Alternating Current

2.1 Core concepts

• Alternating voltage varies like a sine function with time, v = vm sin ωt; the electric mains supply in homes and offices is such a sinusoidal voltage, and most electrical devices today require ac voltage (NCERT §7.1, p. 178).
• The main reason for preferring ac over dc is that ac voltages can be easily and efficiently transformed from one value to another using transformers, enabling economical long-distance transmission (NCERT §7.1, p. 178).
• For an ac source v = vm sin ωt connected across a pure resistor R, Kirchhoff's loop rule gives i = (vm/R) sin ωt = im sin ωt, so voltage and current are exactly in phase — they reach zero, minima and maxima at the same instants (NCERT §7.2, p. 178; Eqs. 7.1–7.3).
• Although the average of an ac current over a cycle is zero, Joule heating depends on i² (always positive); the average power dissipated in a resistor is P = (1/2) im²R, which is recast as P = I²R using rms current I = im/√2 = 0.707 im (NCERT §7.2, pp. 179–180; Eqs. 7.5c–7.7).
• The rms (or effective) voltage is V = vm/√2 = 0.707 vm; for household line voltage of 220 V (rms), the peak is vm = √2 × 220 V = 311 V (NCERT §7.2, p. 180).
• A phasor is a vector that rotates about the origin with angular speed ω; its vertical projection gives the instantaneous value of the sinusoidally varying voltage or current, and its magnitude equals the peak (amplitude) value (NCERT §7.3, p. 181).
• For an ac voltage applied to a pure inductor of self-inductance L, Kirchhoff's loop rule v − L(di/dt) = 0 leads to i = (vm/ωL) sin(ωt − π/2), so the current lags the voltage by π/2 (NCERT §7.4, pp. 181–182; Eqs. 7.10–7.12).
• Inductive reactance XL = ωL has units of ohm and is directly proportional to L and to the frequency; it limits current the way resistance does in a dc circuit, and the average power supplied to a pure inductor over a complete cycle is zero (NCERT §7.4, pp. 182–183; Eqs. 7.13–7.14).
• For an ac voltage applied to a pure capacitor C, q = Cv and i = dq/dt give i = ωCvm cos ωt = im sin(ωt + π/2), so the current leads the voltage by π/2 (NCERT §7.5, p. 184; Eq. 7.16).
• Capacitive reactance XC = 1/ωC has units of ohm and is inversely proportional to both C and frequency; the average power supplied to a pure capacitor over a complete cycle is zero (NCERT §7.5, pp. 184–185; Eqs. 7.17–7.19).
• For a series LCR circuit driven by v = vm sin ωt, the steady-state current is i = im sin(ωt + φ) with im = vm/Z, where the impedance is Z = √(R² + (XC − XL)²) and the phase angle satisfies tan φ = (XC − XL)/R (NCERT §7.6.1, pp. 187–188; Eqs. 7.25–7.27).
• If XC > XL, φ is positive and the circuit is predominantly capacitive (current leads source voltage); if XC < XL, φ is negative and the circuit is predominantly inductive (current lags source voltage) (NCERT §7.6.1, p. 188).
• Because VL and VC are π out of phase with each other, the rms voltages across individual elements can each exceed the source rms voltage; they must be combined using the Pythagorean theorem, not added algebraically (NCERT §7.6, Ex. 7.6, p. 190).
• Resonance occurs in a series LCR circuit when XC = XL; at the resonant angular frequency ω0 = 1/√(LC), the impedance is minimum (Z = R) and the current amplitude is maximum, im = vm/R (NCERT §7.6.2, pp. 188–189; Eq. 7.28).
• Resonance requires the presence of both L and C; with only an RL or RC combination, voltages across L and C cannot cancel each other and so there is no resonance (NCERT §7.6.2, p. 189; Points to Ponder 8, p. 199).
• Tuning circuits of radios/TVs exploit resonance: varying capacitance shifts the resonant frequency to match a particular broadcasting station, making the current at that frequency maximum (NCERT §7.6.2, p. 189).
• The instantaneous power for an LCR circuit averages to P = VI cos φ over a cycle; the quantity cos φ is called the power factor and decides what fraction of the apparent power VI is actually dissipated (NCERT §7.7, pp. 190–191; Eq. 7.30a).
• In a purely inductive or capacitive circuit φ = π/2, cos φ = 0 and no power is dissipated even though current flows — such a current is called wattless current (NCERT §7.7, p. 191).
• At resonance in an LCR circuit, XC = XL, so φ = 0 and cos φ = 1; maximum power is dissipated, all of it in the resistor (NCERT §7.7, p. 191).
• A low power factor in transmission means a larger current must be drawn to deliver the same power VI cos φ, increasing I²R losses; the power factor can be improved by adding a capacitor of appropriate value in parallel to cancel the lagging wattless current (NCERT §7.7, Ex. 7.7, pp. 191–192).
• A transformer has a primary coil of Np turns and a secondary coil of Ns turns wound on a common soft-iron core; an ac voltage on the primary sets up an alternating flux that links the secondary and induces an emf there by mutual induction (NCERT §7.8, p. 194).
• For an ideal transformer, vs/vp = Ns/Np, and assuming 100% efficiency, ipvp = isvs, giving ip/is = Ns/Np; so a step-up transformer (Ns > Np) raises voltage while reducing current, and a step-down transformer (Ns < Np) does the reverse (NCERT §7.8, p. 195; Eqs. 7.33–7.36).
• Real transformers suffer energy losses due to (i) flux leakage, (ii) resistance of the windings (I²R heating, reduced by using thick wire), (iii) eddy currents in the iron core (reduced by using a laminated core), and (iv) hysteresis (reduced by using a magnetic material with low hysteresis loss) (NCERT §7.8, pp. 195–196).
• Large-scale power transmission steps up the generator's voltage so that current — and hence I²R loss in the transmission line — is small; the voltage is later stepped down at sub-stations and again at utility poles before a 240 V supply reaches homes (NCERT §7.8, p. 196).
• The quality factor of a series LCR resonant circuit, Q = ω₀L/R = 1/(ω₀CR), measures sharpness of resonance — large Q means a sharply peaked current-vs-ω curve, so the circuit is highly selective in tuning (NCERT §7.6.2, p. 189; Quantity table p. 198).
• The bandwidth Δω of the resonance curve (full width at the points where im falls to im,max/√2) equals R/L; hence Q = ω₀/Δω — large Q ↔ narrow band ↔ sharper tuning. This explains why radio tuners use high-Q LC tanks (NCERT §7.6.2, p. 189).
• At resonance the rms voltages across L and C individually can be much larger than the source rms voltage (V_L = QV, V_C = QV with V_L and V_C π out of phase) — the so-called voltage magnification, which is why tuned circuits can drive sensitive detectors even with weak input signals (NCERT §7.6.2, p. 189; Ex. 7.6, p. 190).
• For a series LCR circuit the resistance R alone determines the actual energy dissipation, since cos φ = R/Z — at resonance Z = R and the entire source power is dissipated in R; for capacitive or inductive limits the power factor falls and dissipation drops (NCERT §7.7, p. 190).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig. 7.1 (p. 178) — Ac source connected to a pure resistor; current and voltage in phase.
• Fig. 7.2 (p. 178) — Plots of v and i versus ωt for a pure resistor, showing simultaneous zeros, minima and maxima.
• Fig. 7.3 (p. 179) — Relationship between peak current im and rms current I = im/√2 ≈ 0.707 im for sinusoidal ac.
• Fig. 7.4 (p. 181) — Phasor diagram and v–i waveforms for a resistor; V and I phasors aligned.
• Fig. 7.5, 7.6 (pp. 181–183) — Pure inductor circuit and phasor diagram: current phasor I is π/2 behind voltage phasor V.
• Fig. 7.7, 7.8 (pp. 184–185) — Pure capacitor circuit and phasor diagram: current phasor I is π/2 ahead of voltage phasor V.
• Fig. 7.10, 7.11 (pp. 186–187) — Series LCR circuit and the phasor construction giving Z and φ.
• Fig. 7.12 (p. 188) — Impedance diagram: right triangle with Z as the hypotenuse, R and (XC − XL) as legs.
• Fig. 7.14 (p. 189) — Variation of current amplitude im with ω for two values of R (100 Ω and 200 Ω) in an LCR circuit — peak at ω0, sharper peak for smaller R.
• Fig. 7.15 (p. 192) — Phasor decomposition of current into the power component (Ip in phase with V) and wattless component (Iq perpendicular to V).
• Fig. 7.16 (p. 194) — Two arrangements of primary and secondary windings on a soft-iron transformer core.

