ECE 211 · Lecture 13: AC Power and Maximum Power Transfer

§1 · Peak phasors and RMS phasors

What is carried over from Lecture 12

Three results are assumed throughout. A sinusoid $v(t) = V_m\cos(\omega t + \theta)$ is represented by the phasor $\tilde{V} = V_m\angle\theta$ ; each element becomes an impedance; and every DC technique applies unchanged to the resulting complex network.

Z_R = R Z_L = jωL Z_C = −j/(ωC) Z = R + jX, Ṽ = Z Ĩ

Two conventions, one signal

A single sinusoid admits two phasor representations. They share the same angle and differ by a factor of $\sqrt{2}$ in magnitude. Both are in common use, and confusing them corrupts every power result by a factor of two. The two are distinguished here by an explicit subscript, and the same discipline is recommended on paper.

v(t) = V_m\cos(\omega t + \theta_v) \quad\Longrightarrow\quad \tilde{V}_m = V_m\angle\theta_v \quad\text{or}\quad \tilde{V}_{\mathrm{rms}} = \frac{V_m}{\sqrt{2}}\angle\theta_v

peak phasor: Ṽ_m = V_m∠θ_v RMS phasor: Ṽ_rms = (V_m/√2)∠θ_v Ṽ_rms = Ṽ_m/√2, same angle

The RMS value of a periodic signal is defined as the DC value that delivers the same real power to a resistor as the periodic signal does. For a sinusoid the definition evaluates to $V_{\mathrm{rms}} = V_m/\sqrt{2}$ and $I_{\mathrm{rms}} = I_m/\sqrt{2}$ .

The two circuits absorb the same average power in their 10 Ω resistors: on the left a sinusoid of amplitude V_m, on the right a constant V_rms. This equality is the definition of the RMS value, not a consequence of it.

V_m (V, peak) 10

V_rms = 10.0√2 = 7.071 V P = V_rms²10 Ω = 5.000 W

Using V_m where V_rms belongs would give 10.000 W, twice the truth.

Where the choice matters, and where it does not

The distinction is invisible to network analysis and decisive in power analysis, because impedance is a ratio of two phasors, in which the common factor $\sqrt{2}$ cancels, whereas power is a product of two phasors, in which it does not.

Quantity	Peak phasors	RMS phasors	Same answer?
Impedance Z = Ṽ/Ĩ	Ṽ_m/Ĩ_m	Ṽ_rms/Ĩ_rms	yes
KVL, KCL, nodal, mesh, Thevenin	unchanged	unchanged	yes
Real power P	½ V_m I_m cos(θ_v − θ_i)	V_rms I_rms cos(θ_v − θ_i)	yes, with the ½
Complex power S	½ Ṽ_m Ĩ_m*	Ṽ_rms Ĩ_rms*	yes, with the ½

The single most common error in this material

Writing $S = \tilde{V}\tilde{I}^{*}$ with peak phasors, and omitting the factor of one half, doubles every power. The safe procedure is to convert both source phasors to RMS before any power calculation is begun, then to use the unadorned products $S = \tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*}$ , $P = |\tilde{I}_{\mathrm{rms}}|^2 R$ , and $P_{L,\max} = |\tilde{V}_{T,\mathrm{rms}}|^2/(4R_T)$ .

The symbols of this lecture

Seven quantities arrive within two sections, three of which share the same dimensions and differ only in what they measure. The table is worth returning to.

Symbol	Name	Unit	Meaning
p(t)	instantaneous power	W	v(t) · i(t) at one instant
P	real (average) power	W	the average of p(t); the part that does work
Q	reactive power	VAR	energy borrowed and returned each cycle
S	complex power	V·A	P + jQ, one number for both
\|S\|	apparent power	V·A	V_rms I_rms; what a rating plate states
θ_S	power angle	degrees	θ_v − θ_i, equal to the impedance angle
pf	power factor	none	cos θ_S, always with "leading" or "lagging"

