\pi$). Mark three horizontal dashed lines across the curve: The highest line labeled '$I_0$ (Peak = 100%)', a middle line labeled '$I_{rms} \approx 0.707 I_0$', and a slightly lower line labeled '$I_{avg} \approx 0.637 I_0$'. Clean, comparative math graphic on white background #FFFFFF."

A. Mean (Average) Value of AC

Over a complete cycle: The average value of AC over one complete cycle is exactly zero ($I_{avg} = 0$) because the positive half perfectly cancels the negative half.

Derivation: Mean Value over a Half Cycle The mean value of AC over a positive half cycle ($0$ to $T/2$) is that steady direct current which sends the same amount of charge through a circuit as is sent by the AC in the same time.

$I_{avg} = \frac{\int_0^{T/2} I \, dt}{\int_0^{T/2} dt} = \frac{\int_0^{T/2} I_0 \sin\omega t \, dt}{T/2}$
$I_{avg} = \frac{2I_0}{T} \left[ \frac{-\cos\omega t}{\omega} \right]_0^{T/2}$
Substitute $\omega = 2\pi/T$:
$I_{avg} = -\frac{2I_0}{T(2\pi/T)} [ \cos(\frac{2\pi}{T} \cdot \frac{T}{2}) - \cos(0) ]$
$I_{avg} = -\frac{I_0}{\pi} [ \cos\pi - \cos 0 ] = -\frac{I_0}{\pi} [ -1 - 1 ] = -\frac{I_0}{\pi} [-2]$
$$I_{avg} = \frac{2 I_0}{\pi} \approx 0.637 I_0$$

B. Root Mean Square (RMS) Value of AC (Virtual/Effective Value)

Concept: Defined based on the heating effect. The RMS value of AC is that steady DC current which produces the same amount of heat in a given resistance in a given time as is produced by the AC.

Derivation: RMS Value (Important for Boards) Heat produced in small time $dt$ is $dH = I^2 R dt = (I_0 \sin\omega t)^2 R dt$.
Total heat over one cycle ($0$ to $T$):
$H = \int_0^T I_0^2 R \sin^2\omega t \, dt = I_0^2 R \int_0^T \left( \frac{1 - \cos 2\omega t}{2} \right) dt$
The integral of $\cos 2\omega t$ over a full cycle is zero. Thus:
$H = \frac{I_0^2 R}{2} \int_0^T dt = \frac{I_0^2 R T}{2}$

If $I_{rms}$ is the steady current producing the same heat, then $H = I_{rms}^2 R T$.
Equating the two heats: $I_{rms}^2 R T = \frac{I_0^2 R T}{2}$
$I_{rms}^2 = \frac{I_0^2}{2} \implies$ $$I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$$ Similarly, $$V_{rms} = \frac{V_0}{\sqrt{2}}$$

Real-World Note When we say the domestic supply is $220 \text{ V}$, we are ALWAYS talking about the RMS value.
The peak voltage that actually hits your appliances is $V_0 = 220 \times \sqrt{2} \approx \mathbf{311 \text{ V}}$. This is why a $220\text{V}$ AC shock is much more dangerous than a $220\text{V}$ DC shock! All AC ammeters and voltmeters measure RMS values.

Practice Problem 1 Question: An alternating voltage is given by $V = 140 \sin(314 t)$. Find the peak voltage, RMS voltage, and the frequency of the supply.

Solution:
Compare with standard equation $V = V_0 \sin(\omega t)$.
1. Peak Voltage ($V_0$): $\mathbf{140 \text{ V}}$.
2. RMS Voltage ($V_{rms}$): $\frac{V_0}{\sqrt{2}} = \frac{140}{1.414} \approx \mathbf{99 \text{ V}}$.
3. Frequency ($f$): We know $\omega = 314 \text{ rad/s}$.
Since $\omega = 2\pi f \implies 314 = 2 \times 3.14 \times f \implies f = \frac{314}{6.28} = \mathbf{50 \text{ Hz}}$.

3. AC Voltage Applied to Pure Components

Phasor: A vector that rotates about the origin with angular speed $\omega$. The projection of a phasor on the vertical axis gives the instantaneous value of the alternating quantity.

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Three sets of diagrams side-by-side for Pure R, Pure L, and Pure C circuits. Top row: Circuit schematics (AC source connected to R, L, and C respectively). Middle row: Phasor diagrams. For R, V and I arrows overlap. For L, V arrow is $90^\circ$ ahead of I. For C, I arrow is $90^\circ$ ahead of V. Bottom row: Wave graphs. For R, V and I waves peak together. For L, V peaks before I. For C, I peaks before V. White background #FFFFFF."

A. Pure Resistor (R)

Equation: If $V = V_0 \sin\omega t$, then $I = I_0 \sin\omega t$.
Phase Relation: Voltage and Current are perfectly in phase ($\phi = 0^\circ$). They reach zero and peak values simultaneously.
Opposition: Resistance $R$.

