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AC Generator, Transformers and AC Basics

Electromagnetic Induction: AC Generator, Transformers and AC Basics

AC Generator, Transformers and AC Basics

AC Generator, Transformers and AC Basics

What you'll learn

  • Derive the EMF equation of an AC generator ε = NBAω sinωt
  • Define V_rms and I_rms and relate them to peak values
  • Calculate average power in AC circuits using the power factor
  • Find impedance Z for series RLC circuits and identify resonance
  • Determine resonant frequency ω₀ = 1/√(LC) and the Q factor
  • Analyse transformer efficiency including copper and core losses

Key concepts

Level 1 — Foundations

AC Generator: A coil of N turns, area A, rotates with angular velocity ω in a uniform magnetic field B. The flux at time t:

Φ=NBAcos(ωt)\Phi = NBA\cos(\omega t)

By Faraday's law:

ε=dΦdt=NBAωsin(ωt)=ε0sin(ωt)\varepsilon = -\frac{d\Phi}{dt} = NBA\omega\sin(\omega t) = \varepsilon_0\sin(\omega t)

Peak EMF: ε₀ = NBAω. This is alternating with frequency f = ω/(2π).

RMS Values: For a sinusoidal AC signal:

Vrms=V020.707V0V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707\,V_0 Irms=I020.707I0I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707\,I_0

India's domestic supply: V_rms = 220 V (V₀ ≈ 311 V), f = 50 Hz.

Transformer (step-up/step-down):

V2V1=N2N1,I2I1=N1N2\frac{V_2}{V_1} = \frac{N_2}{N_1}, \quad \frac{I_2}{I_1} = \frac{N_1}{N_2}

Step-up (N₂ > N₁): increases voltage, decreases current. Step-down (N₂ < N₁): decreases voltage, increases current. Works only on AC (changing flux needed for induction).

Level 2 — JEE depth

Power in AC circuits: Instantaneous power p = vi = V₀sinωt · I₀sin(ωt − φ), where φ is the phase difference between voltage and current.

Average power:

Pavg=VrmsIrmscosφP_{avg} = V_{rms}\,I_{rms}\cos\varphi

cosφ is the power factor. For pure resistor φ = 0 → P = V_rms I_rms (maximum). For pure inductor or capacitor φ = 90° → P = 0 (reactive circuit).

Impedance in series RLC circuit:

Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

where X_L = ωL (inductive reactance, increases with ω) and X_C = 1/(ωC) (capacitive reactance, decreases with ω).

Current: I_rms = V_rms / Z. Phase angle: tanφ = (X_L − X_C)/R.

Resonance (series RLC): At resonance, X_L = X_C:

ω0L=1ω0C    ω0=1LC\omega_0 L = \frac{1}{\omega_0 C} \implies \omega_0 = \frac{1}{\sqrt{LC}}

At resonance: Z = R (minimum impedance), I = V/R (maximum current), power factor = 1, P = V_rms²/R (maximum power).

Q factor (Quality factor):

Q=ω0LR=1ω0CR=1RLCQ = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}}

High Q → sharp resonance peak (selective circuit, used in radio tuning). Q also equals ω₀/(bandwidth) where bandwidth = R/L.

Transformer efficiency:

η=PoutPin×100%=V2I2V1I1×100%\eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{V_2 I_2}{V_1 I_1} \times 100\%

Losses:

  1. Copper loss (I²R heating in primary and secondary windings) — reduced by using thick copper wire.
  2. Iron/core loss = Eddy current loss + Hysteresis loss — reduced by laminated core (eddy) and soft iron (hysteresis).

Real transformers: η = 95–99% for large power transformers.

JEE traps:

  • V_rms ≠ V₀/2; it is V₀/√2. Similarly I_rms = I₀/√2. Average of a sinusoidal is zero; rms is not.
  • The resonant frequency formula ω₀ = 1/√(LC) has NO R — resistance only affects the sharpness (Q), not the resonant frequency.
  • The phase angle φ: current lags voltage in inductive circuits (X_L > X_C); current leads voltage in capacitive circuits (X_C > X_L).
  • Wattless current: the component I_rms sinφ does not contribute to average power; only I_rms cosφ is the "working" component.
  • At resonance in a series RLC circuit, V_L = V_C (in magnitude) and can individually be much larger than the supply voltage V (voltage magnification = Q × V). This can damage components if Q is high.

Worked example

AC Generator — Peak EMF and V_rms

Given: N = 200 turns, A = 0.05 m², B = 0.3 T, ω = 100π rad/s

Step 1: Peak EMF
  ε₀ = NBAω
  ε₀ = 200 × 0.3 × 0.05 × 100π
  ε₀ = 200 × 0.3 × 0.05 × 314.16
  ε₀ = 200 × 4.712
  ε₀ = 942.5 V

Step 2: RMS voltage
  V_rms = ε₀ / √2
  V_rms = 942.5 / 1.4142
  V_rms ≈ 666.6 V

Step 3: Frequency
  f = ω / (2π) = 100π / (2π) = 50 Hz

Answer: Peak EMF ε₀ ≈ 942.5 V, V_rms ≈ 666.6 V, f = 50 Hz.
EMF equation: ε = 942.5 sin(100πt) V

Step-Up Transformer — Voltage and Current Ratio

Given: Primary: V₁ = 220 V, N₁ = 100 turns
       Secondary: N₂ = 2200 turns

Step 1: Secondary voltage
  V₂/V₁ = N₂/N₁
  V₂ = V₁ × (N₂/N₁)
  V₂ = 220 × (2200/100)
  V₂ = 220 × 22
  V₂ = 4840 V

Step 2: Current ratio (ideal transformer: P₁ = P₂)
  I₂/I₁ = N₁/N₂ = 100/2200 = 1/22

  If primary current I₁ = 22 A:
  I₂ = 22 × (1/22) = 1 A

  Verify: P₁ = 220 × 22 = 4840 W = P₂ = 4840 × 1 = 4840 W ✓

Step 3: This is a step-up transformer (N₂ > N₁)
  Voltage ratio: V₂/V₁ = 22 (voltage increases)
  Current ratio: I₂/I₁ = 1/22 (current decreases proportionally)

Answer: V₂ = 4840 V; current ratio I₁:I₂ = 22:1.

