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Alternating Current Class 12 Physics Case-Based Questions 2026–27 | Chapter 7
Alternating Current Class 12 Physics Case-Based Questions 2026–27
Chapter 7 — 10 competency-focused case studies with 40 questions and answers covering RMS values, R/L/C circuits, reactance, series LCR, resonance, AC power, power factor, wattless current, AC generator and transformer.
Class 12 PhysicsChapter 7CBSE 2026–2710 Case Studies40 QuestionsCompetency Practice
Quick answer: These Class 12 Physics Chapter 7 Alternating Current case-based questions provide 10 original, syllabus-aligned practice cases with 40 linked questions and answers. They target the high-intent areas students search for most often—RMS value, reactance, impedance, LCR resonance, power factor, wattless current, AC generator and transformer. Use them after the Chapter 7 Notes, Important Questions and Numericals.
2026–27 syllabus boundary: Chapter 7 covers alternating current, peak and RMS values, reactance and impedance, series LCR circuit (phasors only), resonance, power in AC circuits, power factor, wattless current, AC generator and transformer. This page stays inside that official boundary. Q-factor and standalone LC oscillations are not used as core Chapter 7 content here. Official CBSE Physics Curriculum 2026–27.
Why the format matters: The official CBSE Class XII Physics 2026–27 SQP has two case study-based questions of 4 marks each in Section D. The SQP also states that there is no change in the question-paper design and assessment pattern for 2026–27. These practice sets therefore use four linked sub-questions per case to build the same kind of extraction, concept-selection, calculation and interpretation skills—without claiming to reproduce an official CBSE question. Official Physics 2026–27 SQP · Official Physics 2026–27 Marking Scheme.
Unit context: Chapter 7 is part of Unit IV: Electromagnetic Induction and Alternating Currents. CBSE assigns 18 marks to Unit IV as a whole for Chapters 6 and 7 together; there is no separate fixed chapter-wise mark allocation in the curriculum. Check the official curriculum.
How to solve a case study: First identify the physical system. Next extract the given quantities and units. Then choose the smallest law/formula that connects them. Finally check the unit, phase relation, limiting behaviour and whether the result makes physical sense.
What you will practise: RMS and peak values; pure R, L and C circuits; inductive and capacitive reactance; series LCR impedance; resonance; phase and power factor; AC power and wattless current; AC generator; transformer voltage/current relations; and integrated competency reasoning. The cases are original practice material, not official CBSE/PYQ reproductions.
1. What These Case Studies Cover
Case 1 — RMS + resistor
RMS/peak conversion, current, power and phase.
RMS/peak conversion, current, power and phase.
Case 2 — Inductor
Inductive reactance, current and frequency dependence.
Inductive reactance, current and frequency dependence.
Case 3 — Capacitor
Capacitive reactance, current and phase.
Capacitive reactance, current and phase.
Case 4 — LCR
Net reactance, impedance, current and power factor.
Net reactance, impedance, current and power factor.
Case 5 — Resonance
Minimum impedance, maximum current and unity power factor.
Minimum impedance, maximum current and unity power factor.
Case 6 — Frequency sweep
Capacitive/inductive regions and current phase.
Capacitive/inductive regions and current phase.
Case 7 — AC power
Power factor and wattless current.
Power factor and wattless current.
Case 8 — Generator
Peak EMF and energy conversion.
Peak EMF and energy conversion.
Case 9 — Transformer
Turns ratio, voltage/current and DC limitation.
Turns ratio, voltage/current and DC limitation.
Case 10 — Integrated
R/L/C comparison and resonance.
R/L/C comparison and resonance.
2. Quick Answer Bank — High-Value Case Study Facts
| Question | Direct answer |
|---|---|
| What is RMS voltage? | For a sinusoidal AC voltage, Vrms = V0/√2. |
| What is inductive reactance? | XL = 2πfL; it increases with frequency. |
| What is capacitive reactance? | XC = 1/(2πfC); it decreases with frequency. |
| What is series-LCR impedance? | Z = √[R² + (XL − XC)²]. |
| When does resonance occur? | XL = XC; then Z = R and power factor = 1. |
| What is AC average power? | P = VrmsIrmscosφ. |
| What is wattless current? | Current in an ideal purely reactive circuit has zero average power over a complete cycle. |
| What is transformer voltage ratio? | Vs/Vp = Ns/Np. |
3. Case-Based Questions with Answers
Case Study 1: AC supply, RMS value and a resistive load
Read the situation carefully:
A laboratory heater is connected to an AC source rated at 220 V RMS and 50 Hz. The heating element can be treated approximately as a pure resistance of 100 Ω. A student records the current and compares the result with the peak value of the supply voltage.
