ECE 211 · The Cumulative Examination Reference

§1 · What the examination covers

The scope, and the three exclusions

Logistics

One sheet of 8.5 × 11 paper is permitted, with anything at all written or typed on both sides. The examination is cumulative.

Three topics are excluded

Mesh analysis is not on the examination; the branch current method is the loop-based method used in this course. Unbalanced three-phase power is not on the examination. Two-port networks are not on the examination.

The order of this page follows the order of the board review: the two systematic methods first, then equivalence, then the theorems, then the amplifier, then the two switched-circuit recipes, then the alternating-current half of the course. Each section states the rule and the procedure only. The circuits that exercise them are worked step by step in Lecture 24.

§2 · The two systematic methods

Both methods begin by counting

The count fixes the number of unknowns, which equals the number of equations that must be written, and each method has one place where its own law must not be applied.

Method	Unknowns	How many	Where the law fails
Branch current	branch currents	# backyards − # current sources	no KVL around a loop containing a current source
Node voltage	node voltages	# nodes − 1 reference − # voltage sources	no KCL at a voltage source, or else use a supernode

The count is the number of KVL equations in the first case and the number of KCL equations in the second. Once the unknowns are solved for, the question that was actually asked is answered from them; solving the network is not the same as answering the question.

Dependent sources change none of this counting. They are carried through the equations like any other source, with one extra relation supplied by the definition of the controlling variable, and they are never turned off.

Labels are written from the element laws

A node voltage adjacent to a known element is written down rather than named, which is what keeps the number of unknowns at the counted minimum. Two cases cover almost every circuit.

Across a voltage source, the label changes by the source value: 5 V is lost in crossing from the plus side to the minus side. Either node may be named first.

Across a resistor carrying a known current, the label changes by RI_a, and the current enters at the plus mark by the passive sign convention.

The passive sign convention

Current flowing into the plus terminal and out of the minus terminal makes the power positive, and positive power means power absorbed. Negative power means power delivered. The convention is applied to every element identically, sources included, so a source with negative absorbed power is supplying the circuit.

Developed in Lectures 2 and 3; dependent sources in Lecture 2, the supernode in Lecture 3.

§3 · The equivalence toolbox

Two circuits are equivalent when their terminal behavior is identical

Two elements are in series when they share one node exclusively, in which case the same current passes through both, and in parallel when they share two nodes, in which case the same voltage appears across both. Series resistances add; parallel conductances add.

Voltage division applies to elements in series: the output takes the fraction of V_s set by R₃ against the sum of the series resistances.

Current division applies to elements in parallel: the branch current through R₂ is set by the other resistance R₁ over the sum.

V_{R_3} = \frac{\tm{vd.r3}{R_3}}{\tm{vd.r1 vd.r2 vd.r3}{\sum R_s}}\, \tm{vd.vs}{V_s} \qquad\qquad I_{R_2} = \frac{\tm{id.r1}{R_1}}{\tm{id.r1 id.r2}{R_1 + R_2}}\, \tm{id.is}{I_s}

The numerator of the current divider is the resistance of the other branch, which is the single most common sign of a divider applied from memory rather than from the derivation. The two-resistor form given here is the only form of the current divider that is safe to memorize; for three or more parallel branches, conductances are used instead.

Y-Δ, when neither series nor parallel applies

Some resistor triples are neither in series nor in parallel; bridge networks are the classic case, and the transformation is what makes any reduction possible at all.

Y (also drawn as a T): three legs meeting at an internal node n

Δ (also drawn as a Π): a triangle over the same three terminals

The two networks are interchangeable because only the three terminals a, b, and c are visible to the rest of the circuit. The internal node of the Y is not a terminal of the network, so it may be created or destroyed freely. The subscripts are assigned by position: $R_1$ , $R_2$ , and $R_3$ sit at terminals a, b, and c respectively, while $R_a$ , $R_b$ , and $R_c$ sit opposite those terminals.

Each Y leg is the product of the two Δ branches that meet at its terminal, divided by the sum of all three Δ branches:

\tm{yd.r1}{R_1} = \frac{\tm{yd.rb}{R_b}\,\tm{yd.rc}{R_c}}{R_a + R_b + R_c}, \qquad \tm{yd.r2}{R_2} = \frac{\tm{yd.rc}{R_c}\,\tm{yd.ra}{R_a}}{R_a + R_b + R_c}, \qquad \tm{yd.r3}{R_3} = \frac{\tm{yd.ra}{R_a}\,\tm{yd.rb}{R_b}}{R_a + R_b + R_c}

The direction of this transformation always reduces the number of elements that are visible at the terminals from three loops to three legs, which is why it is the one usually applied to a bridge.