Common mistakes

MistakeWhy it happensFix
Using V_rms = V₀/2Confusing average of sinusoid with rmsDerive: V_rms² = (1/T)∫V₀²sin²ωt dt = V₀²/2; so V_rms = V₀/√2; average of
Saying resonant frequency depends on RConfusing resonance condition with bandwidthω₀ = 1/√(LC) — no R; but the Q factor = ω₀L/R does depend on R, controlling sharpness
Getting current ratio wrong for transformerThinking both voltage and current increase in step-upPower conservation: V₁I₁ = V₂I₂; if V₂ > V₁ then I₂ < I₁ — always check with P = VI
Computing P = V_rms × I_rms without power factorIgnoring phase difference in non-resistive circuitsP_avg = V_rms I_rms cosφ; for pure L or C, cosφ = 0 and P = 0 even though V_rms I_rms ≠ 0

Quick check

  • Q1: An AC generator has N = 500, A = 0.02 m², B = 0.5 T, f = 50 Hz. Find ε₀ and V_rms.
  • Q2: India's domestic supply is 220 V rms at 50 Hz. Find V₀, I₀ for a 1000 W heater, and the resistance of the heater element.
  • Q3: A series RLC circuit: R = 10 Ω, L = 0.1 H, C = 100 μF. Find ω₀, Z at resonance, and Q factor.
  • Q4: A step-down transformer converts 11 kV to 220 V. If efficiency is 90% and output power is 9 kW, find primary current.
  • Stretch: Q5: In an RLC series circuit at resonance, V_supply = 10 V, Q = 50. Find V_L and V_C and explain why they are larger than the supply voltage.

NCERT Chapter 7 link: Chapter 7 (Class 12 Part I) — "Alternating Current." AC generator derivation is in Section 7.2; rms values in Section 7.3; power in AC circuits in Section 7.7; the series RLC circuit and resonance in Sections 7.5–7.6; transformers in Section 7.8. The phasor diagram technique in Section 7.4 is essential for JEE.

Exam connections: JEE tests: (1) peak vs rms calculation, (2) phasor diagram to find Z and φ for RLC circuits, (3) resonance condition and Q factor as conceptual MCQs, (4) transformer efficiency problems with multiple loss components, (5) graph-based questions: I vs ω showing resonance peak, bandwidth, and effect of changing R.

Study strategy: Draw the phasor diagram for every RLC problem: V_R along the current direction, V_L leading by 90°, V_C lagging by 90°. The resultant voltage phasor gives Z directly. This visual approach eliminates sign errors in X_L − X_C and automatically gives the phase angle direction.

Interactive Exploration Suggestions (Drishti Live Worlds)

  • Use the AC Circuit live world: vary L, C, R and ω; observe the resonance peak in I vs ω; measure bandwidth and compute Q; compare with the formula Q = ω₀L/R.
  • Home activity: Connect an LED (with a 100 Ω resistor) to a phone audio output at different frequencies via an alligator clip; the LED brightness changes with frequency — qualitative resonance effect.
  • AI Mentor voice reflection: "Explain to a family member why transformers are essential for sending electricity across India from a power plant 500 km away."

AI Mentor Prompts (Socratic, Board-Adaptive)

  • "Explain V_rms to a student who only knows DC circuits — why is it not just the average of the AC voltage, and what does it physically mean?"
  • "What is one common mistake students make about the power factor in AC circuits, and how would you catch yourself making it?"
  • Stretch: "How does resonance in an RLC circuit connect to radio tuning, MRI machines, wireless power transfer, and the design of noise-cancelling headphones?"

Gamification, Portfolio & Parent Visibility

  • Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
  • 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
  • Best real-world application stories (anonymised) featured on class or national leaderboard.

Robotics, STEM & Future Skills Bridges

  • Build a simple hand-cranked AC generator: wind 100 turns, two fixed magnets, crank by hand, connect to an oscilloscope app (phone + mic cable) — observe the sinusoidal waveform and measure peak and frequency.
  • Direct link to Green Tech (wind and hydro turbines output AC; transformers step up for transmission; understanding this chain is core to energy literacy) and Money Management (electricity bills use kWh; P_avg × time = energy consumed and billed).
  • Coding extension: Plot ε = ε₀ sin(ωt) and I = I₀ sin(ωt − φ) in Python for φ = 0, π/4, and π/2; shade the area representing instantaneous power and observe how average power changes with phase.

NEP 2020 & Full Education OS Alignment

This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

Portfolio Evidence Idea: Your photo/table/reflection/project + one sentence on "How this helps me in real life or a possible future path."

Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

See curriculum for cross-links and the full future-skills/robotics chapters.

Key Takeaways (TL;DR)

  • What you'll learn
  • Key concepts
  • Worked example
  • Common mistakes

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