B. Pure Inductor (L)

Equation: If $V = V_0 \sin\omega t$, then $I = I_0 \sin(\omega t - \pi/2)$.
Phase Relation: Current lags the voltage by $90^\circ$ ($\pi/2$).
Inductive Reactance ($X_L$): The opposition offered by an inductor to AC.
Derivation concept: $V = L \frac{dI}{dt} \implies X_L = \frac{V_0}{I_0} = \omega L$. $$X_L = \omega L = 2\pi f L$$
Unit: Ohm ($\Omega$). Graph of $X_L$ vs $f$ is a straight line passing through the origin (directly proportional). DC has $f=0$, so an ideal inductor offers zero resistance to DC.

C. Pure Capacitor (C)

Equation: If $V = V_0 \sin\omega t$, then $I = I_0 \sin(\omega t + \pi/2)$.
Phase Relation: Current leads the voltage by $90^\circ$ ($\pi/2$).
Capacitive Reactance ($X_C$): The opposition offered by a capacitor to AC.
Derivation concept: $I = \frac{dq}{dt} = \frac{d(CV)}{dt} \implies X_C = \frac{V_0}{I_0} = \frac{1}{\omega C}$. $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$
Unit: Ohm ($\Omega$). Graph of $X_C$ vs $f$ is a rectangular hyperbola (inversely proportional). For DC ($f=0$), $X_C = \infty$, so a capacitor completely blocks DC.

4. AC Applied to Series LCR Circuit

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Phasor Diagram for Series LCR circuit and Impedance Triangle. Left (Phasor): Current 'I' is taken on the positive x-axis. $V_R$ is parallel to I. $V_L$ points straight UP (+y-axis). $V_C$ points straight DOWN (-y-axis). Show a resultant vector $(V_L - V_C)$ pointing up, and the final net Voltage 'V' forming the diagonal of a rectangle with $V_R$. The angle between V and I is $\phi$. Right (Impedance Triangle): A right-angled triangle with base R, perpendicular $(X_L - X_C)$, and hypotenuse Z. Angle is $\phi$. White background #FFFFFF."

When an Inductor (L), Capacitor (C), and Resistor (R) are connected in series across an AC source $V = V_0 \sin\omega t$, the current $I$ is the same through all elements.

Phasor Diagram Method (Crucial Board Derivation)

1. Take Current $I$ as the reference phasor along the x-axis.
2. Voltage across R: $V_R = I R$ (in phase with I, along x-axis).
3. Voltage across L: $V_L = I X_L$ (leads I by $90^\circ$, along +y-axis).
4. Voltage across C: $V_C = I X_C$ (lags I by $90^\circ$, along -y-axis).

Assuming $V_L > V_C$ (inductive circuit), the net reactive voltage is $(V_L - V_C)$ pointing upwards. The net voltage $V$ is the vector sum of $V_R$ and $(V_L - V_C)$:

$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$ Substitute $V = IZ$, $V_R = IR$, etc.: $$IZ = \sqrt{(IR)^2 + (I X_L - I X_C)^2} = I \sqrt{R^2 + (X_L - X_C)^2}$$

Impedance ($Z$) and Phase Angle ($\phi$) Impedance ($Z$): The total opposition offered by the LCR circuit. $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
Phase Angle ($\phi$): The angle by which the net voltage leads the current. From the impedance triangle: $$\tan\phi = \frac{X_L - X_C}{R}$$

5. Resonance in Series LCR Circuit

A series LCR circuit is said to be in resonance when the current amplitude reaches its maximum value. This happens when the applied frequency naturally matches the natural frequency of the LC system.

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Resonance curve for LCR circuit. X-axis is angular frequency ($\omega$), Y-axis is Current Amplitude ($I_m$). A bell-shaped curve that peaks sharply at $\omega = \omega_r$ (Resonant frequency). Draw a horizontal line at $I_{max}/\sqrt{2}$ cutting the curve at two points, $\omega_1$ and $\omega_2$. Label the gap between them as 'Bandwidth $\Delta\omega$'. Add text '$X_L = X_C$, $Z = R_{min}$' at the peak. White background #FFFFFF."

Characteristics of Resonance 1. Condition: $X_L = X_C$ (Inductive reactance perfectly cancels capacitive reactance).
2. Impedance: Minimum possible. $Z = \sqrt{R^2 + 0} = R$. The circuit behaves like a purely resistive circuit!
3. Current: Maximum possible. $I_{max} = V/R$.
4. Phase: $\tan\phi = 0/R = 0 \implies \phi = 0^\circ$. Voltage and current are perfectly in phase.

Resonant Frequency ($f_r$ or $\omega_r$)

Since $X_L = X_C \implies \omega_r L = \frac{1}{\omega_r C}$

$$\omega_r^2 = \frac{1}{LC} \implies \omega_r = \frac{1}{\sqrt{LC}} \quad \text{and} \quad f_r = \frac{1}{2\pi\sqrt{LC}}$$

Sharpness of Resonance (Q-factor)

The Q-factor (Quality Factor) measures the sharpness of the resonance curve. A higher Q means a sharper, narrower peak (better tuning for radios!).
Definition: The ratio of voltage across L or C to the voltage across R at resonance.

$$Q = \frac{\omega_r L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

It is also defined as the ratio of resonant frequency to Bandwidth ($\Delta\omega$): $Q = \frac{\omega_r}{2\Delta\omega}$.