Each Δ branch is the sum of the three pairwise products of the Y legs, divided by the opposite Y leg:

\tm{yd.ra}{R_a} = \frac{R_1R_2 + R_2R_3 + R_3R_1}{\tm{yd.r1}{R_1}}, \qquad \tm{yd.rb}{R_b} = \frac{R_1R_2 + R_2R_3 + R_3R_1}{\tm{yd.r2}{R_2}}, \qquad \tm{yd.rc}{R_c} = \frac{R_1R_2 + R_2R_3 + R_3R_1}{\tm{yd.r3}{R_3}}

The numerator is the same for all three, so it is computed once. The rule to remember is that the product form belongs to Δ → Y and the sum of products form belongs to Y → Δ.

When all three elements of one network are equal, both formulas collapse to a single factor of three. The Δ carries the larger resistance, which is consistent with the Δ branches being connected between terminals rather than between a terminal and the center.

R_Δ (Ω) 45

R_Y = R_Δ3 = 15.0 Ω R_Δ = 3R_Y in three-phase notation, Z_Δ = 3Z_Y

A numerical pair worth memorizing as a check

The Δ with branches $R_a = 6$ , $R_b = 12$ , $R_c = 18\ \Omega$ transforms into the Y with legs $R_1 = 6$ , $R_2 = 3$ , $R_3 = 2\ \Omega$ . Verification in the forward direction uses $\sum R_\Delta = 36$ : $R_1 = (12)(18)/36 = 6$ , $R_2 = (18)(6)/36 = 3$ , $R_3 = (6)(12)/36 = 2$ . Verification in the reverse direction uses $R_1R_2 + R_2R_3 + R_3R_1 = 18 + 6 + 12 = 36$ : $R_a = 36/6 = 6$ , $R_b = 36/3 = 12$ , $R_c = 36/2 = 18$ ✓. Both routes are carried out on the examination in a minute or less; an answer that fails to transform back is an answer that was computed wrongly.

Developed in Lecture 2, including the bridge that requires the transformation.

§4 · The four theorems

Linearity and superposition

Linearity is a property of the circuit; superposition is the procedure that exploits it. Every circuit in this course is linear, because every element law is linear, so both statements below hold without qualification.

scaling: input × K ⇒ output × K additivity: input V₁ + V₂ ⇒ output V_o1 + V_o2 superposition = additivity applied one source at a time

The superposition procedure

All independent sources but one are turned off.
The circuit is solved for the requested quantity.
Steps 1 and 2 are repeated for every independent source in turn.
The partial answers are added.

A voltage source is turned off by shorting it, because a dead voltage source enforces zero volts across its terminals. A current source is turned off by opening it, because a dead current source enforces zero amperes through its branch.

Dependent sources are never turned off

A dependent source is not an input to the circuit; it is an element whose value happens to be written in terms of another variable. It therefore remains active in every partial circuit, and its controlling variable must be recomputed in each one. Turning a dependent source off is the single most common way to obtain a wrong superposition answer.

Thevenin and Norton

The claim is exact, not approximate: for any two chosen terminals, a linear network is indistinguishable from the two circuits below. Every element inside may be replaced by them without altering a single voltage or current outside.

Thevenin: the source V_T equals the open-circuit voltage V_oc at the port.

Norton: the source I_N equals the short-circuit current, and R_N equals R_T.

\tm{tg.vt}{V_T} = \tm{tg.voc}{V_{oc}}, \qquad \tm{tg.in}{I_N} = I_{sc}, \qquad \tm{tg.rn}{R_N} = \tm{tg.rt}{R_T}, \qquad \boxed{\ \tm{tg.vt}{V_T} = \tm{tg.in}{I_N}\cdot\tm{tg.rt}{R_T}\ }

The boxed relation means that any two of the three quantities determine the third, which supplies both a shortcut and a check.

