Notes on Chapter 7: Alternating Current (AC)

7.1 Introduction

• Direct current (dc) sources and circuits have currents that do not change direction with time.
• Many voltages and currents vary with time in real life; an example is the electric mains supply, which varies sinusoidally in time. This is called alternating voltage (ac voltage) and the current driven by it is called alternating current (ac current).
• Why AC is predominant for electrical energy use:
  • AC voltages can be easily and efficiently transformed from one voltage to another by transformers.
  • AC can be transmitted economically over long distances.
• AC circuits exhibit characteristics that are exploited in devices (e.g., tuning a radio relies on AC properties).
• Terminology note: ac voltage and ac current refer to quantities that display simple harmonic time dependence; the phrases are common but technically redundant with voltage and current.
• Real-world relevance: the chapter introduces how AC voltages are applied to basic components and circuits, and how these ideas underpin power distribution and devices.

7.2 AC Voltage Applied to a Resistor

• Consider a resistor connected to an ac voltage source with a sinusoidal potential difference:
  • Source voltage: v(t)=vm\, ext{sin}(\,oldsymbol{}t) where $v</em>m$ is the amplitude and oldsymbol{} is the angular frequency (ω).
  • Across the resistor, the ac voltage is the same as the source: the current is given by Kirchhoff’s loop rule, yielding
    $i<em>R(t)=i</em>m\sin(\omega t)$
  • Ohm’s law for ac: $i<em>m=\frac{v</em>m}{R}\,,$ so the current through the resistor is also sinusoidal and in phase with the voltage:
    $i(t)=\frac{v_m}{R}\sin(\omega t)$
• Key consequence: for a pure resistor in an ac circuit, the voltage and current are in phase (zero phase difference).
• Power considerations:
  • Instantaneous power: $p(t)=v(t)i(t)=i(t)^2 R=R\left(\frac{v<em>m}{R}\sin(\omega t)\right)^2=\frac{v</em>m^2}{R}\sin^2(\omega t)$
  • Average current over a cycle is zero (the instantaneous current changes sign, but the average over a full cycle is zero).
  • Average power (over a cycle) is the Joule heating:
    $P<em>{ ext{avg}}=\langle p(t)\rangle=I</em>{ ext{rms}}^2 R = \left(\frac{i<em>m}{\sqrt{2}}\right)^2 R = \frac{v</em>m^2}{2R}\,.$
• rms values and peak relationships:
  • $I<em>{ ext{rms}}=\frac{i</em>m}{\sqrt{2}}\quad\text{and}\quad V<em>{ ext{rms}}=\frac{v</em>m}{\sqrt{2}}.$
  • For a household rms voltage of $V<em>{ ext{rms}}=220\,\text{V}$, the peak voltage is $v</em>m=\sqrt{2}\,V_{ ext{rms}}=311\,\text{V}$ (approx).
• Example 7.1: A light bulb rated at 100 W for a 220 V supply.
  • (a) Resistance: $R=\frac{V^2}{P}=\frac{(220)^2}{100}=484\ \,\Omega.$
  • (b) Peak source voltage: $v_m=\sqrt{2}\times 220\approx 311\,\text{V}.$
  • (c) RMS current: $I<em>{ ext{rms}}=\frac{P}{V</em>{ ext{rms}}}=\frac{100}{220}\approx 0.454\,\text{A}.$

7.3 Representation of AC Current and Voltage by Rotating Vectors — Phasors

• Phasor: a rotating vector that represents a sinusoidally varying quantity; the vertical projection corresponds to the instantaneous value at a given time.
• For a resistor, the voltage phasor and current phasor are in the same direction (zero phase difference).
• Phasor diagrams help visualize phase relationships between voltage and current across different circuit elements (R, L, C).
• Important note: phasors represent the steady-state, not initial transients; rotating with angular speed (\omega). The magnitude represents the peak value, while the projection on the vertical axis corresponds to the instantaneous sinusoid at a given time.

