Alternating Current

An alternating voltage supplied to a circuit varies sinusoidally with time: v = v_m sin ωt, where v_m is the peak (maximum) voltage and ω = 2πf is the angular frequency, f being the frequency in hertz (50 Hz for Indian household supply). Unlike direct current, AC reverses direction twice every cycle, which is what allows transformers to work and makes long-distance power transmission efficient.

A convenient way to track phase relationships in AC circuits is the phasor: a vector of length equal to the peak value, rotating counter-clockwise with angular speed ω. Its projection on a fixed (vertical) axis at any instant gives the instantaneous value of the quantity. The angle between the voltage phasor and the current phasor is the phase difference, φ.

For a pure resistor R connected across the AC source, Ohm's law holds at every instant: i = v/R = (v_m/R) sin ωt = i_m sin ωt, where i_m = v_m/R. Since i and v rise and fall together, the current is in phase with the voltage (φ = 0); their phasors point in the same direction at all times.

Because AC swings between +peak and −peak, quoting the peak value alone does not describe "how much" current is effectively present. We use the root-mean-square (rms), or virtual, value instead: square the instantaneous value, average it over a full cycle, then take the square root. Since the average of sin²ωt over a cycle is 1/2, this gives I_rms = i_m/√2 and V_rms = v_m/√2. Ordinary AC ammeters/voltmeters and the household "220 V" rating are all rms values.

Worked example: A 100 Ω resistor is connected to an AC source with peak voltage v_m = 200 V. Find the peak current, rms current and the average power dissipated.
Peak current: i_m = v_m/R = 200/100 = 2 A.
rms current: I_rms = i_m/√2 = 2/1.414 = 1.414 A.
rms voltage: V_rms = v_m/√2 = 200/1.414 = 141.4 V.
Average power: P_avg = I_rms²R = (1.414)² × 100 = 2 × 100 = 200 W (equivalently V_rmsI_rms = 141.4 × 1.414 ≈ 200 W).

Consider an ideal inductor of inductance L (zero resistance) connected to v = v_m sin ωt. By Kirchhoff's voltage law, the applied voltage equals the self-induced back-emf at every instant: v = L(di/dt), so di/dt = (v_m/L) sin ωt. Integrating with respect to time gives i = −(v_m/ωL) cos ωt, which can be rewritten as i = i_m sin(ωt − π/2), with i_m = v_m/ωL.

This shows the current lags the voltage by π/2 (90°). Physically, the self-induced emf always opposes the change in current, so the current cannot build up instantaneously with the voltage — it is delayed by a quarter cycle. A useful mnemonic is "ELI": in an Inductor, the EMF (voltage) leads the current I.

The quantity X_L = ωL is called the inductive reactance — it plays the same role that R plays for a resistor, limiting the peak current to i_m = v_m/X_L. Since X_L is proportional to frequency, an inductor offers negligible opposition to DC (ω = 0 ⇒ X_L = 0, so an ideal inductor behaves like a plain connecting wire for steady current) but increasing opposition as frequency rises.

Worked example: A 25 mH inductor is connected to a 220 V (rms), 50 Hz AC supply. Find the inductive reactance and the rms current.
X_L = 2πfL = 2π × 50 × 0.025 = 7.854 Ω.
I_rms = V_rms/X_L = 220/7.854 ≈ 28.0 A.

Let an ideal capacitor of capacitance C be connected to v = v_m sin ωt. The charge on the capacitor at any instant is q = Cv = Cv_m sin ωt. The current is the rate of change of charge: i = dq/dt = C v_m ω cos ωt, which can be written as i = i_m sin(ωt + π/2), with i_m = v_mωC.

So the current leads the voltage by π/2 (90°): charge (and hence current) must flow to change the voltage across a capacitor, so the current "arrives first." Mnemonic: "ICE" — in a Capacitor, the Icurrent leads the EMF.

