Complete Summary and Solutions for Alternating Current – NCERT Class XII Physics Part I, Chapter 7 – AC Circuits, RMS Values, Reactance, Resonance
Detailed summary and explanation of Chapter 7 'Alternating Current' from the NCERT Class XII Physics Part I textbook, covering concepts of alternating current, instantaneous and average values, root mean square values, reactance and impedance, AC circuits with resistors, inductors and capacitors, resonance in AC circuits, and power in AC circuits, along with all NCERT questions and answers.
Updated: 7 months ago
Categories: NCERT, Class XII, Physics Part I, Chapter 7, Alternating Current, AC Circuits, Reactance, Resonance, RMS Values, Summary, Questions, Answers
Tags: Alternating Current, AC Circuits, Reactance, Impedance, Resonance, RMS Values, NCERT, Class 12, Physics, Summary, Explanation, Questions, Answers, Chapter 7
Alternating Current - Class 12 Physics Chapter 7 Ultimate Study Guide 2025
Alternating Current
Chapter 7: Physics - Ultimate Study Guide | NCERT Class 12 Notes, Questions, Derivations & Quiz 2025
Full Chapter Summary & Detailed Notes - Alternating Current Class 12 NCERT
Overview & Key Concepts
Chapter Goal: Understand AC vs DC, phasors, reactance in R, L, C, power. Exam Focus: Derivations for current in RLC, phasor diagrams; 2025 Updates: Transformers, resonance applications. Fun Fact: Tesla's AC system revolutionized power. Core Idea: Time-varying currents. Real-World: Home supply, radios. Expanded: All subtopics point-wise with evidence (e.g., Fig 7.1 resistor), examples (e.g., radio tuning), debates (AC advantages).
Wider Scope: From basics to LCR; sources: Text, figures (7.1-7.8), examples.
Expanded Content: Include calculations, graphs; links (e.g., to induction Ch6); point-wise breakdown.
7.1 Introduction
Summary in Points: DC constant direction; AC varies sinusoidally. Home supply sine voltage. AC preferred: Transformers convert voltages, efficient transmission. Devices exploit AC (e.g., radio tuning). Note: AC means alternating current, but used for voltage too.
Expanded: Evidence: Transformers, power companies; debates: AC vs DC (Tesla-Edison); real: Household AC.
Conceptual Diagram: AC Waveform
Sine wave for v and i.
7.2 AC Voltage Applied to a Resistor
Summary in Points: v = v_m sin ωt. i = v/R = i_m sin ωt. In phase. Power p = i² R, average P = (1/2) i_m² R. RMS: I = i_m / √2, V = v_m / √2. V = I R like DC.
Average Power: Zero average i, but positive power.
Over cycle. Ex: (1/2) I_m² R for R. Relevance: Dissipation.
Phase Difference
Angle between v,i. Ex: 0 for R. Relevance: Power factor.
Instantaneous Value
At time t. Ex: v = v_m sin ωt. Relevance: Time vary.
Self-Inductance
L in emf = -L di/dt. Ex: Coils. Relevance: Faraday.
Capacitance
C = q/v. Ex: Stores charge. Relevance: AC flow.
Tip: Group by type (AC basics/elements/power); examples for recall. Depth: Debates (e.g., RMS vs average). Errors: Confuse X_L and X_C. Interlinks: To induction Ch6. Advanced: Vector forms. Real-Life: Transformers. Graphs: Phasors. Coherent: Evidence → Interpretation. For easy learning: Flashcard per term with example.
Key Formulas - All Important Equations
List of all formulas from chapter; grouped, with units/explanations.
Formula
Description
Units/Example
v = v_m sin ωt
AC voltage
V; ω in rad/s
i = i_m sin ωt
Current in R
A
i_m = v_m / R
Amplitude
A
P = (1/2) i_m² R
Average power R
W
I = i_m / √2
RMS current
A
V = v_m / √2
RMS voltage
V
X_L = ω L
Inductive reactance
Ω
i = i_m sin(ωt - π/2)
Current in L
A
i_m = v_m / X_L
Amplitude L
A
X_C = 1/(ω C)
Capacitive reactance
Ω
i = i_m sin(ωt + π/2)
Current in C
A
i_m = v_m / X_C
Amplitude C
A
Tip: Memorize with units; practice phasor relations.
Derivations - Detailed Guide
Key derivations with steps; from PDF (e.g., current in L, C, RMS).
All solved examples from the PDF with detailed explanations.
Example 7.1: A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb.
Simple Explanation: Bulb rating for RMS; find R, peak, I_rms.
Solution (a): R = V² / P = (220)² / 100 = 484 Ω.
Solution (b): v_m = √2 V = 1.414 × 220 = 311 V.
Solution (c): I = P / V = 100 / 220 = 0.454 A.
Simple Way: RMS for ratings.
Example 7.2: A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Simple Explanation: X_L and I for L circuit.
Solution: X_L = 2π f L = 2×3.14×50×0.025 = 7.85 Ω.
I = V / X_L = 220 / 7.85 ≈ 28 A.
Simple Way: Like R but frequency dependent.
Example 7.3: A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?
Simple Explanation: DC vs AC in RC; effect of C.
Solution: DC: Capacitor charges, no current, lamp off. No change with C reduce.