Route to R_T	What is done	When it is permitted
Series-parallel	independent sources turned off, then the network reduced by inspection	only when no dependent source is present
Two experiments	V_oc and I_sc computed, then R_T = V_oc / I_sc	always, unless V_T = 0, which makes the ratio indeterminate
Test source	independent sources turned off, the port driven by 1 A or 1 V, then R_T = V_test / I_test	always; this is the only route that survives V_T = 0

Dependent sources stay on, here as well

The independent sources are turned off when a test source is applied. Dependent sources are left alone, exactly as in superposition, which is why a network containing them can present a resistance that is unrelated to any resistor value in it, and can even be negative.

Maximum power transfer

For a fixed network delivering power to an adjustable load,

P_L = I_L^2 R_L = \left(\frac{V_T}{R_T + R_L}\right)^2 R_L \qquad\Longrightarrow\qquad \boxed{\ R_L = R_T\ } \qquad \boxed{\ P_{L,\max} = \frac{V_T^2}{4R_T}\ }

The condition is stated in terms of the Thevenin resistance of the network, so the equivalent must be found before the question can be answered. In an alternating-current circuit the condition becomes $Z_L = Z_T^{\,*}$ .

Matching maximizes load power, not efficiency

At the match, the source resistance dissipates exactly as much power as the load, so the efficiency is fifty per cent. Maximum power transfer is therefore the design goal for signal circuits, where the available power is small and scarce, and is emphatically not the design goal for power delivery, where the line resistance is kept as small as possible instead.

Developed in Lecture 4 (linearity, superposition, and the source equivalents), Lecture 5 (dependent sources at the port, and maximum power transfer), and Lecture 13 (the alternating-current condition).

§5 · Operational amplifiers

Two rules replace the entire device

The ideal model. The input currents are zero and the two input voltages are equal; the output current is whatever the supply rails must provide and is therefore unknown until it is solved for.

i₋ = i₊ = 0 A V_p = V_n i_out is unknown

The first rule follows from infinite input resistance and the second from infinite gain with negative feedback present. Together they reduce every op-amp circuit to a nodal analysis, because the second rule supplies the value of a node voltage without an equation and the first rule removes two branches from every KCL sum.

Where KCL must not be written

No KCL at the output node, unless $i_{\text{out}}$ is itself the quantity being solved for. The output current is supplied by the rails, which are not drawn, so an equation written there introduces an unknown that no other equation constrains.
No KCL at the ground node, ever. The reference node carries the return current of every branch in the circuit, including the op-amp supply returns, so its KCL equation is never independent of the others. This holds for every circuit in the course, but op-amp circuits are where the temptation is strongest, because the ground node often looks like an ordinary node with only two or three visible branches.

Thevenin and Norton with amplifiers present

An op-amp output behaves as an ideal voltage source, so the Thevenin resistance looking back into it is zero and the Norton equivalent does not exist. More generally, the series-parallel route to $R_T$ is unavailable in any circuit containing an amplifier, because the amplifier is an active element; the test-source route is used instead.

Developed in Lecture 6.

§6 · Switched circuits

Energy storage: two element laws and two continuity statements

The element laws are differential, which is what makes the circuit dynamic; the continuity statements are what make the initial condition computable.

Element	Law	At DC steady state	Continuous quantity
Capacitor	i_C = C dv_C/dt	i_C = 0: an open circuit	v_C(0⁻) = v_C(0⁺)
Inductor	v_L = L di_L/dt	v_L = 0: a short circuit	i_L(0⁻) = i_L(0⁺)

The combination rules complete the pair, and they are the opposite of the resistor rules: capacitances add in parallel and inductances add in series.

Continuity holds for one quantity per element, not for all of them

Only $v_C$ and $i_L$ are continuous across the instant of switching. The capacitor current and the inductor voltage may jump, and usually do. When the initial condition asked for is one of those, it is computed from the continuous quantity: $v_C(0^+)$ is known first, and $i_C(0^+)$ then follows from the circuit at $t = 0^+$ with the capacitor replaced by a source of that value.

First order: one storage element, one exponential

The general first-order solution

Every voltage and every current in a first-order DC switched circuit has the same shape, so one formula covers all of them:

f(t) = \left[f(0^+) - f(\infty)\right]e^{-t/\tau} + f(\infty), \qquad \tau = R_T C \quad\text{or}\quad \tau = \frac{L}{R_T}

Three numbers are needed: the initial condition $f(0^+)$ , the final condition $f(\infty)$ , and the time constant. The resistance in the time constant is the Thevenin resistance seen by the storage element, which is found by removing the element, turning off the independent sources, and looking in at the two terminals left behind.