A laboratory heater is connected to an AC source rated at 220 V RMS and 50 Hz. The heating element can be treated approximately as a pure resistance of 100 Ω. A student records the current and compares the result with the peak value of the supply voltage.
Q1. Find the RMS current through the heater.
Key relation: Irms = Vrms/R
Answer: Irms = Vrms/R = 220/100 = 2.2 A.
Q2. Find the peak voltage of the AC source.
Key relation: V0 = √2 Vrms
Answer: V0 = √2 Vrms = √2 × 220 ≈ 311 V.
Q3. Find the average power consumed by the heater.
Key relation: P = VrmsIrms
Answer: For a pure resistor, P = VrmsIrms = 220 × 2.2 = 484 W.
Q4. State the phase relationship between voltage and current.
Answer: For a pure resistive AC circuit, voltage and current are in phase; φ = 0 and power factor = 1.
Case Study 2: Pure inductive circuit and frequency dependence
Read the situation carefully:
An ideal inductor of 0.50 H is connected to a 100 V RMS AC supply. The frequency is first 50 Hz and is then increased to 100 Hz. The resistance of the inductor is neglected.
An ideal inductor of 0.50 H is connected to a 100 V RMS AC supply. The frequency is first 50 Hz and is then increased to 100 Hz. The resistance of the inductor is neglected.
Q5. Calculate the inductive reactance at 50 Hz.
Key relation: XL = 2πfL
Answer: XL = 2πfL = 2π × 50 × 0.50 ≈ 157.1 Ω.
Q6. Find the RMS current at 50 Hz.
Key relation: Irms = Vrms/XL
Answer: Irms = Vrms/XL = 100/157.1 ≈ 0.637 A.
Q7. What happens to XL when the frequency is doubled?
Key relation: XL ∝ f
Answer: Because XL ∝ f, doubling frequency makes the reactance double to about 314.2 Ω.
Q8. What happens qualitatively to the current when the frequency is doubled while the supply voltage remains fixed?
Answer: Since I = V/XL, doubling XL makes the RMS current half its original value.
Case Study 3: Pure capacitor and phase
Read the situation carefully:
A 20 μF ideal capacitor is connected to a 200 V RMS, 50 Hz AC source. The student notices that the capacitor allows AC current even though an ideal capacitor does not provide a steady conducting path for DC.
A 20 μF ideal capacitor is connected to a 200 V RMS, 50 Hz AC source. The student notices that the capacitor allows AC current even though an ideal capacitor does not provide a steady conducting path for DC.
Q9. Calculate the capacitive reactance.
Key relation: XC = 1/(2πfC)
Answer: XC = 1/(2πfC) = 1/[2π × 50 × 20×10−6] ≈ 159.2 Ω.
Q10. Find the RMS current.
Key relation: Irms = Vrms/XC
Answer: Irms = Vrms/XC = 200/159.2 ≈ 1.26 A.
Q11. If the frequency is doubled, what happens to XC?
Key relation: XC ∝ 1/f
Answer: Since XC ∝ 1/f, doubling frequency makes XC half, about 79.6 Ω.
Q12. State the phase relationship between current and voltage.
Answer: In a pure capacitive circuit, current leads voltage by 90°.
Case Study 4: Series LCR circuit and impedance
Read the situation carefully:
A series LCR circuit has resistance R = 30 Ω, inductive reactance XL = 50 Ω and capacitive reactance XC = 10 Ω. It is connected to a 200 V RMS source.
A series LCR circuit has resistance R = 30 Ω, inductive reactance XL = 50 Ω and capacitive reactance XC = 10 Ω. It is connected to a 200 V RMS source.
Q13. Find the net reactance.
Key relation: X = XL − XC
Answer: X = XL − XC = 50 − 10 = 40 Ω inductive.
Q14. Calculate the impedance.
Key relation: Z = √[R² + (XL − XC)²]
Answer: Z = √[R² + (XL − XC)²] = √(30² + 40²) = 50 Ω.
Q15. Find the RMS current.
Key relation: Irms = Vrms/Z
Answer: Irms = 200/50 = 4 A.
Q16. Find the power factor and identify whether current leads or lags.
Key relation: cosφ = R/Z
Answer: cosφ = R/Z = 30/50 = 0.60. The net reactance is inductive, so current lags voltage.
Case Study 5: Resonance in a series LCR circuit
Read the situation carefully:
A series LCR circuit is operated at a frequency at which XL = XC. Its resistance is 20 Ω and the applied voltage is 200 V RMS. The student is asked what is special about this operating point.
A series LCR circuit is operated at a frequency at which XL = XC. Its resistance is 20 Ω and the applied voltage is 200 V RMS. The student is asked what is special about this operating point.
Q17. Find the impedance at resonance.
