Unit 19: Alternating Currents
1. Peak and RMS Values
In alternating current circuits, the instantaneous voltage or current varies sinusoidally with time. Two important measures are used:
- Peak value (V₀ or I₀): Maximum magnitude attained during a cycle.
- RMS value (Vrms or Irms): Root‑mean‑square (effective) value that produces the same heating effect as a DC of equal magnitude.
For a pure sinusoid:
V_rms = V₀ / √2
I_rms = I₀ / √2
Power in AC circuits is calculated using RMS values:
P = V_rms × I_rms × cos φ
AC Through \ R, \ L, \ C \> < /h3> < p>When an AC source is connected to a single passive element, the voltage and current exhibit specific phase relationships: < /p> < ul> < li> < strong>Resistor (R): < /strong> Voltage and current are in phase. Ohm’s law holds instantaneously: V(t) = I(t) R. In phasor form: V̂ = Î R. < /li> < li> < strong>Inductor (L): < /strong> The current lags the voltage by 90° (π/2 rad). Inductive reactance: X_L = ωL, where ω = 2πf. Phasor relation: V̂ = Î jX_L. < /li> < li> < strong>Capacitor (C): < /strong> The current leads the voltage by 90°. Capacitive reactance: X_C = 1/(ωC). Phasor relation: V̂ = Î (-jX_C). < /li> < /ul> < p>These relationships can be visualized using phasor diagrams, where each sinusoidal quantity is represented by a rotating vector (phasor) whose length equals the RMS value and whose angle represents the phase. < /p> < h2>2. Phasor Diagram < /h2> < p>A phasor is a complex number  = A ∠θ that encodes both magnitude (A) and phase (θ) of a sinusoidal quantity a(t) = A√2 cos(ωt + θ). In AC analysis: < /p> < ul> < li>The real part of the phasor corresponds to the instantaneous value at t = 0. < /li> < li>Phasors rotate counter‑clockwise at angular speed ω; however, for steady‑state analysis we consider them stationary, capturing only the relative phase. < /li> < li>Adding or subtracting phasors follows vector addition/subtraction rules, simplifying circuit equations. < /li> < /ul> < p>Example: For a series RL circuit, the voltage phasor V̂ is the vector sum of ÎR (in phase with current) and ÎjX_L (leading current by 90°). The resultant phasor gives the total voltage magnitude and its phase relative to the current. < /p> < h2>3. Series RLC Circuit < /h2> < p>When a resistor, inductor, and capacitor are connected in series with an AC source, the same current flows through each element. The total opposition to current is called impedance (Z). < /p> < h3>Impedance and Phase Angle < /h3> < p>The inductive and capacitive reactances oppose each other. Net reactance: X = X_L – X_C. Impedance magnitude: Z = √[R² + (X_L – X_C)²]. < /p> < p>Phase angle φ between source voltage and current: tan φ = (X_L – X_C)/R. < /p> < p>If X_L > X_C, the circuit is inductive (voltage leads current, φ > 0). If X_C > X_L, it is capacitive (voltage lags current, φ < 0). At resonance X_L = X_C, φ = 0 and the circuit behaves purely resistively. < /p> < h3>Phasor Representation < /h3> < p>In the phasor diagram for a series RLC: < /p> < ol> < li>Draw the current phasor Î along the reference axis (0°). < /li> < li>The voltage across the resistor V̂_R = ÎR lies on the same axis (in phase). < /li> < li>The inductor voltage V̂_L = ÎjX_L leads the current by +90°. < /li> < li>The capacitor voltage V̂_C = Î(-jX_C) lags the current by –90°. < /li> < li>The source voltage V̂ = V̂_R + V̂_L + V̂_C is the vector