Second order: two storage elements, two initial conditions

A circuit containing two independent storage elements, whether two capacitors, two inductors, or one of each, obeys

\frac{d^2y}{dt^2} + A\frac{dy}{dt} + By = K

where $y$ may be any voltage or any current in the circuit. Two initial conditions are required, $y(0^+)$ and $y'(0^+)$ , because two constants appear in the solution.

The process, in the order the board gives it

Initial conditions. At $t = 0^-$ the circuit is at DC steady state, with capacitors open and inductors shorted; $v_C(0^-)$ and $i_L(0^-)$ are read off and carried across the switching instant by continuity. The derivative condition then follows from the element law applied at $t = 0^+$ : $v_C'(0^+) = i_C(0^+)/C$ and $i_L'(0^+) = v_L(0^+)/L$ .
Final condition. At $t = \infty$ the circuit is again at DC steady state. Since all derivatives vanish, the differential equation itself gives $y(\infty) = K/B$ .
The equation for $t > 0$ . Two element equations in two unknowns are written and reduced to a single second-order equation in $y$ .
Solve. The characteristic equation $s^2 + As + B = 0$ is formed, its roots classify the response, the matching solution form is written down, and the two initial conditions fix its two constants.

Roots of s² + As + B = 0	Case	Form of the natural response
real and distinct, s₁ ≠ s₂	overdamped	A₁e^s₁t + A₂e^s₂t
real and repeated, s₁ = s₂ = −α	critically damped	(A₁ + A₂t)e^−αt
complex, −α ± jω_d	underdamped	e^−αt(A₁ cos ω_dt + A₂ sin ω_dt)

The constants are fitted to the complete response

The two initial conditions are applied to $y(t)$ , which is the natural response plus the forced response $y(\infty)$ , and never to the natural response alone. Fitting the constants before the final value has been added is the most common arithmetic error in this topic.

Developed in Lecture 7 (the elements and the instant of switching), Lectures 8 and 9 (first order), and Lectures 10 and 11 (second order).

§7 · Phasors, impedance, and complex power

The steady state of a linear circuit driven at one frequency

A sinusoidal source at a single frequency produces a sinusoidal steady state at that same frequency, so only two numbers, an amplitude and a phase, are unknown in each response. The phasor carries exactly those two and discards the rest.

v(t) = V_m\cos(\omega t + \theta) \qquad\Longleftrightarrow\qquad \tilde{V} = V_m\angle\theta

What the phasor does not carry

The frequency $\omega$ is not represented in the phasor and must be carried separately, because the impedances depend on it. Every source must therefore be at the same frequency before a single phasor circuit may be drawn; if two frequencies are present, the circuit is solved once per frequency and the time-domain answers are added by superposition. The cosine form with a positive amplitude is the reference; a sine or a negative amplitude is converted before the phasor is written.

z = x + jy

Addition and subtraction are performed in this form, term by term. Conversion outward uses $r = \sqrt{x^2 + y^2}$ and $\theta = \operatorname{atan2}(y, x)$ ; the quadrant must be checked, because the principal arctangent alone confuses the second quadrant with the fourth.

z = r\angle\theta

Multiplication, division, powers, and roots are performed in this form: magnitudes multiply or divide, and angles add or subtract. Conversion inward uses $x = r\cos\theta$ and $y = r\sin\theta$ .

z = re^{j\theta} = r\left(\cos\theta + j\sin\theta\right)

Euler's identity is what connects the other two forms, and it is the form in which the phasor transform is derived. On the examination it is used chiefly to justify the polar multiplication rule, since $e^{j\alpha}e^{j\beta} = e^{j(\alpha+\beta)}$ .

Impedance, and the one method that replaces all others

Z = R + jX, \qquad Z_R = R, \qquad Z_L = j\omega L, \qquad Z_C = \frac{1}{j\omega C} = \frac{-j}{\omega C}

Impedances combine exactly as resistances do: in series they add, in parallel their reciprocals add, and both dividers apply unchanged. Every technique of the first half of the course therefore transfers without modification. The procedure is always the same three moves: transform to the phasor domain, solve by any DC technique, and transform the answer back.

