What is Alternating Current?
AC voltage varies sinusoidally with time: v = v_m sin ωt, where v_m is the peak voltage and ω is angular frequency. The current driven by it in a resistor: i = i_m sin ωt.
- In a pure resistor, voltage and current are in phase (phase difference = 0).
- Average current over a complete cycle = 0 (positive and negative halves cancel).
- But average power ≠ 0 because Joule heating ∝ i² (always positive).
RMS (Root Mean Square) Values
- RMS current: I = i_m / √2 = 0.707 i_m
- RMS voltage: V = v_m / √2 = 0.707 v_m
- Average power: P = I²R = V²/R = IV
- India household supply: 220 V rms → peak = 220 × √2 = 311 V
- RMS current is the equivalent DC that produces the same Joule heating.
I = i_m / √2 | V = v_m / √2
P = I²R = V²/R = IV
Peak voltage India: v_m = √2 × 220 = 311 V
Fig 7.1 — AC source connected to resistor (NCERT Figure 7.1)
Fig 7.2 — Voltage and current in phase for pure resistor (NCERT Figure 7.2)
Fig 7.3 — RMS current I = i_m/√2 = 0.707 i_m (NCERT Figure 7.3)
🎯 Practice MCQs — AC & RMS Values
What is a Phasor?
A phasor is a rotating vector that represents a sinusoidally varying quantity. It rotates about the origin with angular speed ω. The vertical component of the phasor represents the instantaneous value.
- Magnitude of phasor = amplitude (peak value) of the quantity.
- Phasors are not real vectors — they represent scalar AC quantities.
- Phasor diagrams show the phase angle between voltage and current.
Phase Summary for AC Elements
| Element | Phase Relation | Phase Angle |
|---|---|---|
| Pure Resistor (R) | V and I in phase | φ = 0 |
| Pure Inductor (L) | I lags V by π/2 | φ = −π/2 |
| Pure Capacitor (C) | I leads V by π/2 | φ = +π/2 |
Fig 7.4 — Phasor diagram for resistor circuit: V and I in same direction (NCERT Figure 7.4)
🎯 Practice MCQs — Phasors
AC Voltage Applied to a Pure Inductor
- v = v_m sin ωt → i = i_m sin(ωt − π/2) → current lags voltage by π/2
- Inductive Reactance: X_L = ωL = 2πνL (unit: Ω)
- i_m = v_m / X_L | I = V / X_L
- X_L ∝ frequency (higher frequency → more opposition)
- Average power = 0 (wattless current)
AC Voltage Applied to a Pure Capacitor
- v = v_m sin ωt → i = i_m sin(ωt + π/2) → current leads voltage by π/2
- Capacitive Reactance: X_C = 1/ωC = 1/2πνC (unit: Ω)
- i_m = v_m / X_C | I = V / X_C
- X_C ∝ 1/frequency (higher frequency → less opposition)
- Average power = 0 (wattless current)
- In DC circuit, capacitor blocks current after charging. In AC, it allows alternating current to pass.
Capacitive: X_C = 1/ωC (Ω) | X_C decreases with frequency
Example: L = 25 mH, ν = 50 Hz:
X_L = 2π × 50 × 25×10⁻³ = 7.85 Ω, I = 220/7.85 = 28 A
Example: C = 15 μF, ν = 50 Hz:
X_C = 1/(2π × 50 × 15×10⁻⁶) = 212 Ω, I = 220/212 = 1.04 A
Fig 7.5 — AC source connected to pure inductor (NCERT Figure 7.5)
Fig 7.6 — Phasor diagram for inductor: I lags V by π/2 (NCERT Figure 7.6)
Fig 7.7 — AC source connected to pure capacitor (NCERT Figure 7.7)
Fig 7.8 — Phasor diagram for capacitor: I leads V by π/2 (NCERT Figure 7.8)
🎯 Practice MCQs — Reactance
Series LCR Circuit
- R, L, C connected in series — same current flows through all at any instant.
- VR parallel to I; VL is π/2 ahead of I; VC is π/2 behind I.
- VL and VC are in opposite phase → net reactive voltage = |V_Lm − V_Cm|
- Current: i = i_m sin(ωt + φ) where i_m = v_m / Z
Current amplitude: i_m = v_m / Z
Phase angle: tan φ = (X_C − X_L) / R
If X_C > X_L: φ > 0 → circuit is capacitive → I leads V
If X_C < X_L: φ < 0 → circuit is inductive → I lags V
Resonant frequency: ω₀ = 1/√(LC)
At resonance: X_C = X_L → Z = R (minimum) → i_m = v_m/R (maximum)
Resonance in LCR Circuit
- Resonance occurs when X_L = X_C → ω₀ = 1/√(LC)
- At resonance: impedance = R (minimum), current = maximum = v_m/R
- Resonance requires BOTH L and C. No resonance possible in RL or RC alone.