2.4 Common confusions / NTA trap points

• "Average current is zero, so no power is dissipated." Wrong — Joule heating depends on i², which is always positive; average power in a resistor is (1/2) im²R = I²R, not zero (NCERT §7.2, p. 179).
• Adding VR, VL, VC algebraically across elements of a series LCR circuit. The voltages are out of phase; the source voltage is the phasor sum, vm = √(vR² + (vC − vL)²) (NCERT Ex. 7.6, p. 190).
• Confusing XL = ωL with XC = 1/(ωC). XL grows with frequency; XC falls with frequency — questions on what happens to current when ν is doubled exploit this (NCERT §§7.4–7.5, pp. 182–185; Ex. 7.4).
• Direction of phase shift: in a pure inductor the current lags voltage by π/2; in a pure capacitor the current leads voltage by π/2. NTA distractors often flip the two (NCERT §§7.4–7.5, pp. 182–185).
• Assuming resonance is possible in any ac circuit. Resonance requires both L and C; an RL or RC circuit alone cannot resonate (NCERT §7.6.2, p. 189; Points to Ponder 8, p. 199).
• For a step-up transformer thinking "voltage and current both go up". The current is reduced by the same factor by which voltage is stepped up — this is what conserves energy (NCERT §7.8, p. 195; Points to Ponder 11, p. 199).
• Using rms values vs peak values inconsistently: P = VI cos φ uses rms values; P = (½) vm im cos φ uses peak values. Mixing them gives an answer off by 2.
• Quality factor and selectivity: smaller R (not larger R) gives larger Q and sharper resonance — easy to invert.
• Power factor in pure L or pure C: cos φ = 0, so even when current is large no power is dissipated; the source supplies energy in one quarter-cycle and the element returns it in the next.
• Forgetting that resonance frequency ω₀ = 1/√(LC) does NOT depend on R — only L and C set it; R controls sharpness, not location, of the resonance peak.
• Soft-iron core misconception: students think the core "carries the current" — actually it only confines the magnetic flux so it links primary and secondary efficiently; the windings carry current.

2.5 Key formulas table