sum of these three. < /li> < p>The resulting phasor V̂ determines the magnitude |V̂| = ÎZ and the phase φ. < /p> < h2>4. Series Resonance and Quality Factor < /h2> < h3>Resonance Condition < /h3> < p>Series resonance occurs when the inductive and capacitive reactances cancel: X_L = X_C. At this frequency the impedance is minimum and purely resistive: Z_min = R. Consequently, the circuit current reaches its maximum value: I_max = V₀ / R. < /p> < h3>Resonant Frequency < /h3> < p>Setting ωL = 1/(ωC) gives the resonant angular frequency: ω₀ = 1/√(LC). In hertz: f₀ = 1/(2π√(LC)). < /p> < h3>Quality Factor (Q) < /h3> < p>The quality factor measures the sharpness of the resonance peak: Q = ω₀L / R = 1/(ω₀CR). A high Q indicates a narrow bandwidth and high selectivity. Bandwidth (Δf) is related by: Δf = f₀ / Q. < /p> < p>Energy perspective: Q also equals the ratio of energy stored in the reactive elements to the energy dissipated per radian in the resistor. < /p> < h2>5. Power in AC Circuits < /h2> < h3>Instantaneous and Average Power < /h3> < p>Instantaneous power: p(t) = v(t) i(t). For sinusoidal quantities, the average (real) power over a cycle is: P = V_rms I_rms cos φ, where cos φ is the power factor. < /p> < h3>Power Factor < /h3> < p>Power factor = cos φ = R / Z. It indicates the fraction of apparent power (S = V_rms I_rms) that is converted into useful work. < /p> < ul> < li>When φ = 0 (purely resistive), power factor = 1 (unity). < /li> < li>When φ = ±90° (purely reactive), power factor = 0 (no real power). < /li> < h3>Power at Resonance < /h3> < p>At series resonance, X_L = X_C ⇒ Z = R ⇒ φ = 0 ⇒ cos φ = 1. Hence, the power factor is unity and the average power delivered to the circuit is maximized: P_max = V_rms² / R. < /p> < h2>6. Illustrative Examples < /h2> < h3>Example 1: RMS Calculation < /h3> < p>A sinusoidal voltage has a peak value of 300 V. Compute its RMS value. < /p> < p>V_rms = 300 V / √2 ≈ 212 V. < /p> < h3>Example 2: Impedance of Series RLC < /h3> < p>Given R = 10 Ω, L = 0.2 H, C = 50 µF, and f = 60 Hz, find Z and φ. < /p> - ω = 2πf = 2π×60 ≈ 377 rad/s.
- X_L = ωL = 377×0.2 ≈ 75.4 Ω.
- X_C = 1/(ωC) = 1/(377×50×10⁻⁶) ≈ 53.1 Ω.
- Net reactance X = X_L – X_C ≈ 22.3 Ω (inductive).
- Z = √[R² + X²] = √[10² + (22.3)²] ≈ √[100 + 497] ≈ √597 ≈ 24.4 Ω.
- tan φ = X/R = 22.3/10 = 2.23 → φ ≈ arctan(2.23) ≈ 65.6° (voltage leads current).
Example 3: Resonant Frequency and Q < /h3> < p>For L = 0.1 H and C = 10 µF: < /p> - f₀ = 1/(2π√(LC)) = 1/(2π√(0.1×10×10⁻⁶)) ≈ 1/(2π√(1×10⁻⁶)) = 1/(2π×0.001) ≈ 159 Hz.
- ω₀ = 2πf₀ ≈ 1000 rad/s.
- Q = ω₀L / R. Assuming R = 5 Ω, Q = (1000×0.1)/5 = 20.
- f₀ = 1/(2π√(LC)) = 1/(2π√(0.1×10×10⁻⁶)) ≈ 1/(2π√(1×10⁻⁶)) = 1/(2π×0.001) ≈ 159 Hz.
- ω₀ = 2πf₀ ≈ 1000 rad/s.
- Q = ω₀L / R. Assuming R = 5 Ω, Q = (1000×0.1)/5 = 20.
Bandwidth Δf = f₀ / Q ≈ 159 Hz / 20 ≈ 8 Hz. < /p>
Summary < /h2>
This chapter has established the core concepts needed to analyse AC circuits: peak and RMS values, phase relationships in R, L, and C elements, phasor techniques for visualising and calculating voltages and currents, impedance and resonance in series RLC circuits, and power calculations including power factor. Mastery of these topics enables the solution of a wide range of problems in alternating‑current theory and prepares students for more advanced studies in electronics and electrical engineering. < /p>