Alternating Current Class 12 Physics Numericals 2026-27 | Solved Numericals
1. Numerical Roadmap
Peak/RMS, instantaneous AC, average value, resistance and inductive reactance.
Capacitive reactance, pure-component current, LCR impedance, phase angle and power.
Resonant frequency, resonance conditions, power factor and frequency-change reasoning.
AC generator, transformer and competency-style circuit calculations, including ideal power relations and efficiency.
2. Essential Formula Map Before You Start
| Concept | Formula |
|---|---|
| RMS value | Vrms = V0/√2; Irms = I0/√2 |
| Angular frequency | ω = 2πf |
| Inductive reactance | XL = ωL = 2πfL |
| Capacitive reactance | XC = 1/(ωC) = 1/(2πfC) |
| Series LCR impedance | Z = √[R² + (XL − XC)²] |
| Phase angle | tanφ = (XL − XC)/R |
| Power factor | cosφ = R/Z |
| Average AC power | P = VrmsIrmscosφ |
| Resonance | XL = XC; ω0 = 1/√(LC) |
| AC generator peak EMF | E0 = NBAω |
| Transformer | Vs/Vp = Ns/Np; ideal VpIp = VsIs |
3. Solved Numericals — Peak, RMS and AC Basics
An AC source has peak voltage 311 V. Find its RMS voltage.
Vrms = V0/√2 = 311/√2 ≈ 219.9 V.
The RMS current in an AC circuit is 5 A. Find the peak current.
I0 = √2 Irms = √2 × 5 ≈ 7.07 A.
An AC voltage is v = 200 sin(100πt) V. Find the peak voltage, angular frequency and frequency.
Comparing with v = V0 sin(ωt): V0 = 200 V and ω = 100π rad s−1. Hence f = ω/(2π) = 50 Hz.
Current is i = 10 sin(100πt) A. Find i at t = 1/600 s.
100πt = 100π/600 = π/6. Therefore i = 10 sin(π/6) = 10 × 1/2 = 5 A.
A sinusoidal current has peak value 20 A. Find its average value over one positive half-cycle.
Iavg, half = 2I0/π = 40/π ≈ 12.73 A.
A 100 Ω resistor is connected to a 220 V RMS AC supply. Find RMS current and average power.
Irms = Vrms/R = 220/100 = 2.2 A. For a pure resistor, P = VrmsIrms = 220 × 2.2 = 484 W.
A 50 Ω resistor is connected to an AC source of peak voltage 100 V. Find the peak and RMS currents.
I0 = V0/R = 100/50 = 2 A. Thus Irms = 2/√2 ≈ 1.414 A.
An inductor of 0.20 H is connected to a 50 Hz AC supply. Find its inductive reactance.
ω = 2πf = 100π rad s−1. XL = ωL = 100π × 0.20 = 20π ≈ 62.83 Ω.
A 0.50 H ideal inductor is connected to 100 V RMS, 50 Hz AC. Find RMS current.
XL = 2πfL = 2π × 50 × 0.50 ≈ 157.08 Ω. Hence Irms = 100/157.08 ≈ 0.637 A.
An inductor has reactance 40 Ω at 50 Hz. What is its reactance at 100 Hz?
Since XL ∝ f, doubling frequency doubles reactance: X′L = 2 × 40 = 80 Ω.
4. Solved Numericals — R, L, C and AC Power
A 20 μF capacitor is connected to a 50 Hz AC source. Find XC.
XC = 1/(2πfC) = 1/[2π × 50 × 20×10−6] ≈ 159.15 Ω.
A 10 μF capacitor is connected to 200 V RMS, 50 Hz AC. Find RMS current.
XC = 1/[2π × 50 × 10×10−6] ≈ 318.31 Ω. Thus Irms = 200/318.31 ≈ 0.628 A.
A capacitor has reactance 200 Ω at 50 Hz. Find its reactance at 100 Hz.
XC ∝ 1/f. Doubling f halves XC: X′C = 200/2 = 100 Ω.
A capacitor has reactance 159.15 Ω at 50 Hz. Find its capacitance.
C = 1/(2πfXC) = 1/[2π × 50 × 159.15] ≈ 20×10−6 F.
A series LCR circuit has R = 30 Ω, XL = 50 Ω and XC = 10 Ω. Find impedance.
Net reactance = 50 − 10 = 40 Ω. Z = √(30² + 40²) = √2500 = 50 Ω.
A series LCR circuit has impedance 50 Ω and is connected to 200 V RMS. Find RMS current.
Irms = Vrms/Z = 200/50 = 4 A.
For a series LCR circuit, R = 40 Ω and XL − XC = 30 Ω. Find the phase angle.
tanφ = (XL − XC)/R = 30/40 = 0.75. Therefore φ = tan−1(0.75) ≈ 36.9°. Positive φ means the circuit is inductive.
A series LCR circuit has R = 30 Ω and Z = 50 Ω. Find power factor.
cosφ = R/Z = 30/50 = 0.6.