§2 · AC power: real, reactive, and complex

Instantaneous power carries a constant term and a 2ω ripple

In a DC circuit $P = VI$ and nothing moves. In an AC circuit the instantaneous power $p(t) = v(t)\,i(t)$ oscillates, and the useful quantity is its average over one period, $P = \frac{1}{T}\int_0^T p(t)\,dt$ . Expanding the product of two cosines separates the two parts:

p(t) = V_m I_m \cos(\omega t + \theta_v)\cos(\omega t + \theta_i) = \frac{V_m I_m}{2}\Big[\underbrace{\cos(2\omega t + \theta_v + \theta_i)}_{\text{average} = 0} + \underbrace{\cos(\theta_v - \theta_i)}_{\text{constant}}\Big]

The first term oscillates at twice the source frequency and averages to zero; the second is constant. The average, or real, power is therefore

\boxed{P = \frac{V_m I_m}{2}\cos(\theta_v - \theta_i) = V_{\mathrm{rms}} I_{\mathrm{rms}} \cos(\theta_v - \theta_i)}

One angle, three pictures

The single slider below drives three views of the same situation: the phasors, the power triangle, and the waveforms. The angle $\theta = \theta_v - \theta_i$ is the impedance angle, and it is the only thing that changes.

The phasors. The voltage is the reference; the current sits θ behind it for an inductive load.

The power triangle. Lagging (θ > 0) points up, leading (θ < 0) points down.

Above: v(t) and i(t) at the selected angle, on separate scales. Below: p(t) = v · i and its average P. Whenever θ ≠ 0 the power curve dips below zero for part of every cycle, and during those intervals energy flows back from the load to the source.

θ = θ_v − θ_i (degrees) 45

v(t), V_m = 10 V i(t), I_m = 2 A p(t) = v · i

P = 7.07 W Q = 7.07 VAR |S| = 10.00 VA pf = cos(45°) = 0.707 lagging, inductive

Complex power collects both parts

Writing the real power as the real part of a product of RMS phasors leads directly to the definition of complex power:

P = \mathrm{Re}\left\{V_{\mathrm{rms}} I_{\mathrm{rms}} e^{j(\theta_v - \theta_i)}\right\} = \mathrm{Re}\left\{\tilde{V}_{\mathrm{rms}}\, \tilde{I}_{\mathrm{rms}}^{*}\right\} \qquad\Longrightarrow\qquad \boxed{S = \tilde{V}_{\mathrm{rms}}\,\tilde{I}_{\mathrm{rms}}^{*} = P + jQ}

S = V_{\mathrm{rms}} I_{\mathrm{rms}}\cos(\theta_v - \theta_i) + j\,V_{\mathrm{rms}} I_{\mathrm{rms}}\sin(\theta_v - \theta_i)

The conjugate on the current is essential: it makes the angle of $S$ equal to $\theta_v - \theta_i$ rather than $\theta_v + \theta_i$ .

The power factor

The angle of $S$ equals the angle of the impedance, and its cosine is the power factor:

\theta_S = \theta_Z = \theta_v - \theta_i \qquad \mathrm{pf} = \cos\theta_S = \frac{P}{|S|} \qquad \theta_S = \cos^{-1}(\mathrm{pf})

Sign of θ_S	Q	Power factor	Current	Load
θ_S > 0	Q > 0	lagging	lags voltage	inductive
θ_S = 0	Q = 0	unity	in phase	resistive
θ_S < 0	Q < 0	leading	leads voltage	capacitive

The power factor alone is ambiguous

Because the cosine is even, a power factor of 0.8 corresponds to $\theta_S = \pm 36.87°$ . The word leading or lagging must always accompany the number, since the sign of $Q$ cannot otherwise be recovered. The convention describes the current relative to the voltage.

Power adds, no matter how the circuit is connected

Complex power obeys conservation exactly as energy does: $\sum P = 0$ and $\sum S = 0$ over all elements under the passive sign convention, so real and reactive power are conserved separately. Apparent power does not add: $\sum |S| \neq 0$ in general, because magnitudes of complex numbers do not sum.

Check yourself

A load draws 4 A rms at 120 V rms with a power factor of 0.5 leading. Find $P$ , $Q$ , and $Z$ .