7.4 AC Voltage Applied to an Inductor

• For an inductor, the voltage across the source is ac: $v= v_m\sin(\omega t)$
• Kirchhoff’s loop rule gives, with the inductive emf:
  $v=L\frac{di}{dt}$
• Integrating (and using the condition that the average current is zero so the constant of integration is zero), the current is
  $i(t)=i_m\sin(\omega t-\tfrac{\pi}{2})$
• Inductive reactance: $X<em>L=\omega L$ and the current amplitude is $i</em>m=\frac{v<em>m}{X</em>L}=\frac{v_m}{\omega L}$
• Therefore, the current lags the voltage by $\pi/2$ (one-quarter cycle).
• Power: the instantaneous power to an inductor is $p(t)=vi=iv=L\frac{di}{dt}i$, which evaluates to a form whose average over a complete cycle is zero; hence, a pure inductor, on average, dissipates no power.
• Phasor relationship: in a purely inductive circuit, the current phasor lags the voltage phasor by 90 degrees ((\pi/2)).
• Example (7.2): If a pure inductor of inductance $L$ is driven by an AC source of amplitude $v<em>m$ at frequency $\omega$, the inductive reactance is $X</em>L=\omega L$ and the current amplitude is $i<em>m= v</em>m/X_L$ with a phase lag of (\pi/2).
• Example 7.2 (Power): the average power in a pure inductor over a cycle is zero.
• Simple numerical example (Example 7.2): If the inductor is excited by suitable values, the inductive reactance and rms current can be computed to illustrate the zero-average-power property.

7.5 AC Voltage Applied to a Capacitor

• For a capacitor, the charge on the capacitor is related to the voltage: $q=Cv\quad\Rightarrow\quad v=\frac{q}{C}$
• Current is the time derivative of charge: $i(t)=\frac{dq}{dt}=C\frac{dv}{dt}$
• With the source voltage $v= v<em>m\sin(\omega t)$, the current becomes $i(t)=i</em>m\sin(\omega t+\tfrac{\pi}{2})$
• Capacitive reactance: $X<em>C=\frac{1}{\omega C}$ and the current amplitude is $i</em>m=\frac{v<em>m}{X</em>C}=\omega C v_m$
• Thus, the current leads the voltage by 90 degrees ((\pi/2)).
• The rms current is related by: the current magnitude relation can be written as $i(t)=i<em>m\sin(\omega t+\tfrac{\pi}{2})=v</em>m/(\omega C)\sin(\omega t+\tfrac{\pi}{2}).$
• Power for a capacitor: instantaneous power is $p(t)=vi=i<em>m v</em>m\cos(\omega t)\sin(\omega t)=\tfrac{1}{2}i<em>m v</em>m\sin(2\omega t)$, and the average power over a complete cycle is zero.
• Key takeaway: current leads voltage by 90° in a pure capacitor; there is no average power dissipated in a pure capacitor.
• Example 7.3 and 7.4 illustrate how dc vs ac connections affect a capacitor-connected device; in dc, the capacitor eventually blocks current; in ac, the capacitor presents a reactance and allows current to flow with a phase lead.
• Example 7.4: A 15.0 µF capacitor at 220 V, 50 Hz.
  • Capacitive reactance: $X_C=\frac{1}{2\pi f C}=\frac{1}{2\pi(50)(15.0\times 10^{-6})}\approx 212\,\Omega.$
  • RMS current: $I<em>{\text{rms}}=\frac{V</em>{\text{rms}}}{X_C}=\frac{220}{212}\approx 1.04\text{ A}.$
  • Peak current: $I<em>m=\sqrt{2}\,I</em>{\text{rms}}\approx 1.47\text{ A}.$

7.6 AC Voltage Applied to a Series LCR Circuit

• Consider a series circuit containing a resistor (R), an inductor (L), and a capacitor (C) driven by a sinusoidal source: $v(t)=v_m\sin(\omega t).$
• Kirchhoff’s loop rule for the instantaneous voltages across the elements gives $v(t)=v<em>L(t)+v</em>R(t)+v_C(t)$ with driving relations:
  • $v_R= iR$
  • $v<em>L= iX</em>L= i(\omega L)$
  • $v<em>C= i(-X</em>C)= -i\frac{1}{\omega C}$
• Phasor/analytical approach yields the impedance and current:
  • Impedance magnitude: $Z=\sqrt{R^2+(X<em>L-X</em>C)^2}$ where $X<em>L=\omega L,\quad X</em>C=\frac{1}{\omega C}.$
  • Current amplitude: $i<em>m=\frac{v</em>m}{Z}$ and the rms current $I<em>{\text{rms}}=\frac{V</em>{\text{rms}}}{Z}.$ (Since $V<em>{\text{rms}} = v</em>m/\sqrt{2}$, this is consistent with $i<em>m= v</em>m/Z$.)
  • Phase angle: $\phi=\tan^{-1}\left(\frac{X<em>L-X</em>C}{R}\right)$
• Interpretation of phase:
  • If $X<em>C>X</em>L\, (XC>XL)$, the circuit is predominantly capacitive and the current leads the voltage (negative or positive phase depending on sign convention).
  • If $X<em>L> X</em>C$, the circuit is predominantly inductive and the current lags the voltage.
• Phasor diagram (Fig. 7.11) and impedance diagram (Fig. 7.12) summarize the relationships between VR, VL, VC, and I.
• Steady-state vs transient solutions:
  • Phasor analysis gives the steady-state solution, assuming transients have died out.
  • The full solution includes a transient component that decays with time; the long-time behaviour is governed by the steady-state solution.

