Alternating Current

A direct current flows in one direction with a fixed value. An alternating current reverses direction periodically and, in the case we study, varies sinusoidally with time. The household supply in India is an alternating voltage of 220 V at 50 Hz. The word "ac" is used loosely for both alternating current and alternating voltage; the phrase "ac voltage" is standard even though it is redundant.

Why ac and not dc? Three reasons the examiner likes to see. First, ac voltages can be stepped up or down easily and efficiently with a transformer, which is what makes long-distance transmission practical. Second, most electrical energy is generated by ac generators, which are simpler than dc machines. Third, ac can be converted to dc where needed with a rectifier, while the reverse is harder.

The whole chapter rests on one input: a sinusoidal voltage v = v_m sin ωt applied to a circuit element, where v_m is the amplitude (peak value) and ω = 2πν is the angular frequency. We ask, for each element and then for the series LCR combination, what current flows, how it is related in phase to the voltage, and how much power is absorbed on average. Keep the NCERT symbols: lowercase v and i for instantaneous values, v_m and i_m for peak values, uppercase V and I for rms values. Mixing them up is the commonest source of lost marks in the numericals.

Connect a resistor R across v = v_m sin ωt. By Kirchhoff's loop rule v_m sin ωt = iR, so i = (v_m/R) sin ωt = i_m sin ωt with i_m = v_m/R. Ohm's law holds at every instant and the current is in phase with the voltage: both are zero together and peak together. On the graph of v and i against ωt the two sine curves reach their maxima at the same instants; in the phasor diagram the phasors v and i lie along the same line.

The average of a sine over a full cycle is zero, so the average current is zero. That does not mean no energy is dissipated, because Joule heating goes as i², which is always positive. The instantaneous power is p = i²R = i_m²R sin² ωt. Using sin² ωt = ½(1 − cos 2ωt) and the fact that cos 2ωt averages to zero over a cycle, the average power is P = ½ i_m²R.

To write this like the dc formula, define the root mean square (rms) or effective current I = √(average of i²) = i_m/√2 = 0.707 i_m. Then P = I²R. The rms current is the steady dc current that would produce the same heating in the same resistor in the same time. Similarly V = v_m/√2, and V = IR, P = IV = V²/R. The 220 V of the mains is an rms value; the peak is v_m = √2 × 220 = 311 V. Ac ammeters and voltmeters read rms values, which is why ac circuit calculations use rms quantities exactly as dc formulas use steady values.

In an inductor or a capacitor the current is not in phase with the voltage, and adding sinusoids of different phases by algebra is clumsy. The phasor is the tool that makes it easy. A phasor is a vector that rotates anticlockwise about the origin with angular speed ω. Its length is the amplitude of the quantity (v_m or i_m) and its projection on the vertical axis at time t gives the instantaneous value: v_m sin ωt for a voltage phasor at angle ωt.

Two quantities with a phase difference φ are represented by two phasors with the angle φ between them; since both rotate at the same ω, the angle between them never changes. For a resistor, the v and i phasors lie along the same direction. For an inductor, the current phasor is π/2 behind the voltage phasor (measured against the sense of rotation). For a capacitor, the current phasor is π/2 ahead.

Because the projections of phasors obey the same addition rules as the phasors themselves, voltages across elements in series can be added as vectors. That is exactly what we do for the LCR circuit in Section 7.6. When the examiner asks for a phasor diagram, draw the reference phasor along the horizontal, mark the angle between the phasors clearly with an arrow showing the sense of rotation, and label each phasor with its peak value. A labelled diagram with the phase angle marked usually carries one full mark on its own.

A pure inductor of inductance L (zero resistance) is connected to v = v_m sin ωt. The changing current induces a back emf L(di/dt), and since there is no resistor the loop rule gives v − L(di/dt) = 0, that is di/dt = (v_m/L) sin ωt. Integrating, i = −(v_m/ωL) cos ωt (the integration constant is zero because the current has no steady part). Since −cos ωt = sin(ωt − π/2),

i = i_m sin(ωt − π/2), with i_m = v_m/X_L and X_L = ωL.