- Application: Radio/TV tuning — capacitance varied to match station frequency.
- Metal detector at airport works on resonance principle.
Important Note on Voltages in LCR
In a series LCR circuit, the algebraic sum of V_R, V_L, V_C may exceed the source voltage. This is NOT a paradox — voltages across R, L, C are out of phase and must be added as phasors (using Pythagorean theorem), not algebraically.
Fig 7.10 — Series LCR circuit connected to AC source (NCERT Figure 7.10)
Fig 7.11 — Phasor diagram: VL, VR, VC and their relation (NCERT Figure 7.11)
Fig 7.12 — Impedance triangle diagram (NCERT Figure 7.12)
Fig 7.14 — Variation of current amplitude i_m with ω for two values of R (NCERT Figure 7.14)
🎯 Practice MCQs — LCR Circuit & Resonance
Average Power in AC Circuit
For a series LCR circuit with voltage v = v_m sinωt and current i = i_m sin(ωt + φ):
Power factor: cos φ = R/Z
Phase angle: φ = tan⁻¹[(X_C − X_L)/R]
Case 1 (Pure R): φ = 0, cosφ = 1, P = VI (maximum power)
Case 2 (Pure L or C): φ = π/2, cosφ = 0, P = 0 (wattless current)
Case 3 (LCR): 0 < cosφ < 1, partial power dissipation
Case 4 (Resonance): X_C = X_L, φ = 0, cosφ = 1, P = I²R (maximum)
Wattless Current
In a purely inductive or capacitive circuit, even though current flows, no power is dissipated. This current is called wattless current. Energy is alternately stored and returned — never consumed.
- Low power factor → high current needed to supply same power → large I²R losses in transmission lines.
- Power factor can be improved by adding a capacitor in parallel.
Fig 7.15 — Power factor improvement using capacitor — phasor components I_p and I_q (NCERT Figure 7.15)
🎯 Practice MCQs — Power Factor
Principle and Construction
- Works on mutual induction principle.
- Two coils (primary: N_p turns, secondary: N_s turns) wound on a soft iron core.
- AC in primary → changing flux → EMF induced in secondary.
- Ideal transformer: 100% efficiency (P_input = P_output).
Current ratio: I_s / I_p = N_p / N_s
Power: I_p V_p = I_s V_s (ideal transformer)
Step-up: N_s > N_p → V_s > V_p, I_s < I_p
Step-down: N_s < N_p → V_s < V_p, I_s > I_p
Example: 100 turns primary, 200 turns secondary:
220V at 10A → 440V at 5A (step-up)
Energy Losses in Real Transformers
- Flux leakage: Not all flux links both coils. Reduced by winding one coil over the other.
- Resistance of windings: I²R heating in wire. Minimised by using thick wire for high-current coils.
- Eddy currents: Induced currents in iron core cause heating. Reduced by using laminated core.
- Hysteresis: Energy lost in repeatedly magnetising/demagnetising the core. Minimised by using low-hysteresis material.
- Well-designed transformer efficiency: > 95%
Role in Power Transmission
- Generator output stepped UP (very high voltage, low current) → transmitted long distances → less I²R loss.
- Stepped down at sub-stations near consumers.
- Finally stepped down to 240V for home supply.
Fig 7.16 — Two arrangements for winding of transformer coils: (a) one on top of other, (b) separate limbs (NCERT Figure 7.16)
🎯 Practice MCQs — Transformers
🎯 Mixed Rapid-Fire MCQs
📋 Chapter Summary
∿ AC Basics
v = v_m sinωt. RMS: I = i_m/√2, V = v_m/√2. P = IV = I²R. India: 220V rms → 311V peak.
🔁 Phasors
Rotating vectors representing AC quantities. Phase angle φ between V and I. Magnitude = amplitude.
🌀 Inductor
X_L = ωL. I lags V by π/2. Average power = 0. X_L ∝ frequency.
⚡ Capacitor
X_C = 1/ωC. I leads V by π/2. Average power = 0. X_C ∝ 1/frequency.
🔗 LCR Series
Z = √[R²+(X_C−X_L)²]. tanφ = (X_C−X_L)/R. Resonance: ω₀ = 1/√(LC), Z_min = R.
📊 Power Factor
P = VIcosφ. cosφ = R/Z. Pure L/C: cosφ=0 (wattless). Resonance: cosφ=1 (max power).
🔌 Transformer
V_s/V_p = N_s/N_p. I_s/I_p = N_p/N_s. Losses: flux leakage, I²R, eddy currents, hysteresis.
📡 Resonance App
Radio tuning, metal detectors. At resonance: max current, min impedance, max power dissipation.
Ace the AAI ATC Exam!
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