Answer. $|S| = (120)(4) = 480$ VA and $\theta_S = -60°$ , since leading means $\theta_S < 0$ . Therefore $P = 480(0.5) = 240$ W and $Q = 480\sin(-60°) = -415.7$ VAR. The impedance follows from $Z = |V|/|I| \angle\theta_S = 30\angle -60° = 15 - j25.98$ Ω, whose negative reactance confirms a capacitive load.

§3 · Four quantities, two at a time

Given any two of $S$ , $\tilde{V}$ , $\tilde{I}$ , and $Z$ , the other two follow

Ohm's law $\tilde{V} = Z\tilde{I}$ and the definition $S = \tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*}$ together give four equivalent expressions for the complex power. Which one to use is decided by what is already known, not by preference.

\boxed{\;S = \tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*} = |\tilde{I}_{\mathrm{rms}}|^2 Z = \frac{|\tilde{V}_{\mathrm{rms}}|^2}{Z^{*}} = |S|\angle\theta_S = P + jQ \;}

S = \tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*} \qquad Z = \frac{\tilde{V}_{\mathrm{rms}}}{\tilde{I}_{\mathrm{rms}}}

The direct route. Conjugating the current is the only step that is easy to forget, and forgetting it flips the sign of $Q$ .

S = |\tilde{I}_{\mathrm{rms}}|^2 Z \qquad \tilde{V}_{\mathrm{rms}} = Z\,\tilde{I}_{\mathrm{rms}}

The preferred route for series elements, which share one current. Because $|\tilde{I}|^2$ is real, $P$ and $Q$ are read directly from the real and imaginary parts of $Z$ . This is the route used in Section 4.

S = \frac{|\tilde{V}_{\mathrm{rms}}|^2}{Z^{*}} \qquad \tilde{I}_{\mathrm{rms}} = \frac{\tilde{V}_{\mathrm{rms}}}{Z}

The preferred route for parallel elements, which share one voltage. Note the conjugate in the denominator: using $Z$ instead of $Z^{*}$ flips the sign of $Q$ and reports an inductor as a capacitor.

|\tilde{I}_{\mathrm{rms}}| = \sqrt{\frac{|S|}{|Z|}} \qquad |\tilde{V}_{\mathrm{rms}}| = \sqrt{|S|\,|Z|}

The route used when a load is specified by its rating rather than by its terminal quantities, as in Section 5. Angles follow from $\theta_S = \theta_Z$ , with the voltage or current angle assigned by whichever is taken as the reference.

§4 · Example D: the complete power balance

Find every power, and show that they sum to zero

A single loop driven by a 10∠0° V rms source, containing 40 Ω, −j10 Ω, and j40 Ω. All phasors here are RMS phasors, as the source label states.

The loop impedance is $Z = 40 - j10 + j40 = 40 + j30 = 50\angle 36.87°$ Ω, so one application of Ohm's law delivers the current:

\tilde{I}_{\mathrm{rms}} = \frac{\tm{xd.vs}{10\angle 0°}}{40 + j30} = \tfrac15 \angle -36.87°\ \text{A}_{\mathrm{rms}}

The current lags by $36.87°$ , which is expected: the net reactance is $+30$ Ω, so the loop is inductive and the power factor is $\cos 36.87° = 0.8$ lagging.

The source power carries a minus sign, because its current leaves the positive terminal:

S_{VS} = -\tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*} = -(10\angle 0°)\left(\tfrac15\angle 36.87°\right) = -2\angle 36.87° = -1.6 - j1.2\ \text{V·A}

Each passive element takes $S = |\tilde{I}_{\mathrm{rms}}|^2 Z$ with the same current, since the elements are in series:

Element	S = \|I\|² Z	P + jQ (V·A)
10∠0° V source	−(10∠0°)(0.2∠36.87°)	−1.6 − j1.2
40 Ω	(0.2)² · 40	+1.6 + j0
−j10 Ω	(0.2)² · (−j10)	0 − j0.4
j40 Ω	(0.2)² · (j40)	0 + j1.6
Total		0 + j0 ✓

\sum S = (-1.6 + 1.6) + j(-1.2 - 0.4 + 1.6) = 0 + j0 \ \checkmark

Both columns balance independently, which is the content of the conservation statement. The division of labour is worth noting: the resistor is the only element consuming real power, the inductor absorbs 1.6 VAR, the capacitor supplies 0.4 VAR, and the source makes up the 1.2 VAR difference.

