Alternating Current

In this chapter: AC voltage and rms values, AC through a resistor, phasors, AC through an inductor and a capacitor, reactance, the series LCR circuit and impedance, resonance, power in AC circuits and power factor, wattless current, the transformer.

AC voltage and rms values

An AC source gives v = v_m sin ωt, where v_m is the peak value and ω = 2πν the angular frequency. Across a resistor, i = (v_m/R) sin ωt, so the current is in phase with the voltage. The average current over a cycle is zero, but the average power is not, because i² is never negative:

The rms value is the steady DC value that would give the same average power in a resistor. Meters read rms values, and when we say household supply is 220 V, that is the rms voltage; its peak is about 311 V.

Phasors

A phasor is a vector that rotates anticlockwise about the origin with angular speed ω. Its length is the peak value, and its projection on the vertical axis gives the instantaneous value. Drawing the voltage and current phasors together shows the phase difference between them at a glance.

AC through an inductor

For a pure inductor, v = L di/dt, which gives i = i_m sin(ωt − π/2). The current lags the voltage by π/2 (a quarter cycle).

The average power over a cycle is zero: energy stored in the magnetic field in one quarter cycle is returned to the source in the next.

AC through a capacitor

For a capacitor, q = Cv and i = dq/dt, which gives i = i_m sin(ωt + π/2). The current leads the voltage by π/2.

So a capacitor blocks DC but passes high-frequency AC, while an inductor does the opposite. Average power is again zero.

Series LCR circuit

In series, the same current flows through R, L and C, so take the current phasor as the reference. V_R is along I, V_L is 90° ahead and V_C is 90° behind. V_L and V_C are opposite, so they subtract, and the source voltage is the vector sum.

NCERT writes the current as i = i_m sin(ωt + φ) with tan φ = (X_C − X_L)/R, which is the same statement seen from the current's side. If X_L > X_C the circuit is inductive (current lags); if X_C > X_L it is capacitive (current leads).

Worked example: A series circuit has R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω, connected to a 220 V (rms) supply. Find the impedance, the current, the power factor and the average power.
Solution: Z = √(30² + 40²) = 50 Ω. I = 220/50 = 4.4 A. Power factor cos φ = R/Z = 30/50 = 0.6, so φ ≈ 53° with the voltage leading. Power P = VI cos φ = 220 × 4.4 × 0.6 = 580.8 W; check with I²R = 19.36 × 30 = 580.8 W.

Resonance

Since X_L rises and X_C falls with frequency, there is one frequency at which they are equal. Then Z = R, the current is maximum and in phase with the voltage.

At resonance V_L and V_C are equal and opposite, and each can be much larger than the source voltage. Resonance needs both L and C in the circuit; an RL or RC circuit cannot resonate. A smaller resistance gives a sharper resonance, which is what makes a radio tuner select one station.

Worked example: A 230 V variable-frequency source is connected to L = 5.0 H, C = 80 μF and R = 40 Ω in series. Find the resonant angular frequency and the rms current at resonance.
Solution: ω₀ = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(4 × 10⁻⁴) = 1/0.02 = 50 rad s⁻¹. At resonance Z = R, so I = 230/40 = 5.75 A.

Power in an AC circuit

• Purely resistive circuit or LCR at resonance: φ = 0, cos φ = 1, maximum power.
• Purely inductive or capacitive circuit: φ = π/2, cos φ = 0, no power is consumed even though current flows. Such a current is called wattless current.
• The current can be split into a component I cos φ in phase with V (which does the work) and I sin φ at 90° to V (wattless).

Industrial loads are mostly inductive (motors). A low power factor means a larger current for the same power and more loss in transmission lines, so capacitors are connected in parallel to bring the power factor close to 1.

Transformer

A transformer has two coils, the primary and the secondary, wound on a laminated soft iron core. An alternating current in the primary produces a changing flux in the core, which induces an emf in the secondary. It works only with AC. For an ideal transformer (no losses, all flux linked with both coils):

A step-up transformer (N_s > N_p) raises the voltage and lowers the current in the same ratio; a step-down transformer does the reverse. Power is not increased. Real transformers lose energy by:

• Flux leakage: not all the primary flux passes through the secondary; reduced by winding one coil over the other.
• Resistance of the windings: I²R heating; reduced by using thick wire for the high-current (low-voltage) winding.
• Eddy currents: reduced by a laminated core.
• Hysteresis: reduced by using a core material with a narrow hysteresis loop.

Power stations generate at a moderate voltage, step it up for long-distance transmission (lower current, less I²R loss), and step it down in stages near the consumer.

Common mistakes: (1) Adding V_R, V_L and V_C as plain numbers; they must be added as phasors. (2) Mixing peak and rms values in the same formula. (3) Mixing up lead and lag: in an inductor current lags, in a capacitor current leads. (4) Writing P = VI for an AC circuit that has reactance; include cos φ. (5) Thinking a step-up transformer increases power.

JEE and NEET focus

• rms and peak values, and the reading of AC meters.
• Reactances and how they change with frequency.
• Impedance, phase angle and voltage across each element in a series LCR circuit.
• Resonant frequency, current at resonance and the effect of R on sharpness (Q factor).
• Power factor, wattless current and average power.
• Transformer ratios, efficiency and sources of energy loss.

Practice questions