Defining the capacitive reactance X_C = 1/(ωC), the peak current is i_m = v_m/X_C. As ω → 0 (DC), X_C → ∞, consistent with the fact that a capacitor blocks steady current once fully charged. As frequency increases, X_C falls, so a capacitor passes high-frequency AC more easily than low-frequency AC.

Worked example: A 15 µF capacitor is connected to a 220 V (rms), 50 Hz AC supply. Find the capacitive reactance and the rms current.
X_C = 1/(2πfC) = 1/(2π × 50 × 15×10⁻⁶) = 1/(4.712×10⁻³) ≈ 212.2 Ω.
I_rms = V_rms/X_C = 220/212.2 ≈ 1.04 A.

When R, L and C are connected in series to v = v_m sin ωt, the same current i flows through all three, but the voltages across them have different phases relative to i: V_R is in phase with i, V_L leads i by 90°, and V_C lags i by 90° — so V_L and V_C are exactly opposite (180° apart) and partially cancel. Adding the three voltage phasors, the resultant peak voltage is v_m = i_m√[R² + (X_L − X_C)²]. The quantity Z = √[R² + (X_L − X_C)²] is the circuit's impedance, and i_m = v_m/Z. The phase angle between current and voltage satisfies tanφ = (X_L − X_C)/R.

Three cases arise: if X_L > X_C the circuit is net inductive (current lags voltage); if X_L < X_C it is net capacitive (current leads voltage); if X_L = X_C the reactive parts cancel completely and the circuit behaves as a pure resistor — this special condition is called resonance. It occurs at the angular frequency ω₀ = 1/√(LC), where impedance is minimum (Z = R) and current is maximum, in phase with the voltage.

How narrow and tall the resonance peak is depends on the quality factor Q = ω₀L/R = (1/R)√(L/C). A larger Q (smaller R, or larger L/C ratio) gives sharper, more selective resonance. Resonance can occur only when both L and C are present — a pure LR or pure RC series circuit never shows this cancellation.

Worked example: A series LCR circuit has R = 50 Ω, L = 0.2 H, C = 20 µF, connected to a 200 V (rms), 50 Hz supply.
ω = 2π × 50 = 314.16 rad/s.
X_L = ωL = 314.16 × 0.2 = 62.83 Ω.
X_C = 1/(ωC) = 1/(314.16 × 20×10⁻⁶) = 159.15 Ω.
X_L − X_C = −96.32 Ω (circuit is net capacitive).
Z = √[50² + 96.32²] = √11778 ≈ 108.5 Ω.
I_rms = 200/108.5 ≈ 1.84 A.
tanφ = −96.32/50 = −1.926 ⇒ φ ≈ −62.6° (current leads voltage).
For the same L and C, the resonant angular frequency is ω₀ = 1/√(LC) = 1/√(0.2×20×10⁻⁶) = 1/√(4×10⁻⁶) = 500 rad/s (f₀ ≈ 79.6 Hz), and Q = ω₀L/R = (500×0.2)/50 = 2.

The instantaneous power delivered to an AC circuit is p = vi. For a series LCR circuit with v = v_m sin ωt and i = i_m sin(ωt − φ), multiplying and averaging over one full cycle (using the fact that the average of sinωt·sin(ωt−φ) over a cycle is (1/2)cosφ) gives the average power: P_avg = (v_mi_m/2) cosφ = V_rmsI_rms cosφ. The factor cosφ is called the power factor; it equals R/Z for a series LCR circuit.

The power factor ranges from 0 to 1. For a pure resistor, φ = 0 and cosφ = 1, so all the supplied power is dissipated. For a pure inductor or pure capacitor, φ = ±90° and cosφ = 0, so the average power is zero even though current flows — this current is called wattless (or idle) current, since energy is only exchanged back and forth between source and the reactive element, never permanently dissipated. In general, the current can be split into a component I_rmscosφ (in phase with voltage, does real work) and a component I_rmssinφ (90° out of phase, wattless).

