Alternating Current
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Direct answer
Indian mains supply is 220 V rms at 50 Hz — an actual peak of about 311 V — and this chapter tracks what resistors, inductors and capacitors each do to that oscillation: the resistor keeps current in phase, the inductor drags current 90° behind voltage, the capacitor pushes it 90° ahead. The series LCR circuit binds them through Z = sqrt(R^2 + (X_L - X_C)^2), collapsing to pure R at resonance, ω = 1/sqrt(LC), and the transformer closes the chapter with AC's one-line case for high-voltage transmission.
What you must remember
- rms conversion: I_rms = I_0/sqrt(2); AC meters read rms; "220 V, 50 Hz" means V_0 ≈ 311 V and ω = 2π × 50 = 100π rad/s.
- Pure R: current and voltage in phase. Pure L: current lags by 90°, reactance X_L = ωL grows with frequency. Pure C: current leads by 90°, X_C = 1/(ωC) falls with frequency.
- Series LCR: Z = sqrt(R^2 + (X_L - X_C)^2), phase tan φ = (X_L - X_C)/R; the circuit is net inductive or net capacitive according to which reactance wins.
- Resonance: X_L = X_C at ω0 = 1/sqrt(LC); Z falls to R, current peaks, power factor hits 1; sharpness Q = ω0L/R — the acceptor circuit behind radio tuning.
- Average power P = V_rms I_rms cos φ with power factor cos φ = R/Z; a pure L or C carries wattless current (φ = 90°, zero average power).
- LC oscillations trade capacitor energy and inductor energy at f = 1/(2π sqrt(LC)), an electromagnetic twin of the mass-spring oscillator, damped by any resistance.
- Ideal transformer: V_s/V_p = N_s/N_p = I_p/I_s; real losses are copper (I^2R), eddy currents (met by lamination), hysteresis (met by soft cores) and flux leakage.
- A choke coil limits AC current through inductive reactance while dissipating almost no power — a resistor doing the same job would waste it as heat.
Tuning a radio the LCR way
A station broadcasts at 1 MHz and the set's inductance is 10 μH. Resonance demands ω0 = 1/sqrt(LC), so C = 1/(ω0^2 L) = 1/(3.94 × 10^13 × 10^-5) ≈ 2.5 nF — you tune by varying exactly this capacitance. With a coil resistance of 2 Ω the quality factor is Q = ω0L/R ≈ 31, so the resonance curve is a spike about 32 kHz wide: the wanted station drives a large current, stations 100 kHz away barely register. At resonance, also note the bookkeeping: the supply sees only R, the power factor is 1, and yet the inductor and capacitor individually support large voltages, Q times the supply voltage each — which is why series-resonant components need voltage ratings far above the source. The same physics run backwards is a rejector: a parallel LC trap passes everything except its resonant frequency.
Where students slip
Peak versus rms is the harvest field: a bulb labelled 220 V is an rms rating, but the insulation of the wiring must survive the 311 V peak — questions ask which value fuses, meters and insulation care about, and the answers differ. Second, the leads-and-lags assignment: current leads voltage in a capacitor and lags in an inductor; "ELI the ICE man" encodes it if mnemonics help, but the exam wants the direction stated correctly, and the wrong pairing is always an available option. Third, resonance changes the current, not the supply voltage — the current peaks because Z bottoms out at R while V is whatever the mains provide. Fourth, transformers run on AC only, because mutual induction needs a changing current; a transformer fed DC gives one transient and then a shorted winding, not a stepped-down voltage. Fifth, high-voltage transmission: for fixed power, raising V lowers I, and line loss I^2R falls as the square — students who halve the reasoning by forgetting the square lose the numerical.
Frequently asked questions
What is the peak voltage of the Indian mains?
220√2 ≈ 311 V, at a frequency of 50 Hz; the rms value 220 V is what meters read and ratings quote.
What is wattless current?
Current in a purely reactive element (ideal L or C): the phase difference is 90°, so the average power V_rms I_rms cos 90° is zero even though current flows.
What is the condition for resonance in a series LCR circuit?
X_L = X_C, giving ω0 = 1/sqrt(LC); impedance falls to R, current is maximum and in phase with the voltage, and the power factor is unity.
Why is electrical power transmitted at high voltage?
At fixed delivered power, higher voltage means lower current, and line loss I^2R falls with the square of the current; transformers make the voltage change possible only because the supply is AC.
What does a choke coil do that a resistor cannot?
It limits AC current through reactance ωL while dissipating almost no average power (φ near 90°), whereas a resistor limiting the same current wastes it entirely as heat.