A circuit draws 5 A RMS from a 240 V RMS supply at power factor 0.8. Find average power.
P = VrmsIrmscosφ = 240 × 5 × 0.8 = 960 W.
An AC circuit has Vrms = 100 V, Irms = 2 A and average power 100 W. Find power factor.
cosφ = P/(VI) = 100/(100×2) = 0.5. The numerical data alone give the magnitude of power factor; the circuit type requires the phase/lead-lag information.
5. Solved Numericals — LCR Circuit and Resonance
A series LCR circuit has L = 0.20 H and C = 50 μF. Find the resonant angular frequency and frequency.
ω0 = 1/√(LC) = 1/√(0.20×50×10−6) ≈ 316.23 rad s−1. Therefore f0 = ω0/(2π) ≈ 50.33 Hz.
A series LCR circuit contains L = 0.50 H and operates at 50 Hz. Find C for resonance.
C = 1/(ω²L) = 1/[(2π×50)²×0.50] ≈ 20.26 μF.
A series LCR circuit has C = 100 μF and resonates at 50 Hz. Find L.
L = 1/(ω²C) = 1/[(2π×50)²×100×10−6] ≈ 0.1013 H.
A series LCR circuit has resistance 20 Ω and is connected to 200 V RMS at resonance. Find current and power factor.
At resonance XL = XC, so Z = R = 20 Ω. Hence I = 200/20 = 10 A and cosφ = 1.
A circuit resonates at 100 Hz. If its inductance becomes four times while C is unchanged, find the new resonant frequency.
f0 ∝ 1/√L. If L becomes 4L, f′0 = f0/2 = 50 Hz.
For a series LCR circuit, XL = 30 Ω and XC = 50 Ω. State whether the circuit is inductive or capacitive and whether current leads or lags voltage.
Since XC > XL, net reactance is capacitive. Therefore current leads the supply voltage.
A series LCR circuit has impedance 100 Ω and power factor 0.6. Find resistance.
cosφ = R/Z, so R = Z cosφ = 100×0.6 = 60 Ω.
A circuit has resistance 24 Ω and power factor 0.8. Find its impedance.
cosφ = R/Z, hence Z = R/cosφ = 24/0.8 = 30 Ω.
A series LCR circuit is connected to 200 V RMS. Its impedance is 50 Ω and resistance is 30 Ω. Find RMS current and average power.
I = V/Z = 200/50 = 4 A. Power factor = R/Z = 30/50 = 0.6. Thus P = VI cosφ = 200×4×0.6 = 480 W.
A series LCR circuit has XL = 80 Ω and XC = 30 Ω. What capacitive reactance would make the circuit purely resistive at the same frequency?
For zero net reactance, XL = XC. Therefore the required capacitive reactance is 80 Ω.
6. Solved Numericals — AC Generator and Transformer
A coil of 100 turns and area 0.02 m² rotates at 50 revolutions per second in a uniform magnetic field of 0.5 T. Find the peak induced EMF.
ω = 2πf = 100π rad s−1. Peak EMF E0 = NBAω = 100×0.5×0.02×100π ≈ 314.16 V.
An AC generator coil rotates at angular speed 200π rad s−1. Find the frequency of generated AC.
f = ω/(2π) = 200π/(2π) = 100 Hz.
An AC generator produces peak EMF E0. If the magnetic field is doubled while all other quantities remain unchanged, find the new peak EMF.
E0 = NBAω, so E0 is directly proportional to B. Doubling B doubles E0.
An ideal transformer has 500 primary turns and 2500 secondary turns. If the primary voltage is 200 V, find the secondary voltage.
Vs/Vp = Ns/Np = 2500/500 = 5. Hence Vs = 5×200 = 1000 V.
An ideal transformer reduces 2200 V to 220 V. If the secondary current is 10 A, find primary current.
For an ideal transformer, VpIp = VsIs. Thus Ip = (220×10)/2200 = 1 A.
An ideal transformer has 1000 primary turns and produces 440 V from a 220 V primary. Find secondary turns.
Ns/Np = Vs/Vp = 440/220 = 2. Hence Ns = 2×1000 = 2000 turns.
An ideal transformer receives 240 V at 2 A on the primary and delivers 120 V on the secondary. Find the secondary current.
Input power = 240×2 = 480 W. For an ideal transformer, output power is also 480 W. Hence Is = 480/120 = 4 A.
A transformer takes 1000 W input and delivers 920 W output. Find efficiency and power loss.
η = (Pout/Pin)×100 = 920/1000×100 = 92%. Power loss = 1000−920 = 80 W.
An ideal step-up transformer has turns ratio Ns/Np = 4. If Vp = 110 V and Ip = 8 A, find Vs and Is.
Vs = 4×110 = 440 V. Input power = 110×8 = 880 W. Thus Is = 880/440 = 2 A.
A series LCR circuit is connected to 100 V RMS. Its resistance is 40 Ω and its net reactance is 30 Ω inductive. Find impedance, current, power factor and average power.