Practice Problem 2 Question: A series LCR circuit with $R = 20 \, \Omega$, $L = 1.5 \text{ H}$, and $C = 35 \text{ \mu F}$ is connected to a variable-frequency $200 \text{ V}$ AC supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Solution:
When supply frequency = natural frequency, the circuit is in Resonance.
At resonance, $X_L = X_C$, so Impedance $Z = R = 20 \, \Omega$.
Current at resonance: $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{20} = 10 \text{ A}$.
Average Power $P = I_{rms}^2 R = (10)^2 \times 20 = 100 \times 20 = \mathbf{2000 \text{ W}}$.

6. Power in AC Circuits

Average Power Dissipation Formula For any LCR circuit, the average power dissipated over one cycle is: $$P_{avg} = V_{rms} I_{rms} \cos\phi$$ Where $\cos\phi$ is the Power Factor.

Power Factor ($\cos\phi = R/Z$):

Pure R (or Resonance): $\phi = 0^\circ \implies \cos 0^\circ = 1$. Power dissipation is Maximum ($P = VI$).
Pure L or Pure C: $\phi = 90^\circ \implies \cos 90^\circ = 0$. Power dissipation is exactly Zero!

Wattless Current: In a purely inductive or capacitive circuit, the current flowing does not consume any real power. The component of current ($I_{rms} \sin\phi$) perpendicular to the voltage is called wattless current.

7. LC Oscillations

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Four sequential panels showing LC oscillations compared to a block-spring mechanical system. Panel 1: Capacitor fully charged (Electric field E max, $I=0$), compared to a fully stretched spring (Potential energy max, $v=0$). Panel 2: Capacitor discharging (Current I max, magnetic field B forming in inductor), compared to block passing through equilibrium (Kinetic energy max). Panels 3 and 4 continue the cycle of energy bouncing between the E-field of the capacitor and B-field of the inductor. White background #FFFFFF."

When a fully charged capacitor ($C$) is connected across an inductor ($L$), the electrical energy stored in the capacitor starts discharging through the inductor, converting into magnetic energy. Once fully discharged, the collapsing magnetic field of the inductor recharges the capacitor in the opposite direction. This continuous swapping of energy between the Electric Field and Magnetic Field produces electrical oscillations of frequency $\omega = 1/\sqrt{LC}$.

8. Transformers (High Probability for Board Long Answers)

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "A detailed schematic of a Step-Up Transformer. A rectangular laminated soft iron core. On the left side, the Primary coil ($N_p$) has a few turns of thick wire connected to an AC source. On the right side, the Secondary coil ($N_s$) has many turns of thin wire connected to a load. Show dashed magnetic flux lines ($\Phi$) looping entirely through the rectangular iron core linking both coils. Label 'Input Voltage $V_p$' and 'Output Voltage $V_s$'. White background #FFFFFF."

Principle: Based on Mutual Induction. An alternating current in the primary coil generates a continuously changing magnetic flux, which links with the secondary coil via the soft iron core, inducing an alternating EMF in the secondary.

Theory and Working

Assuming no flux leakage, the flux $\Phi$ linked with each turn is the same for both coils.
By Faraday's law:
Primary EMF: $V_p = -N_p \frac{d\Phi}{dt}$
Secondary EMF: $V_s = -N_s \frac{d\Phi}{dt}$
Dividing the two equations gives the Transformer Equation:

Transformer Equation $$\frac{V_s}{V_p} = \frac{N_s}{N_p} = k$$ Where $k$ is the Transformation Ratio.

Step-Up Transformer: $N_s > N_p$ ($k > 1$). Increases voltage, decreases current.
Step-Down Transformer: $N_s < N_p$ ($k < 1$). Decreases voltage, increases current.

For an ideal transformer (100% efficiency), Power In = Power Out $\implies V_p I_p = V_s I_s \implies \frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$. Note that voltage and current are inversely related!

Energy Losses in a Real Transformer (Must-Know)

Copper Loss: Heat lost as $I^2R$ in the primary and secondary copper windings. Solution: Use thick copper wires.
Eddy Current (Iron) Loss: Alternating flux induces eddy currents in the iron core, causing heating. Solution: Use a laminated iron core.
Hysteresis Loss: Energy lost due to the continuous reversal of magnetization of the core every AC cycle. Solution: Use Soft Iron, which has a narrow hysteresis loop.
Flux Leakage: Not all flux generated by the primary links with the secondary. Solution: Wind the primary and secondary coils directly over each other.

Practice Problem 3 Question: A power transmission line feeds input power at $2300 \text{ V}$ to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at $230 \text{ V}$?

Solution:
Given: $V_p = 2300 \text{ V}$, $N_p = 4000 \text{ turns}$, $V_s = 230 \text{ V}$.
Using the transformer equation: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$\frac{230}{2300} = \frac{N_s}{4000}$
$\frac{1}{10} = \frac{N_s}{4000}$
$N_s = \frac{4000}{10} = \mathbf{400 \text{ turns}}$.