The peak-phasor route to the same numbers

Had the source been stated as $v_s(t) = 14.14\cos(\omega t)$ V, the peak phasors would be $\tilde{V}_m = 14.14\angle 0°$ V and $\tilde{I}_m = 0.283\angle -36.87°$ A, and the factor of one half would be mandatory:

S_{VS} = -\tfrac12 \tilde{V}_m \tilde{I}_m^{*} = -\tfrac12 (14.14)(0.283)\angle 36.87° = -2\angle 36.87°\ \text{V·A} \ \checkmark

Omitting the one half would have returned $-4\angle 36.87°$ V·A, twice the true value.

§5 · Power factor correction

A poor power factor is paid for in current

A load specified by its real power and its power factor draws a line current $I_{\mathrm{rms}} = P/(V_{\mathrm{rms}}\cdot\mathrm{pf})$ . The real power is fixed by what the load does; the power factor is not. At pf 0.6 the same 1.2 kW requires 67 % more current than at unity, and that current heats the supply conductors, whose loss is $I^2R$ and is paid for by the utility.

Most industrial loads are inductive, so $Q > 0$ . A capacitor placed in parallel with the load supplies reactive power without consuming real power, which cancels part of $Q$ and leaves $P$ untouched.

The shunt capacitor is placed across the load terminals, so the load voltage is unchanged and only the line current falls.

The real power leg is fixed. Correction shortens the reactive leg from Q₁ to Q₂, and the highlighted segment is the Q_C supplied by the capacitor.

target power factor 0.9

Q₂ = 581 VAR Q_C = 1019 VAR C = 46.9 μF I_line 8.33 → 5.56 A rms (−33.3 %)

The design equation

The correction is designed on the power triangle. The real power is unchanged, so only the reactive leg moves:

Q_1 = P\tan\theta_1 \qquad Q_2 = P\tan\theta_2 \qquad Q_C = Q_1 - Q_2 = P\left(\tan\theta_1 - \tan\theta_2\right)

A capacitor across an RMS voltage $V_{\mathrm{rms}}$ supplies $Q_C = \omega C V_{\mathrm{rms}}^2$ , from which the required capacitance follows:

\boxed{C = \frac{Q_C}{\omega V_{\mathrm{rms}}^2} = \frac{P\left(\tan\theta_1 - \tan\theta_2\right)}{\omega V_{\mathrm{rms}}^2}}

For the values above, $P = 1.2$ kW at 240 V rms and 60 Hz, correcting from pf 0.6 lagging ( $\theta_1 = 53.13°$ ) to pf 0.9 lagging ( $\theta_2 = 25.84°$ ):

Q_C = 1200(1.3333 - 0.4843) = 1019\ \text{VAR} \qquad C = \frac{1019}{(377)(240)^2} = 46.9\ \mathrm{\mu F}

Correction does not reduce the load current

The capacitor is placed across the load, so the load voltage and therefore the load current are unchanged. What falls is the line current upstream of the capacitor, because the reactive component now circulates between the capacitor and the load instead of travelling back to the source. Over-correcting past unity is possible and makes the power factor leading, which restores the current penalty with the opposite sign.

§6 · Example C: an equivalent in the phasor domain

The port experiments, unchanged

Maximum power transfer requires a Thevenin equivalent, so one is constructed here. The procedure of Lecture 5 carries over without modification: the open-circuit voltage and the short-circuit current are computed, and their ratio is the Thevenin impedance. Dependent sources remain active throughout.

Reading the schematic

The controlling mesh on the left and the controlled network on the right share no wire. They are coupled only through the two dependent sources: the voltage source $\tilde{V}_0/3$ in the left mesh is controlled by the port voltage $\tilde{V}_0$ , and the current source $4\tilde{I}_0$ on the right is controlled by the mesh current $\tilde{I}_0$ .

