What is alternating current?
Direct current (DC) flows one way, like from a cell. Alternating current (AC) keeps changing direction. In India it changes direction 100 times each second (50 full cycles, so f = 50 Hz).
We write V = V₀ sin ωt and I = I₀ sin ωt. V₀ and I₀ are peak values (the highest values). ω = 2πf is the angular frequency. One full cycle takes time T = 1/f.
Peak, mean and rms values
Over one full cycle AC is positive half the time and negative half the time, so its average is zero. That tells us nothing useful.
So we use the rms value (root mean square): square the values, take the mean, then take the square root. For a sine wave: Irms = I₀/√2 ≈ 0.707 I₀ and Vrms = V₀/√2.
Meaning: the rms value is the steady DC that would heat a resistor at the same rate. Meters and bulbs show rms values. The mean over a half cycle is 2I₀/π ≈ 0.637 I₀.
Phasors: a spinning arrow for a wave
A phasor is an arrow of length V₀ that spins anticlockwise with angular speed ω. Its projection on the vertical axis at any time is V₀ sin ωt. So the spinning arrow draws the wave.
We draw the voltage phasor and the current phasor together. The angle between them is the phase difference φ.
AC through R, L and C: reactance
Resistor only
V and I are in phase (φ = 0). V₀ = I₀ R.
Inductor only
The inductor fights every change of current, so the current lags the voltage by 90°. Its opposition is the inductive reactance Xʟ = ωL (ohms). Higher frequency → bigger Xʟ. For DC (ω = 0), Xʟ = 0.
Capacitor only
The capacitor must charge before its voltage builds up, so the current leads the voltage by 90°. Capacitive reactance Xᴄ = 1/(ωC). Higher frequency → smaller Xᴄ. A capacitor blocks DC (Xᴄ → ∞).
Series LCR circuit and impedance
In series, the same current flows through R, L and C. Draw the current phasor first. Vʀ is along it, Vʟ is 90° ahead, Vᴄ is 90° behind. Vʟ and Vᴄ point opposite ways, so they partly cancel.
V² = Vʀ² + (Vʟ − Vᴄ)², which gives the impedance Z = √(R² + (Xʟ − Xᴄ)²). Then I = V/Z.
Phase angle: tan φ = (Xʟ − Xᴄ)/R. If Xʟ > Xᴄ the circuit acts inductive (current lags); if Xᴄ > Xʟ it acts capacitive (current leads).
Resonance in a series LCR circuit
As frequency changes, Xʟ rises and Xᴄ falls. At one frequency they are equal: ω₀L = 1/(ω₀C), so ω₀ = 1/√(LC) and f₀ = 1/(2π√LC).
At resonance Z = R (smallest), the current is largest, and V and I are in phase. Vʟ and Vᴄ may each be much larger than the supply voltage, but they cancel.
Sharpness: a small R gives a tall, narrow peak. The quality factor Q = ω₀L/R = 1/(ω₀CR) measures this. Radios need a sharp peak to pick one station. Resonance needs both L and C; an RL or RC circuit has no resonance.
Power and power factor
Average power in an AC circuit: P = Vrms Irms cos φ. The term cos φ = R/Z is the power factor.
- Pure R: φ = 0, cos φ = 1, full power.
- Pure L or pure C: φ = 90°, cos φ = 0, no average power.
- LCR at resonance: cos φ = 1.
Only the resistor uses up energy. An ideal L or C takes energy for a quarter cycle and gives it back in the next quarter.
A low power factor means a large current for the same useful power, which wastes energy as heat (I²R) in the supply wires. Adding a suitable capacitor raises the power factor.
Wattless current
Split the current phasor into two parts: I cos φ along the voltage and I sin φ at 90° to it.
The part Irms sin φ does no average work, because it is 90° out of phase with the voltage. It is called the wattless current. In a pure inductor or capacitor the whole current is wattless. A choke coil in a fan regulator uses this idea: it lowers the current with almost no energy loss.