7.6.1 Phasor-diagram solution (Key relations)

• For the series LCR circuit, with current phasor I and element voltages VR, VC, VL:
  • VR is in phase with I: VR = I R
  • VC lags I by 90°: VC = I X_C (96° phase difference with respect to I?) Actually, VC is along the same line but opposite direction to VL in the impedance diagram, and the lag/lead is captured by the net reactance.
  • VL leads I by 90°: VL = I X_L
• The impedance: $Z=\sqrt{R^2+(X<em>L-X</em>C)^2}$ and the current amplitude is $i<em>m=\frac{v</em>m}{Z}.$
• The phase angle between the source voltage and circuit current is given by $\phi=\tan^{-1}\left(\frac{X<em>L-X</em>C}{R}\right).$
• Resonance in a series LCR circuit occurs when $X<em>L=X</em>C\,,$ i.e. when $\omega<em>0L=\frac{1}{\omega</em>0C}$, which gives the resonant angular frequency
  $\omega<em>0=\frac{1}{\sqrt{LC}}\quad\Rightarrow\quad f</em>0=\frac{\omega_0}{2\pi}=\frac{1}{2\pi\sqrt{LC}}.$
• At resonance, the impedance is purely resistive: $Z=R$ and the current amplitude is $i<em>m=\frac{v</em>m}{R}.$
• Example 7.6/7.9 illustrates calculation of XL, XC, Z, current, and resonance conditions for given L, C, R and frequency f.

7.6.2 Resonance

• Concept: Resonance occurs in a series LCR circuit when the reactive effects cancel (i.e., XL=XC\$), leaving only the resistive part.
• At resonance, the current is maximum (for a given voltage): Im=\frac{vm}{R}$; the impedance is minimum:$Z_\text{min}=R.
• Practical implications and examples:
  • Example: with L=1.00 mH and C=1.00 nF, the resonant angular frequency is \omega0=\frac{1}{\sqrt{LC}}=1.0\times 10^6\ \text{rad/s}$and the resonant frequency is$f0\approx 159.1\ \text{kHz}.
  • In tuning circuits (e.g., radio receivers), changing C tunes the resonant frequency toward the signal to be received.
• Important note: resonance requires both L and C; an RL or RC circuit alone cannot exhibit resonance in the same sense.

7.7 POWER IN AC CIRCUIT: THE POWER FACTOR

• General power relation for an AC circuit with current i(t) and voltage v(t):
  P = \langle v(t)\,i(t)\rangle = V{ ext{rms}}\,I{ ext{rms}}\cos\phi,
  where \phi is the phase difference between voltage and current.
• In terms of impedance: P= I^2 Z\cos\phi = V^2/Z\cos\phi = VI\cos\phi.
• The power factor is defined as \text{pf}=\cos\phi and is a measure of how close the circuit is to delivering maximum power.
• Case discussions:
  • Case (i) Purely resistive network: \phi=0\Rightarrow \cos\phi=1; maximum power dissipation.
  • Case (ii) Purely inductive or purely capacitive network: \phi=\tfrac{\pi}{2}\Rightarrow \cos\phi=0; current flows without net energy dissipation (wattless current).
  • Case (iii) LCR series circuit: power dissipated is P=VI\cos\phi$where$\phi=\tan^{-1}\left(\frac{XC-XL}{R}\right).
  • Case (iv) Resonance: at resonance, XL=XC\Rightarrow \phi=0,$so$P=I^2R; maximum power dissipated through the resistive element at a given source amplitude.
• Power-factor improvement: can be achieved by adding a capacitor in parallel to partially cancel the wattless current component. Decompose current into:
  • I_p: in-phase (power) component with the voltage.
  • I_q: wattless component perpendicular to the voltage.
  • To improve pf, neutralize Iq with a leading wattless current Iq', so that the net wattless component cancels and the current in phase with the voltage is maximized.
• Example 7.8 (resistive-capacitive-inductive circuit): given a series LCR circuit, compute impedance, phase, power, and power factor.
  • L = 25.48 mH, C = 796 µF, R = 3 Ω, f = 50 Hz, V_peak = 283 V.
  • Compute:
    XL = 2π f L ≈ 8.0 Ω,
    XC = 1/(2π f C) ≈ 4.0 Ω, Z = \sqrt{R^2 + (XL - XC)^2} ≈ 5.0 Ω, Vrms = Vpeak/√2 ≈ 200 V, Irms = Vrms/Z ≈ 40 A, P = Irms^2 R ≈ 4800 W,
    φ = tan^{-1}((XL - XC)/R) ≈ tan^{-1}(4/3) ≈ 53.13°,
    pf = cosφ ≈ 0.60.
• Example 7.9 (resonance investigation): with the same circuit, vary frequency to find resonance where ω0 = 1/√(LC) and observe changes in impedance, current, and power at resonance.
• Example 7.10 (metal detector): relies on resonance; passage through a coil in a tuned circuit changes the circuit’s impedance, triggering an alarm when resonance is perturbed by metal.

