Alternating Current · Walkthrough Walkthrough · § 1 / 9
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Class XII · Physics · Unit 6 · Walkthrough

Alternating Current

Walk through AC one step at a time — every idea comes alive in the live panel on the right. A wave reverses on an oscilloscope, current leads and lags through a capacitor and coil, the resonance peak rises like a radio finding its station, and a transformer hands power across the country.

A direct current (DC), like the one from a dry cell, flows steadily one way — on an oscilloscope it is just a flat horizontal line. An alternating current (AC), the kind in your wall socket, smoothly reverses direction over and over, tracing a sine wave.

  • Time period T — time for one complete cycle (one up-and-down swing).
  • Frequency f — cycles per second, in hertz: f = 1/T. Pakistan's mains runs at 50 Hz.
  • Angular frequency ω — ω = 2πf, so the instantaneous values are V = V₀ sin ωt and I = I₀ sin ωt.
Why AC? Only AC can be stepped up and down by transformers, which is what makes long-distance power transmission practical.

An AC waveform spends most of its time below its maximum, so its peak value I₀ overstates its real heating effect. The root-mean-square (rms) value is the steady DC that would dissipate the same power in a resistor — the truly useful, "effective" value.

rms (effective) valuesI_rms = I₀ / √2 = 0.707 I₀ · V_rms = V₀ / √2
average power: P = I_rms² R = V_rms I_rms
worked — the mains socket
V_rms = 230 V → peak voltage?
V₀ = √2 × 230 = ≈ 325 V — the meter reads rms, the wire actually swings to ±325 V
Remember: every AC value quoted — "230 V mains", "5 A appliance" — is an rms value unless stated otherwise.

A resistor does not care which way the current flows — it obeys Ohm's law at every instant. So when the voltage peaks, the current peaks; when the voltage is zero, the current is zero. The two waves rise and fall exactly together: the phase difference is zero.

Pure resistorI = V/R at every instant · phase difference φ = 0
power is always positive → the resistor steadily dissipates heat

This is why a resistive load — a heater, a filament bulb, a kettle — behaves the same on AC as on DC: it just gets hot.

A capacitor opposes AC not with resistance but with capacitive reactance X_C. Charge must flow in before the voltage across the plates can build up, so the current reaches its peak a quarter-cycle before the voltage — the current leads by 90°.

Capacitive reactanceX_C = 1 / (2πfC) = 1 / (ωC) · unit: ohm (Ω)
higher frequency → smaller X_C (capacitor "passes" high frequencies)
worked — reactance of a capacitor
C = 10 µF at f = 50 Hz?
X_C = 1 / (2π × 50 × 10×10⁻⁶) = ≈ 318 Ω
DC blocked: at f = 0 (DC), X_C → ∞ — a charged capacitor blocks steady current completely.

An inductor (a coil) opposes AC with inductive reactance X_L. By Lenz's law it resists any change in current, inducing a back-emf, so the current can only build up after the voltage — the current lags the voltage by 90°.

Inductive reactanceX_L = 2πfL = ωL · unit: ohm (Ω)
higher frequency → larger X_L (inductor "blocks" high frequencies)
worked — reactance of a coil
L = 0.2 H at f = 50 Hz?
X_L = 2π × 50 × 0.2 = ≈ 63 Ω
Mirror image of C: capacitor → current leads, X_C falls with f; inductor → current lags, X_L rises with f.

In a series R-L-C circuit the three oppositions cannot just be added — they point in different directions on a phasor diagram. R is along the current; X_L points up, X_C points down (they oppose each other). Their combination is the impedance Z, the total opposition to AC.

Impedance & phaseZ = √[ R² + (X_L − X_C)² ] · unit: ohm (Ω)
tan φ = (X_L − X_C) / R · V₀ = I₀ Z
worked — series circuit
R = 40 Ω, X_L = 60 Ω, X_C = 30 Ω?
Z = √[40² + (60−30)²] = √(1600+900) = 50 Ω

As frequency rises, X_L grows while X_C shrinks. At one special resonant frequency f₀ they become equal and cancel, so the impedance drops to just R — its minimum — and the current shoots up to a sharp peak.

Resonance conditionX_L = X_C → 2πf₀L = 1/(2πf₀C)
resonant frequency: f₀ = 1 / (2π√(LC)) · at resonance Z = R (minimum), I = maximum

Turning a radio tuning dial changes C so that f₀ matches one station's frequency: that station resonates loudly while all the others stay weak. This selective response is the heart of every tuner.

worked — tuning circuit
L = 2 mH, C = 8 nF?
f₀ = 1/(2π√(2×10⁻³ × 8×10⁻⁹)) = ≈ 40 kHz

A transformer is two coils on a shared iron core. Changing AC flux in the primary induces a voltage in the secondary; the voltages share the turns ratio. This only works with AC — and it is exactly why the grid uses AC.

Ideal transformerV_s / V_p = N_s / N_p = I_p / I_s
power conserved: V_p I_p = V_s I_s (ideal, no loss)
  • Step-up at the plant — raise V to hundreds of kV, so the current I is tiny.
  • Low loss in the wires — line loss is I²R, so small I means almost no heat wasted over hundreds of kilometres.
  • Step-down at your street — a local transformer drops it back to a safe 230 V.
The big idea: transmit at high voltage / low current to beat the I²R loss, then step down for safe use.

Every piece fits together: the mains delivers a 50 Hz sine wave whose rms value is what your meter reads; resistors stay in phase, capacitors lead and inductors lag; together they set the impedance Z and a resonant frequency f₀; and transformers move that AC power efficiently across the country.

  1. AC reverses direction as a sine wave; f = 1/T, ω = 2πf; Pakistan mains = 50 Hz.
  2. I_rms = I₀/√2; V_rms = V₀/√2 — quoted values are rms.
  3. Resistor: V and I in phase (φ = 0); power always dissipated.
  4. Capacitor: X_C = 1/(2πfC), current leads 90°; inductor: X_L = 2πfL, current lags 90°.
  5. Impedance Z = √[R² + (X_L − X_C)²]; tan φ = (X_L − X_C)/R.
  6. Resonance at f₀ = 1/(2π√(LC)): Z minimum, I maximum — radio tuning.
  7. Transformer: V_s/V_p = N_s/N_p; transmit high-V / low-I to cut I²R losses.
Applications: household mains & meters, radio/TV tuning circuits, the national grid, induction motors, and switch-mode chargers. Next: 📝 Practice · 📖 Full lecture
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