A low power factor is undesirable in practice: inductive loads such as motors and transformers draw a larger current than necessary for the real power they use, increasing I²R losses in transmission lines. Power utilities often ask industrial consumers to improve their power factor (commonly by adding capacitors in parallel with inductive loads) to reduce this wasted current.

Worked example (continuing the LCR circuit above): R = 50 Ω, Z = 108.5 Ω, I_rms = 1.84 A, V_rms = 200 V.
Power factor: cosφ = R/Z = 50/108.5 ≈ 0.461, so φ ≈ 62.6°.
Average power: P_avg = V_rmsI_rmscosφ = 200 × 1.84 × 0.461 ≈ 169.8 W (checks against I_rms²R = 1.84² × 50 ≈ 169.8 W).
Wattless current component: I_rmssinφ = 1.84 × sin(62.6°) ≈ 1.84 × 0.888 ≈ 1.63 A.

Suppose a capacitor charged to q_m is connected across an inductor L (no resistance in the circuit). As the capacitor discharges through the inductor, its electric-field energy is converted into the inductor's magnetic-field energy, and then back again, repeatedly — this is called LC oscillation. Since there is no resistance to dissipate energy, the total energy stays constant: U = q²/(2C) + (1/2)Li² = q_m²/(2C). Differentiating this energy-conservation equation with respect to time and using i = dq/dt leads to d²q/dt² = −(1/LC)q, the equation of simple harmonic motion. Its solution is q(t) = q_m cos(ωt + θ), with angular frequency ω = 1/√(LC) — exactly the same expression as the resonant frequency of a driven series LCR circuit.

The analogy with SHM is close: charge q corresponds to displacement x, current i = dq/dt corresponds to velocity, L corresponds to mass (inertia opposing change of current), and 1/C corresponds to the spring constant (restoring effect). The current i = −ωq_m sin(ωt + θ) is 90° out of phase with the charge, just as velocity leads displacement by 90° in SHM. In any real LC circuit some resistance is always present, so the oscillations are damped and gradually die out as energy is dissipated as heat, unless continuously resupplied.

A transformer is a device that changes an AC voltage using mutual induction. It has a primary coil of N_p turns and a secondary coil of N_s turns, both wound on a common laminated soft-iron core so that nearly all the flux produced by the primary links the secondary. The changing flux (only possible with AC) induces an emf in the secondary coil. For an ideal transformer (no losses), V_s/V_p = N_s/N_p, and since input and output power are equal, I_p/I_s = N_s/N_p as well, giving V_s/V_p = N_s/N_p = I_p/I_s. A step-up transformer has N_s > N_p (raises voltage, lowers current); a step-down transformer has N_s < N_p.

Real transformers are not perfectly efficient: energy is lost as heat due to resistance of the windings (copper loss, I²R), and due to hysteresis and eddy currents in the iron core (core loss). Using a laminated core (thin iron sheets insulated from each other) greatly reduces eddy-current losses and improves efficiency, η = (P_out/P_in) × 100%, typically well above 90% in good transformers.

Worked example (LC oscillations): L = 20 mH, C = 50 µF.
ω = 1/√(LC) = 1/√(0.02 × 50×10⁻⁶) = 1/√(1×10⁻⁶) = 1000 rad/s.
f = ω/2π ≈ 159.2 Hz; period T = 2π√(LC) = 2π×1×10⁻³ ≈ 6.28 ms.

Worked example (transformer): An ideal transformer has N_p = 2000 turns and N_s = 200 turns; the primary is connected to 220 V (rms) AC.
V_s = V_p × (N_s/N_p) = 220 × (200/2000) = 22 V (a step-down transformer).
If the secondary supplies a current of 3 A, the primary current is I_p = I_s × (N_s/N_p) = 3 × 0.1 = 0.3 A.
Check: P_in = 220 × 0.3 = 66 W; P_out = 22 × 3 = 66 W — power is conserved, as expected for an ideal transformer.