Why do electric wall sockets deliver 220V AC reversing direction 100 times every second instead of smooth steady DC, and why can an electrical transformer step up voltage to 400,000 volts across high-voltage cross-country transmission grids? AC circuit impedance and resonance power the grid.
Why This Chapter Matters
In Class 12 Physics, "Alternating Current" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
Electromagnetic induction from Chapter 6.
Trigonometric sinusoidal functions.
Ohm's law.
What You Will Learn (Core Objectives)
Define Alternating Voltage ($V = V_0\sin\omega t$) and Alternating Current ($I = I_0\sin\omega t$).
Calculate Peak, Mean, and RMS values: $I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$ and $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0$.
Analyze AC through pure Resistor, Inductor (current lags by $90^\circ, X_L = \omega L$), and Capacitor (current leads by $90^\circ, X_C = \frac{1}{\omega C}$).
Analyze Series LCR Circuit using Phasor Diagrams: Impedance ($Z = \sqrt{R^2 + (X_L - X_C)^2}$) and Resonant Frequency ($\omega_0 = \frac{1}{\sqrt{LC}}$).
Compute Average AC Power: $P = V_{\text{rms}} I_{\text{rms}}\cos\phi$ (Power Factor $\cos\phi$) and explain Wattless Current.
Explain working principle, construction, turns ratio ($V_s/V_p = N_s/N_p$), and energy losses of a Transformer.
Chapter Roadmap & Progression
11. Peak vs. RMS Values & Phasors
22. Series LCR Circuit & Resonance
33. Transformers: Stepping Voltages
Complete Concept Guide (100% Curriculum Coverage)
1. Peak vs. RMS Values & Phasors
Household $220\text{ V}$ is the RMS (Root Mean Square) value. Peak voltage is: $$\mathbf{V_0 = \sqrt{2} V_{\text{rms}} = \sqrt{2}(220) \approx 311\text{ Volts}!}$$ (AC at $220\text{ V}$ is far more dangerous than $220\text{ V}$ DC because its peak hits $311\text{ V}$!). Reactances: Inductive $X_L = \omega L$ (chokes high frequencies); Capacitive $X_C = \frac{1}{\omega C}$ (blocks DC where $\omega = 0$).
2. Series LCR Circuit & Resonance
Total opposition to current is Impedance ($Z$): $$\mathbf{Z = \sqrt{R^2 + (X_L - X_C)^2}} \quad \text{and} \quad \mathbf{\tan\phi = \frac{X_L - X_C}{R}}$$ Electrical Resonance: When $X_L = X_C \implies \omega L = \frac{1}{\omega C}$: $$\mathbf{\omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}}$$ At resonance, impedance is minimum ($Z = R$) and current is maximum! (How radio receivers tune into a specific station!).
3. Transformers: Stepping Voltages
Based on Mutual Induction: $$\mathbf{\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}}$$ High voltage transmission over thin cables minimizes $I^2R$ thermal joule heating losses across national grids!
Alternating Current - Key Conceptual & Analytical Model
Chapter Summary & 10 Key Takeaways
Takeaway 1
RMS Value: Effective thermal equivalent direct current ($I_{rms} = I_0 / \sqrt{2}$).
Takeaway 2
Inductive Lag vs Capacitive Lead: Current lags voltage by $90^\circ$ in $L$; leads by $90^\circ$ in $C$.
Power Factor: $\cos\phi = R/Z$ measuring real electrical power dissipation efficiency.
Takeaway 5
Transformer Voltage Transformation: $V_s/V_p = N_s/N_p$ stepping voltages for grid transmission.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Why is alternating current preferred over direct current for long-distance electrical power transmission?
Reveal Answer & Explanation
Answer: AC voltage can be stepped up efficiently to extremely high voltages (hundreds of kilovolts) using transformers. At high voltage, current $I$ is reduced proportionally ($P = VI$), drastically minimizing joule heating losses ($I^2R$) in long transmission cables. Transformers step up voltage, reducing current and I^2R cable losses.
2
A $100\ \Omega$ resistor is connected to a $220\text{ V}, 50\text{ Hz}$ AC supply. (a) Find the rms value of current. (b) What is the net power consumed over a full cycle?
Reveal Answer & Explanation
Answer: (a) $I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{220}{100} = 2.2\text{ Amperes}$. (b) In a pure resistor, power factor $\cos\phi = 1$. Net power $P = V_{\text{rms}} I_{\text{rms}} = 220 \times 2.2 = 484\text{ Watts}$. (a) 2.2 A, (b) 484 W.
3
What is meant by 'Wattless Current' in an AC circuit? Under what condition does it occur?
Reveal Answer & Explanation
Answer: When an AC circuit contains only a pure inductor or a pure capacitor, the phase difference between voltage and current is $\phi = \pi/2$. The average power consumed is $P = V_{\text{rms}} I_{\text{rms}} \cos(\pi/2) = 0$. The current flowing without consuming any net electrical power is called Wattless Current. Current consuming zero net power (when phase angle is 90°).
4
In a series LCR circuit, derive the condition for electrical resonance and write the expression for resonant frequency.
Reveal Answer & Explanation
Answer: At resonance, inductive reactance equals capacitive reactance: $X_L = X_C \implies \omega L = \frac{1}{\omega C} \implies \omega^2 = \frac{1}{LC} \implies \omega_0 = \frac{1}{\sqrt{LC}}$. Resonant frequency is $f_0 = \frac{1}{2\pi\sqrt{LC}}$. At this frequency, $Z = R$ is minimum and current is maximum. X_L = X_C, giving f0 = 1 / (2π √(LC)).
5
Explain the working principle and turns ratio formula of a transformer. Mention two causes of energy loss in practical transformers.
Reveal Answer & Explanation
Answer: Principle: Mutual induction between primary and secondary coils linked by soft iron core. Turns ratio: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. Energy loss causes: (1) Joule heating in copper windings ($I^2R$), (2) Eddy current losses in the iron core, (3) Hysteresis losses. Mutual induction; losses from resistance, eddy currents, and hysteresis.
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