The problem

The source is $6\cos(10t)$ V, so $\omega = 10$ rad/s and $\tilde{V}_s = 6\angle 0°$ V. Find the Thevenin equivalent at a and b.

Impedances at ω = 10 rad/s

\tm{xc.cc}{Z_C = \frac{-j}{10 \cdot \frac{1}{20}} = -j2\ \Omega} \qquad \tm{xc.lc}{Z_L = j(10)(1) = j10\ \Omega}

Experiment 1: short the port

A short across a-b forces $\tilde{V}_0 = 0$ , which switches the dependent voltage source off and shorts out both reactive elements. The left mesh is then a bare 4 Ω loop:

\tilde{I}_0 = \frac{6\angle 0°}{4} = 1.5\angle 0°\ \text{A} \;\Longrightarrow\; \boxed{\tilde{I}_N = 4\tilde{I}_0 = 6\angle 0°\ \text{A}}

Experiment 2: open the port, and write two equations

With the port open, $\tilde{V}_T = \tilde{V}_0$ . One KVL around the left mesh and one KCL at the top node of the right network suffice:

\text{KVL:}\quad -6\angle 0° + \tm{xc.r4}{4\tilde{I}_0} + \tm{xc.dvs}{\frac{\tilde{V}_0}{3}} = 0

\text{KCL:}\quad -\tm{xc.dcs}{4\tilde{I}_0} + \tm{xc.cc}{\frac{\tilde{V}_0}{-j2}} + \tm{xc.lc}{\frac{\tilde{V}_0}{j10}} = 0

Solve for the open-circuit voltage

The KCL equation gives $4\tilde{I}_0 = \tilde{V}_0\left(\frac{j}{2} - \frac{j}{10}\right) = j0.4\,\tilde{V}_0$ . Substituting into the KVL equation:

\tilde{V}_0\left(\tfrac13 + j0.4\right) = 6 \;\Longrightarrow\; \boxed{\tilde{V}_T = 11.52\angle -50.19°\ \text{V} = 7.38 - j8.85\ \text{V}}

The impedance is the ratio of the two experiments

Z_T = \frac{\tilde{V}_T}{\tilde{I}_N} = \frac{11.52\angle -50.19°}{6\angle 0°} \;\Longrightarrow\; \boxed{Z_T = 1.92\angle -50.19° = 1.23 - j1.48\ \Omega}

The reactance is negative, so the equivalent is capacitive at this frequency.

Both equivalents follow from the same two experiments: the Thevenin source behind Z_T on the left, and the Norton source across Z_N = Z_T on the right. Each is valid only at ω = 10 rad/s. This equivalent is loaded in Section 7.

Independent check: the test-source method

The same $Z_T$ must follow from driving the dead circuit. With the independent source set to zero and a test voltage $\tilde{V}_x$ applied at the port, the left mesh gives $4\tilde{I}_0 + \tilde{V}_x/3 = 0$ , so $\tilde{I}_0 = -\tilde{V}_x/12$ . KCL at the port, with $\tilde{I}_x$ entering from the test source:

\tilde{I}_x + 4\tilde{I}_0 = j0.4\,\tilde{V}_x \;\Longrightarrow\; \tilde{I}_x = \tilde{V}_x\left(\tfrac13 + j0.4\right)

Z_T = \frac{\tilde{V}_x}{\tilde{I}_x} = \frac{1}{\frac13 + j0.4} = 1.23 - j1.48\ \Omega \ \checkmark

The equivalent is tied to one frequency

$Z_T$ was evaluated at $\omega = 10$ rad/s. At any other frequency the capacitor and the inductor present different impedances, and the whole equivalent must be recomputed. A phasor-domain equivalent is a single-frequency object.