Key relation: At resonance: XL = XC, Z = R
Answer: At resonance, XL − XC = 0, so Z = R = 20 Ω.
Q18. Find the RMS current.
Key relation: Irms = Vrms/R
Answer: Irms = Vrms/R = 200/20 = 10 A.
Q19. Find the power factor.
Key relation: cosφ = R/Z
Answer: At resonance φ = 0, so cosφ = 1.
Q20. What happens to the individual inductor and capacitor voltages at resonance?
Answer: They can each be substantial and can be equal in magnitude, but their phasor contributions are opposite, so their net reactive contribution cancels.
Case Study 6: Changing frequency around resonance
Read the situation carefully:
A series LCR circuit has a fixed R, L and C. At its resonant frequency f0, XL = XC. The frequency is then changed first below and then above f0.
A series LCR circuit has a fixed R, L and C. At its resonant frequency f0, XL = XC. The frequency is then changed first below and then above f0.
Q21. What is the nature of the circuit below resonance?
Answer: Below resonance, XC > XL, so the circuit is capacitive.
Q22. What is the nature of the circuit above resonance?
Answer: Above resonance, XL > XC, so the circuit is inductive.
Q23. How does current phase change below resonance?
Answer: In the capacitive region, current leads the supply voltage.
Q24. How does current phase change above resonance?
Answer: In the inductive region, current lags the supply voltage.
Case Study 7: AC power and wattless current
Read the situation carefully:
An AC load draws 5 A RMS from a 240 V RMS source. Its power factor is 0.8. A second ideal purely reactive load draws the same RMS current from the same source.
An AC load draws 5 A RMS from a 240 V RMS source. Its power factor is 0.8. A second ideal purely reactive load draws the same RMS current from the same source.
Q25. Calculate the average power of the first load.
Key relation: P = VrmsIrmscosφ
Answer: P = VI cosφ = 240 × 5 × 0.8 = 960 W.
Q26. Find the phase-angle magnitude of the first load.
Answer: cosφ = 0.8, so |φ| = cos−1(0.8) ≈ 36.9°.
Q27. What is the average power of the ideal purely reactive load?
Answer: For a purely reactive circuit, cosφ = 0, so average power over a complete cycle is zero.
Q28. Why can current exist when average power is zero?
Answer: Energy is alternately stored and returned by the reactive element; the net energy transfer over a complete cycle is zero. This is described as wattless current in the ideal case.
Case Study 8: AC generator
Read the situation carefully:
An AC generator has a coil of N = 100 turns and area A = 0.02 m². It rotates at 50 revolutions per second in a uniform magnetic field of 0.5 T. The coil rotates about an axis perpendicular to the field.
An AC generator has a coil of N = 100 turns and area A = 0.02 m². It rotates at 50 revolutions per second in a uniform magnetic field of 0.5 T. The coil rotates about an axis perpendicular to the field.
Q29. Find the angular speed of rotation.
Key relation: ω = 2πf
Answer: ω = 2πf = 2π × 50 = 100π rad s−1.
Q30. Calculate the peak induced EMF.
Key relation: E0 = NBAω
Answer: E0 = NBAω = 100 × 0.5 × 0.02 × 100π ≈ 314 V.
Q31. If the magnetic field is doubled while all other quantities remain unchanged, what happens to peak EMF?
Answer: Because E0 ∝ B, the peak EMF becomes twice its original value.
Q32. What energy conversion takes place in an AC generator?
Answer: It converts mechanical energy into electrical energy through electromagnetic induction.
Case Study 9: Ideal transformer and power transmission
Read the situation carefully:
An ideal transformer has 1000 primary turns and 5000 secondary turns. The primary is supplied with 220 V AC. The transformer feeds a load on the secondary side. Ignore losses.
An ideal transformer has 1000 primary turns and 5000 secondary turns. The primary is supplied with 220 V AC. The transformer feeds a load on the secondary side. Ignore losses.
Q33. Find the secondary voltage.
Key relation: Vs/Vp = Ns/Np
Answer: Vs/Vp = Ns/Np = 5000/1000 = 5, so Vs = 1100 V.
Q34. If the secondary current is 2 A, find the primary current.
Key relation: VpIp = VsIs
Answer: For an ideal transformer, VpIp = VsIs. Thus Ip = (1100×2)/220 = 10 A.
Q35. Is this a step-up or step-down transformer?
Answer: Since Ns > Np and Vs > Vp, it is a step-up transformer.
Q36. Why is a transformer not normally operated from steady DC?
Answer: Transformer action requires changing magnetic flux. A steady DC current does not maintain continuously changing flux after the initial transient.
