Alternating Current
1. Introduction
Alternating current (AC) reverses direction periodically. AC is the standard form of electrical power because it can be easily transformed between voltage levels.
2. Representation of AC
i = I₀ sin ωt (or I₀ cos ωt), where I₀ is the peak value and ω = 2πf.
2.1 RMS Value
I_rms = I₀/√2, V_rms = V₀/√2.
2.2 Average Value (over half cycle)
I_avg = 2I₀/π (over positive half cycle).
3. Phasor Diagrams
A phasor is a rotating vector representing an AC quantity. The length represents amplitude, the angle represents phase.
4. AC Circuits
4.1 Purely Resistive Circuit
V and I are in phase. I₀ = V₀/R.
4.2 Purely Inductive Circuit
I lags V by 90°. Inductive reactance: X_L = ωL = 2πfL. I₀ = V₀/X_L.
4.3 Purely Capacitive Circuit
I leads V by 90°. Capacitive reactance: X_C = 1/ωC = 1/(2πfC). I₀ = V₀/X_C.
5. LCR Series Circuit
Impedance: Z = √(R² + (X_L - X_C)²) Phase angle: tan φ = (X_L - X_C)/R
5.1 Resonance
When X_L = X_C, the circuit is in resonance. ω₀ = 1/√(LC), f₀ = 1/(2π√(LC)). At resonance: Z = R (minimum), I = V/R (maximum).
Q-factor: Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C).
6. Power in AC Circuits
P = V_rms × I_rms × cos φ, where cos φ = R/Z is the power factor.
7. Transformer
A device that changes AC voltage using mutual inductance.
7.1 Working Principle
Based on mutual induction. Vₛ/Vₚ = Nₛ/Nₚ = k (turns ratio). Step-up: k > 1. Step-down: k < 1.
7.2 Efficiency
η = (VₛIₛ)/(VₚIₚ) × 100%. Iₛ/Iₚ = Nₚ/Nₛ = 1/k.
7.3 Energy Losses
Copper losses (I²R), iron losses (eddy currents, hysteresis), flux leakage.
8. Worked Problems
Problem 1: An AC source of 200V, 50Hz is connected to a 0.5H inductor. Find reactance and current. Solution: X_L = 2πfL = 2π×50×0.5 = 157 Ω. I = V/X_L = 200/157 = 1.27 A. I lags V by 90°.
Problem 2: A series LCR circuit has R = 10Ω, L = 0.1H, C = 10 μF. Find resonant frequency. Solution: f₀ = 1/(2π√(LC)) = 1/(2π√(0.1×10^{-5})) = 1/(2π√(10^{-6})) = 1/(2π×10^{-3}) = 159.15 Hz.
Problem 3: A transformer has 500 primary turns and 50 secondary turns. Primary voltage is 220V. Find secondary voltage. Solution: Vₛ = Vₚ × Nₛ/Nₚ = 220 × 50/500 = 22 V.
9. Applications of AC Circuits
9.1 Choke Coil
An inductor used to control current in AC circuits without wasting power (ideally). Since power factor of a pure inductor is 0, power consumed is zero.
9.2 Tuning Circuit (Radio Receiver)
An LCR circuit is used as a tuning circuit. By varying the capacitor, the resonant frequency is adjusted to match the desired station's frequency.
9.3 Power Factor Correction
Industrial loads (motors) have a lagging power factor. Capacitors are added in parallel to improve the power factor, reducing power losses in transmission.
10. Common Mistakes
'Students often confuse phase relationships. Remember: L - I lags V (ELI). C - I leads V (ICE). Think ELI the ICE man.'
'Power factor is cos φ, not sin φ. A power factor of 1 means maximum power transfer.'
10. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| RMS and phasors | 3 | 2 |
| LCR circuits | 5 | 3 |
| Resonance | 3 | 2 |
| Transformer | 3 | 2 |
11. Self-Test Questions
- Derive an expression for the impedance of a series LCR circuit.
- A coil of resistance 50Ω and inductance 0.2H is connected to 230V, 50Hz supply. Find current and power factor.
- Explain resonance in a series LCR circuit. How does it vary with frequency?
- Describe the working of a transformer. Why is its core laminated?
- An AC voltage V = 200 sin 314t is applied to a 50 μF capacitor. Find the reactance and RMS current.