Z = √(40²+30²) = 50 Ω. I = 100/50 = 2 A. cosφ = R/Z = 40/50 = 0.8. P = VI cosφ = 100×2×0.8 = 160 W.
7. NCERT + CBSE Competency Alignment
NCERT Chapter 7 alignment: The practice set follows the same conceptual progression students meet in the official Alternating Current chapter: AC and RMS values → resistor/inductor/capacitor → series LCR → resonance → power factor → transformer. It uses original numbers and wording rather than reproducing textbook questions.
CBSE competency alignment: The set also includes application-style calculations where students must infer a circuit quantity from a rating, ratio, frequency change or physical condition. CBSE's Physics Learning Framework includes competency-style AC items involving resistor-capacitor circuits and multi-step reasoning. CBSE Physics Learning Framework.
8. Quick Answer Bank — Most-Searched Formulas
Vrms = V0/√2 and Irms = I0/√2.
XL = 2πfL; XC = 1/(2πfC).
Z = √[R² + (XL − XC)²].
XL = XC; Z = R; power factor = 1.
P = VrmsIrmscosφ = Irms²R.
Vs/Vp = Ns/Np = Ip/Is for an ideal transformer.
9. Additional Competency-Style Numericals
A 100 W, 100 V bulb is to be operated from a 200 V, 50 Hz AC supply by connecting a suitable capacitor in series. Assuming the bulb behaves as a resistor, find the capacitance required so that the bulb operates at its rated current.
Bulb resistance R = 100²/100 = 100 Ω. Rated current I = P/V = 100/100 = 1 A. Required series impedance Z = 200/1 = 200 Ω. Hence XC = √(200² − 100²) = √30000 ≈ 173.21 Ω. Therefore C = 1/(2π×50×173.21) ≈ 18.38 μF.
A series LCR circuit has R = 40 Ω, L = 5 H and C = 80 μF and is connected to a 230 V RMS supply at resonance. Find the RMS current and the magnitudes of the voltages across the inductor and capacitor.
ω0 = 1/√(LC) = 1/√(5×80×10−6) = 50 rad s−1. Thus I = 230/40 = 5.75 A. XL = 50×5 = 250 Ω. Therefore VL = IXL = 5.75×250 = 1437.5 V. At resonance VC has the same magnitude and opposite phasor direction.
A series LCR circuit is capacitive at a particular frequency because XC = 60 Ω and XL = 40 Ω. If the frequency is increased, explain what happens to the two reactances and identify the direction of the circuit's movement toward resonance.
Increasing frequency increases XL and decreases XC. Starting from XL < XC, the two values therefore move toward equality. The circuit moves toward its series-resonance condition as frequency is increased.
An ideal pure inductor draws 5 A RMS from a 230 V RMS AC supply. Find the average power consumed by the inductor.
For a pure inductor, cos90° = 0. Hence P = 230×5×0 = 0 W. Current flows, but the average power transferred over a complete cycle is zero.
A transformer receives 2.5 kW of input power and operates at 92% efficiency. Find the output power and total power loss.
Pout = ηPin = 0.92×2.5 kW = 2.30 kW. Total loss = 2.50 − 2.30 = 0.20 kW = 200 W.
10. Common Numerical Traps
11. A Reliable 5-Step Numerical Method
Write what is given and what is required.
Select the relation that directly connects the known and unknown quantities.
Put SI units into the formula before calculating.
Show substitution and retain sensible significant figures.
Check units, limiting behaviour and whether the result is physically reasonable.
12. Frequently Asked Numerical Questions
Start with RMS/peak conversion, XL, XC, LCR impedance, resonance, power factor, AC-generator EMF and transformer ratios. These are central numerical relationships in the NCERT Chapter 7 sequence and are also reflected in CBSE competency-oriented material. Official NCERT Physics Part-I Textbook · CBSE Physics Learning Framework.
For an inductor, XL doubles. For a capacitor, XC becomes half. This is one of the most useful frequency-dependence checks.
XL = XC, so the net reactance is zero, Z = R, current is maximum for a fixed RMS supply, and power factor is unity.
No. They are syllabus-aligned practice problems built around recurring concepts and numerical formats; they are not claims about exact future CBSE questions.
13. Continue the Chapter 7 Practice Sequence
Read concepts: Alternating Current Notes
Practise theory: Alternating Current Important Questions
Practise objective questions: Alternating Current MCQs
Build the foundation: Chapter 6 — Electromagnetic Induction Notes
Next in the Chapter 7 workflow: Case-Based Questions (to be linked only after publication).
14. Final Numerical Readiness Check
✓ I can convert peak values to RMS values and back.
✓ I can calculate XL and XC and explain their frequency dependence.
✓ I can calculate series-LCR impedance, current and phase angle.
✓ I can identify resonance and calculate resonant frequency.
✓ I can calculate power factor and average AC power.
✓ I can calculate AC-generator peak EMF.
✓ I can use transformer voltage, turns and ideal-current relations.
✓ I show units and a clear calculation path in every board-style numerical.
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