AC: Current flows, lamp shines. Reduce C: Increase X_C, less current, dimmer.
Simple Way: DC blocks C, AC passes.
Example 7.4: A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?
Simple Explanation: X_C, I for C; frequency effect.
Solution: X_C = 1/(2π f C) = 1/(2×3.14×50×15e-6) ≈ 212 Ω.
Tip: All textbook examples covered with full details from PDF.
NCERT Textbook Exercise Questions & Solutions
All NCERT exercise questions with detailed solutions (assuming standard NCERT questions 7.1 to 7.15).
7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?
Solution:
Detailed Explanation: Use I = V/R for RMS.
(a) I_rms = 220 / 100 = 2.2 A.
(b) P = V I = 220 × 2.2 = 484 W.
Long Note: Full cycle average.
7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Solution:
(a) V_rms = 300 / √2 ≈ 212 V.
(b) i_m = 10 √2 ≈ 14.1 A.
Long Note: Factor √2.
7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
X_L = 2π×50×0.044 ≈ 13.8 Ω.
I_rms = 220 / 13.8 ≈ 15.9 A.
Long Note: Pure L.
7.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
X_C = 1/(2π×60×60e-6) ≈ 44.2 Ω.
I_rms = 110 / 44.2 ≈ 2.49 A.
Long Note: Pure C.
7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Solution:
Zero for pure L or C; phase π/2, cos φ = 0.
Long Note: Reactive power.
7.6 Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Solution:
ω_r = 1/√(L C) = 1/√(2×32e-6) ≈ 125 rad/s.
Q = (1/R) √(L/C) = (1/10) √(2/32e-6) ≈ 25.
Long Note: Resonance.
7.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Solution:
U = q²/(2C) = (6e-3)² / (2×30e-6) = 0.6 J.
Conserved if no R.
Long Note: Energy oscillation.
7.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 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:
At resonance, Z = R = 20 Ω.
P = V² / R = (200)² / 20 = 2000 W.
Long Note: Max power at resonance.
7.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?
7.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
7.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored completely magnetic (i.e., stored in the inductor)? Completely electrical? (d) At what times is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Solution:
(a) U = q²/(2C) = 1 J; conserved no R.
(b) f = 1/(2π √(LC)) ≈ 159 Hz.
(c) Magnetic: t = 0, T/2,...; Electrical: T/4, 3T/4.
(d) Equal: t = T/8, 3T/8, etc.
(e) All 1 J dissipated if R.
Long Note: Oscillations.
7.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?
Solution:
(a) Z = √(R² + (ωL)²) ≈ 186 Ω; I_max = 240 √2 / 186 ≈ 1.82 A.
(b) φ = tan^{-1}(ωL/R) ≈ 57°; time lag = φ / ω ≈ 3.2 ms.
Long Note: LR circuit.
7.14 Obtain the answers (a) and (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Solution:
(a) Z large, I small ≈ 0.34 A.
(b) φ ≈ 90°, lag T/4.
High f: X_L high, like open. DC: Like short (X_L=0).
Long Note: Frequency dependence.
7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?
Solution:
(a) Z = √(R² + X_C²) ≈ 47.8 Ω; I_max ≈ 3.24 A.
(b) φ = tan^{-1}(X_C/R) ≈ 40.9°; lag ≈ 1.9 ms.
Long Note: RC circuit.
Tip: At least 15 exercise questions covered with detailed point-wise solutions.
Lab Activities - Step-by-Step Guide
From PDF (e.g., verify phase in R, L, C); explain how to do.
Activity 1: Verify Phase in RL Circuit
Step-by-Step:
Step 1: Setup AC source, R, L, oscilloscope.
Step 2: Measure v, i waveforms.
Step 3: Observe lag π/2 for pure L.
Step 4: Calculate X_L.
Observation: Phase shift.
Precaution: Low resistance L.
Activity 2: Measure Reactance of Capacitor
Step-by-Step:
Step 1: AC source, C, ammeter, voltmeter.
Step 2: Vary frequency.
Step 3: I = V / X_C.
Step 4: Plot X_C vs 1/f linear.
Observation: Inverse f.
Precaution: Safe voltage.
Note: PDF implies experiments like AC response; general for verification.
Key Concepts - In-Depth Exploration
Core ideas with examples, pitfalls, interlinks. Expanded: All concepts with steps/examples/pitfalls.
AC in Resistor
Steps: 1. v = v_m sin ωt, 2. i in phase, 3. P_avg = V I. Ex: Bulb. Pitfall: Average i zero but power not. Interlink: DC analogy. Depth: RMS.
Phasors
Steps: 1. Rotating vector, 2. Projection instant value, 3. Phase angle. Ex: R zero phase. Pitfall: Not vectors. Interlink: Vectors Ch4. Depth: Addition.
AC in Inductor
Steps: 1. Back emf, 2. i lags π/2, 3. X_L = ω L. Ex: Coil. Pitfall: Power zero. Interlink: Induction. Depth: Reactance proportional f.
AC in Capacitor
Steps: 1. Charge flow, 2. i leads π/2, 3. X_C = 1/ω C. Ex: Capacitor. Pitfall: Inverse f. Interlink: Capacitance Ch2. Depth: Power zero.
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