§7 · Maximum power transfer

The conjugate match

A Thevenin equivalent $\tilde{V}_{T,\mathrm{rms}}$ behind $Z_T = R_T + jX_T$ drives a load $Z_L = R_L + jX_L$ . The load current and the real power delivered to the load are

\tilde{I}_L = \frac{\tilde{V}_{T,\mathrm{rms}}}{(R_T + R_L) + j(X_T + X_L)} \qquad P_L = \frac{|\tilde{V}_{T,\mathrm{rms}}|^2 R_L}{(R_T + R_L)^2 + (X_T + X_L)^2}

The reactances appear only in the denominator, and only as the sum $X_T + X_L$ , so making that sum vanish is free power. With $X_L = -X_T$ the expression reduces to the resistive case of Lecture 5, whose maximum sits at $R_L = R_T$ . Together:

\boxed{Z_L = Z_T^{*}, \quad\text{that is}\quad R_L = R_T \ \text{ and } \ X_L = -X_T} \qquad P_{L,\max} = \frac{|\tilde{V}_{T,\mathrm{rms}}|^2}{4R_T}

The equivalent Z_T = 100 + j50 Ω driving the adjustable load.

P_L against R_L at the selected X_L. The ceiling is reached only when X_L = −50 Ω.

R_L (Ω) 100

X_L (Ω) -50

Z_L = 100 + j(−50) Ω P_L = 0.2500 W (100.0 % of the ceiling; best R_L here: 100.0 Ω)

With $\tilde{V}_{T,\mathrm{rms}} = 10\angle 30°$ V rms and $Z_T = 100 + j50$ Ω, the matched load is $100 - j50$ Ω, the loop becomes purely resistive at 200 Ω, and $P_L = (10/200)^2(100) = 0.25$ W.

Applied to the Section 6 equivalent

The equivalent constructed in Section 6 was given as a peak phasor, so it must be converted before the ceiling formula is used:

\tilde{V}_{T,\mathrm{rms}} = \frac{11.52}{\sqrt{2}}\angle -50.19° = 8.15\angle -50.19°\ \text{V}_{\mathrm{rms}} \qquad Z_T = 1.23 - j1.48\ \Omega

Z_L = Z_T^{*} = 1.23 + j1.48\ \Omega \qquad P_{L,\max} = \frac{(8.15)^2}{4(1.23)} = \boxed{13.5\ \text{W}}

The check by current confirms it: the matched loop is resistive at $1.23 + 1.23 = 2.46$ Ω, so $|\tilde{I}_{\mathrm{rms}}| = 8.15/2.46 = 3.31$ A rms and $P_L = (3.31)^2(1.23) = 13.5$ W.

Every quantity here is RMS

Inserting the peak phasor $11.52\angle-50.19°$ V into $|\tilde{V}_T|^2/(4R_T)$ returns 27.0 W, exactly twice the truth. Either divide by $\sqrt{2}$ first, as above, or use $P_{L,\max} = |\tilde{V}_{T,m}|^2/(8R_T)$ .

When the load is forced to be resistive

If the load is constrained to $Z_L = R_L + j0$ , the reactance $X_T$ can no longer be cancelled and survives in the denominator:

P_L = \frac{|\tilde{V}_{T,\mathrm{rms}}|^2 R_L}{(R_T + R_L)^2 + X_T^2}

Setting $dP_L/dR_L = 0$ now yields a different condition, in which the leftover reactance participates:

\boxed{R_L = \sqrt{R_T^2 + (X_T + X_L)^2} \;\xrightarrow{\;X_L = 0\;}\; R_L = \sqrt{R_T^2 + X_T^2} = |Z_T|}

For $Z_T = 100 + j50$ Ω, $R_L = \sqrt{100^2 + 50^2} = 111.8$ Ω and

P_L = \frac{10^2 (111.8)}{(211.8)^2 + 50^2} = \frac{11180}{47360} = \boxed{0.236\ \text{W}}

The resistive load recovers about 94 % of the conjugate-matched ceiling. The sliders confirm both results: setting $X_L = 0$ moves the peak of the curve to 111.8 Ω and lowers it to 0.236 W.

Why the derivative gives

R_L = |Z_T|

Write $D = (R_T + R_L)^2 + X_T^2$ and $P_L = |V|^2 R_L / D$ . Then

\frac{dP_L}{dR_L} = |V|^2\,\frac{D - R_L\cdot 2(R_T + R_L)}{D^2}

The numerator vanishes when $(R_T + R_L)^2 + X_T^2 = 2R_L(R_T + R_L)$ , that is $R_T^2 + X_T^2 = R_L^2$ , hence $R_L = |Z_T|$ . Setting $X_T = 0$ recovers the resistive result $R_L = R_T$ of Lecture 5, so the two statements are consistent.