Case Study 10: Integrated competency case: choosing and interpreting an AC circuit
Read the situation carefully:
A student compares three ideal AC loads connected separately to the same 200 V RMS source at the same frequency: a resistor, an inductor and a capacitor. The student then connects R, L and C in series and varies the frequency to locate resonance.
A student compares three ideal AC loads connected separately to the same 200 V RMS source at the same frequency: a resistor, an inductor and a capacitor. The student then connects R, L and C in series and varies the frequency to locate resonance.
Q37. Which ideal load consumes average power continuously from the source?
Answer: The pure resistor consumes average power. For ideal pure L and pure C, average power over a cycle is zero.
Q38. What is the power factor of the pure resistor, ideal inductor and ideal capacitor?
Answer: For the pure resistor, PF = 1. For ideal pure inductor and capacitor, PF = 0.
Q39. What condition identifies resonance in the series LCR circuit?
Key relation: ω0 = 1/√(LC)
Answer: Resonance occurs when XL = XC, equivalently ω0 = 1/√(LC).
Q40. At resonance, what happens to impedance and current for a fixed applied RMS voltage?
Key relation: Irms = Vrms/R at resonance
Answer: Impedance becomes minimum, Z = R, and the RMS current becomes maximum for the fixed supply voltage.
4. Common Case-Study Traps
Trap 1 — “220 V AC”: In standard AC-supply questions, treat the stated supply voltage as an RMS value unless the question explicitly says peak voltage.
Trap 2 — L versus C: XL rises with frequency; XC falls with frequency. Reversing these relationships changes the entire answer.
Trap 3 — Resonance: Resonance does not mean every individual voltage in the LCR circuit is zero. The net reactive contribution is zero because the inductor and capacitor effects cancel in phase.
Trap 4 — Power: Use P = VI cosφ for a general AC circuit. P = VI directly is valid when the power factor is unity.
Trap 5 — Transformer: In an ideal transformer, increasing voltage means decreasing current so that input and output power remain equal.
5. Case-Study Answer Strategy
Step 1 — Extract
Underline the physical quantities and units given in the passage.
Underline the physical quantities and units given in the passage.
Step 2 — Classify
Decide whether the situation is R, L, C, LCR, resonance, generator or transformer.
Decide whether the situation is R, L, C, LCR, resonance, generator or transformer.
Step 3 — Relate
Choose the smallest formula that directly connects the known quantities.
Choose the smallest formula that directly connects the known quantities.
Step 4 — Verify
Check units, phase direction, limiting behaviour and the sign of reactance.
Check units, phase direction, limiting behaviour and the sign of reactance.
6. Frequently Asked Questions
What are case-based questions in Class 12 Physics?
They present a physical situation, passage, data set or application and then ask linked questions that test concept selection, interpretation and calculation.Which Chapter 7 topics are useful for case-study practice?
RMS values, R/L/C behaviour, reactance, series LCR impedance, resonance, power factor, AC power, wattless current, generator and transformer are all within the current Chapter 7 scope.Are these exact CBSE board questions?
No. These are original, syllabus-aligned practice case studies designed around current assessment patterns and recurring Chapter 7 concepts. They are not claims about future questions.What should I do after these case studies?
Use the Chapter 7 MCQs for rapid recall and the Chapter 7 Numericals for step-by-step calculation practice.How many marks are the case-study questions in the CBSE Class 12 Physics 2026–27 SQP?
The official 2026–27 Physics SQP has two case study-based questions in Section D, each carrying 4 marks, for 8 marks in that section.Are these Alternating Current case-based questions official CBSE questions?
No. They are original Learn Revise Hub practice questions aligned to the current CBSE syllabus and assessment structure. Use the official SQP and marking scheme for the authoritative paper format.7. Continue the Chapter 7 Resource Chain
Concepts: Alternating Current Class 12 Physics Notes
Important practice: Alternating Current Important Questions
Rapid recall: Alternating Current MCQs
Calculation practice: Alternating Current Numericals
Foundation: Chapter 6 — Electromagnetic Induction Notes
Next workflow component: Assertion–Reason Questions will be linked here only after that page is published and verified.
8. Final Case-Study Readiness Check
✓ I can identify RMS versus peak values.
✓ I know how frequency changes XL and XC.
✓ I can calculate LCR impedance and current.
✓ I can identify capacitive, inductive and resonant conditions.
✓ I can calculate average AC power and power factor.
✓ I understand wattless current.
✓ I can solve generator and transformer data-based questions.
✓ I can extract formulas and physical meaning from a case passage.
Research and scope note: The case-study design follows the current CBSE 2026–27 Chapter 7 syllabus and the official Class XII SQP/MS framework. CBSE's Physics competency-focused material also uses application-based AC/LCR situations, supporting the use of interpretation plus calculation rather than formula-only questions. The cases and answers on this page are original practice content.
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