7.8 TRANSFORMERS

• Purpose: to transform alternating voltage from one value to another using mutual induction.
• Structure: two coils—primary (Np turns) and secondary (Ns turns)—wrapped on a soft-iron core.
• For an ideal transformer (no losses, negligible winding resistance):
  • The induced emf in the secondary with Ns turns is
    vs= -\frac{d\phi}{dt}Ns\quad\Rightarrow\quad vs = \frac{Ns}{Np}\,vp. This yields
    \frac{vs}{vp}=\frac{Ns}{Np}.
  • The corresponding relation for currents (with power conserved, pin ≈ pout):
    ,ip vp = is vs \Rightarrow \frac{is}{ip}=\frac{Np}{Ns}. Using RMS values, the same relationships hold for rms voltages and currents:
    Vs = \frac{Ns}{Np} Vp,
    Is = \frac{Np}{Ns} Ip.
• Step-up vs step-down:
  • If Ns>Np, the transformer steps up the voltage (Vs > Vp) and current is reduced (Is < Ip).
  • If Ns
• Efficiency and losses (real transformers):
  • Real transformers have nonzero losses due to
    (i) flux leakage, (ii) winding resistance, (iii) eddy currents in the core, (iv) hysteresis losses in the core.
  • These losses can be mitigated by design choices (e.g., laminated cores, proper winding design).
  • In practice, transformers operate with efficiencies typically exceeding 95%.
• Practical note: The large-scale transmission and distribution of electrical energy rely on transformers to step up the generator voltage for transmission and then step it down for safe distribution and use in homes.

7.9 SUMMARY (Key results and formulas)

• AC voltage across a resistor in a circuit: v(t)=vm\sin(\omega t),\quad i(t)=\frac{vm}{R}\sin(\omega t),\quad
  \text{phase difference} = 0.
• rms quantities and power for a resistor in AC:
  • I{\text{rms}}=\frac{im}{\sqrt{2}},\quad V{\text{rms}}=\frac{vm}{\sqrt{2}}.
  • Average power: P=I{\text{rms}}^2 R=\frac{V{\text{rms}}^2}{R}.
• For an inductor: v= vm\sin(\omega t),\quad i= im\sin(\omega t-\tfrac{\pi}{2}),\quad XL=\omega L, with im=\frac{vm}{XL},$current lags by$\pi/2 and average power is zero.
• For a capacitor: v= vm\sin(\omega t),\quad i= im\sin(\omega t+\tfrac{\pi}{2}),\quad XC=\frac{1}{\omega C}, with im=\omega C v_m,$current leads by$\pi/2 and average power is zero.
• For a series LCR circuit (R, L, C in series) driven by v(t)=v_m\sin(\omega t):
  • Impedance: Z=\sqrt{R^2+(XL-XC)^2},\quad XL=\omega L,\quad XC=\frac{1}{\omega C}.
  • Current amplitude: im=\frac{vm}{Z},\quad \text{rms current} = \frac{V_{\text{rms}}}{Z}.
  • Phase angle: \phi=\tan^{-1}\left(\frac{XL-XC}{R}\right).
  • At resonance: XL=XC\Rightarrow Z=R\Rightarrow Im=\frac{vm}{R}.
• Power factor and energy flow:
  • P=VI\cos\phi\quad\text{with}\quad \text{pf}=\cos\phi.
  • In a purely inductive or purely capacitive circuit, \cos\phi=0\Rightarrow P=0\ (wattless\ current).
  • At resonance in an LCR circuit, all the voltage drop occurs across R (phase alignment yields P = I^2 R).
• Transformers (ideal):
  • Vs = \frac{Ns}{Np} Vp,\quad Is = \frac{Np}{Ns} Ip,\quad P{in}=P{out}.
  • Step-up: Ns>Np\Rightarrow Vs>Vp, Is<Ip.
  • Step-down: Ns
• Practical notes and “points to ponder”:
  • The rms value is typically used for ac quantities.
  • The phase relationships must be accounted for when adding voltages across series or parallel combinations due to phase differences.
  • Phasor diagrams provide a convenient steady-state representation; actual circuits may exhibit transient behavior before reaching steady-state.
  • In ac devices, the transformer voltage/current relationships do not violate energy conservation; currents adjust so that input and output powers balance (accounting for losses).