X_L is the inductive reactance. It has the dimensions of resistance and its SI unit is the ohm, and it plays the role of resistance in limiting the current amplitude. It is directly proportional to frequency: X_L = 2πνL, so the graph of X_L against ν is a straight line through the origin. At dc (ν = 0) the reactance is zero, so an ideal inductor is a short circuit for dc.

The current lags the voltage by π/2, or one-quarter of a cycle. In the phasor diagram the current phasor is 90° behind the voltage phasor. On the v–i graph the current peaks one quarter period after the voltage does.

Power. The instantaneous power is p = vi = v_m sin ωt × i_m sin(ωt − π/2) = −v_mi_m sin ωt cos ωt = −(v_mi_m/2) sin 2ωt. Over a full cycle the average of sin 2ωt is zero, so the average power is zero. Physically, in one quarter cycle the source builds up the magnetic field of the inductor and in the next quarter the collapsing field returns that energy to the source. This is why a choke coil (an inductor) is used instead of a resistor to reduce the current in a fluorescent-tube circuit: it limits the current without wasting energy as heat.

A capacitor C is connected to v = v_m sin ωt. In a dc circuit a capacitor charges once, current flows only for a moment, and then it blocks the current. With ac the plates are charged, discharged and charged in the opposite sense every half cycle, so a current flows continuously in the wires even though no charge crosses the gap.

At any instant q = Cv = Cv_m sin ωt, and i = dq/dt = ωCv_m cos ωt. Writing cos ωt = sin(ωt + π/2),

i = i_m sin(ωt + π/2), with i_m = v_m/X_C and X_C = 1/ωC.

X_C is the capacitive reactance, in ohms. It is inversely proportional to frequency: X_C = 1/(2πνC), so the graph of X_C against ν is a rectangular hyperbola. At dc, ω → 0 and X_C → ∞: the capacitor blocks dc. At high frequency it offers almost no opposition. This is why a lamp in series with a capacitor glows brighter when the frequency of the supply is raised, and does not glow at all on dc.

The current leads the voltage by π/2: the current phasor is 90° ahead of the voltage phasor, and on the graph the current peaks a quarter period before the voltage.

Power. p = vi = v_m sin ωt × i_m sin(ωt + π/2) = v_mi_m sin ωt cos ωt = (v_mi_m/2) sin 2ωt, whose average over a cycle is zero. The capacitor is charged in one quarter cycle and returns the stored electric energy in the next. A pure inductor and a pure capacitor both pass current and both absorb no net power; the difference between them is only the sign of the phase shift.

A resistor R, an inductor L and a capacitor C are in series with v = v_m sin ωt. The same current i = i_m sin(ωt + φ) flows through all three, so the current phasor I is the natural reference. The board asks for the phasor-diagram solution, which goes in four steps.

Step 1: voltage phasors. The voltage across R, V_R, is in phase with I and has amplitude i_mR. The voltage across L, V_L, leads I by π/2 and has amplitude i_mX_L. The voltage across C, V_C, lags I by π/2 and has amplitude i_mX_C. So V_L and V_C are opposite to each other, along a line perpendicular to V_R.

Step 2: add them. By the loop rule the applied voltage is the phasor sum V = V_R + V_L + V_C. Adding the two collinear ones first gives a phasor of length i_m|X_C − X_L| perpendicular to V_R; by Pythagoras, v_m² = (i_mR)² + (i_mX_C − i_mX_L)².

Step 3: impedance. Hence i_m = v_m/Z where Z = √(R² + (X_C − X_L)²) is the impedance of the circuit, in ohms.

Step 4: phase. From the diagram tan φ = (X_C − X_L)/R. If X_C > X_L, φ is positive and the circuit is capacitive: current leads. If X_L > X_C, φ is negative and the circuit is inductive: current lags. The same result holds for an RL or RC circuit by setting X_C or X_L to zero.