Try it: a practical
Try it: In the 3D, choose "LCR series" and start at 30 Hz. Before you move the slider, predict: will the current go up or down as you go to 50 Hz? Now slide slowly and watch Irms. Note the frequency where the purple "Resonance" tag appears and check it against f₀ = 1/(2π√LC) = 1/(2π√(0.2 × 50×10⁻⁶)) ≈ 50 Hz. Then pick "L only" and see P = 0 W in the readout.
Key formulas and definitions
- V = V₀ sin ωt, I = I₀ sin(ωt − φ), ω = 2πf
- I_rms = I₀/√2 ≈ 0.707 I₀; V_rms = V₀/√2
- Inductive reactance: Xʟ = ωL = 2πfL
- Capacitive reactance: Xᴄ = 1/(ωC) = 1/(2πfC)
- Impedance: Z = √(R² + (Xʟ − Xᴄ)²)
- Phase: tan φ = (Xʟ − Xᴄ)/R
- Resonance: ω₀ = 1/√(LC), f₀ = 1/(2π√LC), Z = R
- Quality factor: Q = ω₀L/R
- Average power: P = V_rms I_rms cos φ; power factor cos φ = R/Z
- Wattless current = I_rms sin φ
Worked examples
1. The mains voltage is 230 V rms. Find its peak value.
V₀ = √2 × V_rms = 1.414 × 230 ≈ 325 V.
2. A 100 Ω resistor is connected to 220 V, 50 Hz AC. Find I_rms, I₀ and the power.
I_rms = 220/100 = 2.2 A. I₀ = √2 × 2.2 ≈ 3.11 A. P = V_rms I_rms = 220 × 2.2 = 484 W (φ = 0).
3. Find the reactance of a 0.5 H inductor at 50 Hz and the rms current from 220 V.
Xʟ = 2πfL = 2 × 3.14 × 50 × 0.5 ≈ 157 Ω. I_rms = 220/157 ≈ 1.4 A. Power = 0 (current lags by 90°).
4. Find the reactance of a 20 µF capacitor at 50 Hz.
Xᴄ = 1/(2πfC) = 1/(2 × 3.14 × 50 × 20 × 10⁻⁶) = 1/0.00628 ≈ 159 Ω.
5. In a series LCR circuit R = 30 Ω, Xʟ = 80 Ω, Xᴄ = 40 Ω, V_rms = 200 V. Find Z, I, φ and the power factor.
Z = √(30² + (80 − 40)²) = √(900 + 1600) = 50 Ω. I = 200/50 = 4 A. tan φ = 40/30, φ ≈ 53° (current lags). cos φ = 30/50 = 0.6.
6. For the circuit above, find the average power and the wattless current.
P = V I cos φ = 200 × 4 × 0.6 = 480 W (check: I²R = 16 × 30 = 480 W). sin φ = 0.8, wattless current = 4 × 0.8 = 3.2 A.
7. L = 0.1 H and C = 10 µF are in series with R = 10 Ω. Find the resonant frequency and the current at resonance from 100 V.
ω₀ = 1/√(LC) = 1/√(10⁻⁶) = 1000 rad/s. f₀ = 1000/(2π) ≈ 159 Hz. At resonance Z = R = 10 Ω, so I = 100/10 = 10 A.
8. For the resonance circuit above, find Q and the voltage across the inductor at resonance.
Q = ω₀L/R = 1000 × 0.1/10 = 10. Xʟ = 100 Ω, Vʟ = I Xʟ = 10 × 100 = 1000 V, ten times the supply! Vᴄ is also 1000 V and they cancel.
Common mistakes
- Adding voltages in an LCR circuit as plain numbers (Vʀ + Vʟ + Vᴄ). They are out of phase; add them as phasors.
- Using peak values in P = VI cos φ. Use rms values (or write P = ½ V₀ I₀ cos φ).
- Thinking an inductor has a large reactance for DC. For DC ω = 0, so Xʟ = 0; it is the capacitor that blocks DC.
- Saying the current is zero at resonance. At resonance Z is smallest, so the current is largest.