§8 · Your turn

Practice: from the phasor domain to the power triangle

The circuit exercises the whole lecture: the phasor convention, the power calculation, and both maximum power conditions. It should be attempted on paper before the hints are opened, in order.

A source $v_s(t) = 20\cos(1000t)$ V in series with 30 Ω and 40 mH feeds the terminals a and b, where a load $Z_L$ is connected. Find (a) the Thevenin impedance at a-b, (b) the load absorbing maximum average power and that power, (c) the best purely resistive load and its power, and (d) the complex power delivered by the source and its power factor under the matched load of part (b).

Hint 1: fix the frequency and the phasor convention first

The source frequency is $\omega = 1000$ rad/s, and the source is stated as a cosine amplitude, so $20\angle 0°$ V is a peak phasor. Every answer below is an average power, so the RMS phasor $\tilde{V}_{T,\mathrm{rms}} = (20/\sqrt{2})\angle 0° = 14.14\angle 0°$ V rms should be formed before part (b) is started.

Hint 2: the Thevenin impedance needs no test source

There is no dependent source, so $Z_T$ is read off by turning the independent source into a short and combining what remains in series. Convert the inductor first: $Z_L = j\omega L = j(1000)(0.04)$ .

Hint 3: which condition applies to which part

Part (b) is the conjugate match, $Z_L = Z_T^{*}$ , so the ceiling $|\tilde{V}_{T,\mathrm{rms}}|^2/(4R_T)$ may be used directly. Part (c) is the constrained case, $R_L = |Z_T|$ , and the general expression for $P_L$ must be evaluated because the reactance no longer cancels.

Solution

(a) At $\omega = 1000$ rad/s the inductor presents $j\omega L = j(1000)(0.04) = j40$ Ω. With the source shorted, the resistor and the inductor remain in series:

\boxed{Z_T = 30 + j40\ \Omega = 50\angle 53.13°\ \Omega}

(b) The conjugate match is $Z_L = Z_T^{*} = 30 - j40$ Ω. With the RMS Thevenin voltage $14.14\angle 0°$ V rms:

P_{L,\max} = \frac{|\tilde{V}_{T,\mathrm{rms}}|^2}{4R_T} = \frac{200}{120} = \boxed{1.667\ \text{W}}

The check by current: the loop is resistive at $30 + 30 = 60$ Ω, so $\tilde{I}_{\mathrm{rms}} = 14.14/60 = 0.2357$ A rms and $P_L = (0.2357)^2(30) = 1.667$ W ✓.

(c) With the load forced resistive, $R_L = |Z_T| = 50$ Ω and

P_L = \frac{(14.14)^2 (50)}{(30 + 50)^2 + 40^2} = \frac{10000}{8000} = \boxed{1.25\ \text{W}}

which is 75 % of the matched ceiling. The penalty is larger here than in Section 7 because the reactance of this source is comparatively large.

(d) Under the matched load the current is $\tilde{I}_{\mathrm{rms}} = 0.2357\angle 0°$ A rms, in phase with the source voltage because the loop reactance cancels. Therefore

S_{\text{src}} = \tilde{V}_{\mathrm{rms}}\tilde{I}_{\mathrm{rms}}^{*} = (14.14)(0.2357)\angle 0° = \boxed{3.333 + j0\ \text{V·A}} \qquad \mathrm{pf} = 1

The source supplies 3.333 W, of which the load receives 1.667 W, so the efficiency at the matched load is 50 %, exactly as in the resistive case of Lecture 5. The reactive powers also balance: the inductor absorbs $(0.2357)^2(40) = 2.222$ VAR and the load capacitance supplies the same 2.222 VAR, leaving the source with $Q = 0$ .

A trap worth naming

Inserting the peak phasor $20\angle 0°$ V into $P_{L,\max} = |\tilde{V}_T|^2/(4R_T)$ without conversion returns $400/120 = 3.33$ W, exactly twice the truth. This is the error the subscripts of Section 1 exist to prevent.