Resonance. Since X_L rises with ω and X_C falls, there is one angular frequency at which X_L = X_C: ω₀L = 1/ω₀C, so ω₀ = 1/√(LC) and ν₀ = 1/(2π√(LC)). At this resonant frequency the impedance is minimum, Z = R, the current amplitude is maximum, i_m = v_m/R, and φ = 0, so the circuit behaves as pure resistance. The graph of i_m against ω is a peak at ω₀; a smaller R gives a taller, sharper peak. V_L and V_C are equal and opposite at resonance and cancel, so the whole applied voltage appears across R; each of V_L and V_C can individually exceed the source voltage. Resonance needs both an inductor and a capacitor; an RL or RC circuit cannot resonate. Tuning a radio uses exactly this: the capacitor is varied until ω₀ matches the frequency of the station wanted, and that signal alone drives a large current.

With v = v_m sin ωt and i = i_m sin(ωt + φ) in a series LCR circuit, the instantaneous power is p = vi = v_mi_m sin ωt sin(ωt + φ). Using 2 sin A sin B = cos(A − B) − cos(A + B), p = (v_mi_m/2)[cos φ − cos(2ωt + φ)]. The second term averages to zero over a cycle, so the average power is

P = (v_mi_m/2) cos φ = VI cos φ = I²Z cos φ.

P is the true (average) power, VI is the apparent power, and cos φ is the power factor. From the phasor diagram cos φ = R/Z, so P = I²R: on average, only the resistance dissipates energy. Four cases the examiner expects you to state:

• Pure resistor: φ = 0, cos φ = 1, P = VI. Maximum power.
• Pure inductor or pure capacitor: φ = ±π/2, cos φ = 0, P = 0. Current flows, but no power is consumed. Such a current is called wattless (or idle) current.
• Series LCR off resonance: 0 < cos φ = R/Z < 1, P = VI cos φ.
• Series LCR at resonance: X_L = X_C, φ = 0, cos φ = 1 and P = VI, maximum.

Wattless current in general: the current I can be resolved into I cos φ in phase with the voltage, which carries all the power, and I sin φ at 90° to the voltage, which carries none. The component I sin φ is the wattless current. A low power factor in a transmission line means a larger current for the same delivered power, so more I²R loss in the cables; that is why supply companies want cos φ close to 1 and add capacitors to inductive loads (motors) to raise it.

A transformer changes an alternating voltage from one value to another. Principle: mutual induction. A changing current in one coil induces an emf in a neighbouring coil linked by the same magnetic flux.

Construction: two coils, the primary (N_p turns) and the secondary (N_s turns), insulated from each other and wound on a laminated soft-iron core. Either the coils are wound one over the other on one limb, or on separate limbs of a rectangular core. The core is a closed loop so that almost all the flux from the primary links the secondary.

Working and turns ratio: an alternating voltage v_p across the primary sets up an alternating flux φ in the core. If every turn of both coils links the same flux, the induced emf per turn is the same, dφ/dt. So the secondary emf is ε_s = −N_s dφ/dt and the primary back emf is ε_p = −N_p dφ/dt. For an ideal transformer with negligible primary resistance, the applied voltage equals the back emf, giving v_s/v_p = N_s/N_p. If N_s > N_p it is a step-up transformer (v_s > v_p); if N_s < N_p it is a step-down transformer. For an ideal transformer there is no loss, so the power input equals the output: i_pv_p = i_sv_s. Combining, i_s/i_p = v_p/v_s = N_p/N_s. A step-up transformer raises the voltage and lowers the current in the same ratio. It does not create energy.

Energy losses in a real transformer (efficiency is still often above 90%):

• Flux leakage: not all the primary flux links the secondary. Reduced by winding one coil over the other on a well-designed core.
• Resistance of the windings: I²R heating in the copper. Reduced by using thick copper wire, especially in the high-current winding.
• Eddy currents: the changing flux induces currents in the iron core, heating it. Reduced by laminating the core, thin sheets insulated from each other.
• Hysteresis: the core is magnetised and demagnetised every cycle and warms up. Reduced by using soft iron, which has a narrow hysteresis loop.

Transmission: power P sent along a line of resistance R_c at voltage V carries a current P/V and loses (P/V)²R_c as heat. Raising V by a step-up transformer at the generating station (say 440 V to 132 kV or more) reduces the loss enormously; step-down transformers at substations and near homes bring it back to 220 V. A transformer works only on ac: a steady dc gives constant flux, no dφ/dt, and no secondary voltage.