7.10 Points to Ponder (highlights and conceptual checks)

• RMS values: ac quantities are usually specified as rms values; peak values are related by vm=\sqrt{2}\,V{\text{rms}},\quad im=\sqrt{2}\,I{\text{rms}}.
• The power in an ac circuit is never negative; the average power depends on the phase relationship between voltage and current (the power factor).
• In RC, RL, and RLC circuits, resonance requires both L and C; pure L or pure C cannot resonate.
• In a real transformer, losses (flux leakage, winding resistance, eddy currents, hysteresis) reduce efficiency, but high-quality transformers achieve efficiencies well above 95%.

7.11 Exercises (selected representative items from the set)

• 7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
  • (a) What is the rms current? I{\text{rms}}=\frac{V{\text{rms}}}{R}=\frac{220}{100}=2.2\,\text{A}.
  • (b) What is the net power consumed over a full cycle? P{\text{avg}}=I{\text{rms}}^2 R= (2.2)^2\times 100=968\,\text{W}(\approx 968\text{ W}).
• 7.2 (a) If the peak voltage of an ac supply is 300 V, what is the rms voltage? V{\text{rms}}=\frac{vm}{\sqrt{2}}=\frac{300}{\sqrt{2}}\approx 212.1\,\text{V}.
  • (b) If the rms current is 10 A, what is the peak current? im=\sqrt{2}\,I{\text{rms}}=\sqrt{2}\times 10\approx 14.14\,\text{A}.
• 7.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms current in the circuit.
  • X_C=\frac{1}{2\pi f C}=\frac{1}{2\pi\times 60\times 60\times10^{-6}}\approx 44.1\,\Omega.
  • If there were only the capacitor, the current magnitude: I{\text{rms}}=\frac{V{\text{rms}}}{X_C} \approx \frac{110/\sqrt{2}}{44.1}\approx 1.77\,\text{A}.
• 7.6 A series LCR circuit with R = 20 Ω, L = 1.5 H, C = 35 μF is connected to a 200 V, 50 Hz source. When the source frequency equals the natural frequency, what is the average power transferred to the circuit in one complete cycle? (Hints: compute the resonant frequency, impedance, and current at resonance.)
• 7.8 A series LCR circuit with L = 5.0 H, C = 80 μF, R = 40 Ω, driven by a 230 V source at some frequency. (a) Determine the resonant frequency; (b) Determine the impedance and current at resonance; (c) Compute rms voltage drops across R, L, and C and show that the LC drop cancels at resonance.

7.12 Connections to prior principles and real-world relevance

• The ac analysis framework (rms values, phasors, impedance) generalizes dc concepts to time-varmiscus, enabling convenient comparison across components and simple prediction of voltages, currents, and power in AC circuits.
• The same core ideas underpin everyday devices: radio tuning, transformers powering the grid, and impedance matching in communications circuits.
• Ethical and practical implications: understanding power factor and transformer efficiency informs energy usage, utility design, and device efficiency improvements to reduce losses and environmental impacts.

7.13 Quick reference: Core formulas (recap)

• Resistor in AC: v(t)=vm\sin(\omega t),\quad i(t)=\frac{vm}{R}\sin(\omega t)\,;\quad X=R
• Inductor: XL=\omega L,\quad im=\frac{vm}{XL},\quad i(t)=i_m\sin(\omega t-\tfrac{\pi}{2})
• Capacitor: XC=\frac{1}{\omega C},\quad im=\omega C vm,\quad i(t)=im\sin(\omega t+\tfrac{\pi}{2})
• Series LCR impedance: Z=\sqrt{R^2+(XL-XC)^2},\quad \phi=\tan^{-1}\left(\frac{XL-XC}{R}\right)
• Resonant frequency: \omega0=\frac{1}{\sqrt{LC}},\quad f0=\frac{\omega_0}{2\pi}
• Power in AC: P=VI\cos\phi,\quad \text{pf}=\cos\phi
• Transformer relations (ideal): Vs=\frac{Ns}{Np}Vp,\quad Is=\frac{Np}{Ns}Ip,\quad P{in}=P{out}.$$