X > 0 ⇒ inductive X < 0 ⇒ capacitive X = 0 ⇒ resistive

Complex power

Average power is what a resistor dissipates and what a utility bills for. Written with rms phasors, it is the real part of a single complex quantity that carries both the average power and the reactive power:

S = \tilde{V}_{\text{rms}}\tilde{I}_{\text{rms}}^{\,*} = \frac{|\tilde{V}_{\text{rms}}|^2}{Z^*} = |\tilde{I}_{\text{rms}}|^2 Z = |S|\angle\theta_S = \tm{pt.p}{P} + j\,\tm{pt.q}{Q}

P = V_{\text{rms}}I_{\text{rms}}\cos(\theta_v - \theta_i)\ \ [\text{W}], \qquad Q = V_{\text{rms}}I_{\text{rms}}\sin(\theta_v - \theta_i)\ \ [\text{VAR}], \qquad |S|\ \ [\text{VA}]

The power triangle of an inductive load. The real power is the base, the reactive power the vertical leg, and the apparent power the hypotenuse. The angle of $S$ is the angle of the impedance, and its cosine is the power factor. A capacitive load mirrors the triangle about the base, since $Q$ is then negative.

Q > 0: inductive, current lags, pf lagging Q < 0: capacitive, current leads, pf leading ΣS = 0: complex power is conserved

The conjugate is taken on the current, which is what makes the angle of $S$ come out as $\theta_v - \theta_i$ rather than its negative.

Developed in Lecture 12 (phasors and impedance) and Lectures 13 and 14 (alternating-current power, the power triangle, and power factor correction).

§8 · Balanced three-phase

One phase is solved; the other two follow by symmetry

A balanced system has three sources of equal magnitude spaced $120^\circ$ apart and three equal load impedances. Because the neutral carries no current, the system is solved on the per-phase equivalent below, and the remaining two phases are obtained by shifting the answer by $\mp120^\circ$ .

The per-phase equivalent of a balanced Y-Y system. The source drives the line impedance in series with one load leg. The neutral connection between N and n carries no current when the system is balanced, so it may be drawn as an ideal short whether or not the physical system has a neutral wire.

Connection	Voltages	Currents
Y load	Ṽ_ab = √3 ∠30° · Ṽ_an	line current = phase current
Δ load	line voltage = phase voltage = Ṽ_ab	Ĩ_a = √3 ∠−30° · Ĩ_ab
Conversion	Z_Δ = 3Z_Y, so a Δ load is converted to Y before the per-phase circuit is drawn

The factor $\sqrt3$ and the $\pm30^\circ$ shift appear once in each row, and which quantity carries them is decided by which connection is present. In both cases the quantity that is larger by $\sqrt3$ is the one measured between two lines for a Y and the one flowing in a line for a Δ.

S_{1\phi} = \tilde{V}_\phi\tilde{I}_\phi^{\,*} = |\tilde{I}_a|^2 Z_Y = \frac{|\tilde{V}_\phi|^2}{Z_\phi^*}, \qquad \boxed{\ S_{3\phi} = 3S_{1\phi} = P_{3\phi} + jQ_{3\phi}\ }

Two different power factors

The power factor at the load is $\cos\theta_{Z_Y}$ , computed from the load impedance alone. The power factor at the source is $\cos\theta_{(Z_L + Z_Y)}$ , computed from the line impedance in series with the load. The two differ whenever the line is not ideal, and a question that does not say which one is wanted is answered by stating both.

Developed in Lectures 15 and 16. The examination covers the balanced case only.

§9 · Coupling, matching, and the frequency response

Magnetically coupled circuits

Two coils sharing a magnetic flux are described by a mutual inductance $M$ , and the dot convention fixes the sign of the coupled term: a current entering the dotted terminal of one coil produces a voltage that is positive at the dotted terminal of the other. An ideal transformer is the limiting case in which the coupling is perfect, and it reduces to two ratios with $n = N_2/N_1$ :

v_2 = \pm n\,v_1, \qquad i_1 = \pm n\,i_2, \qquad Z_{\text{reflected into primary}} = \frac{Z_L}{n^2}

The reflected-impedance relation is what makes a transformer a matching device. For maximum power transfer through an ideal transformer, the load and the source resistance are related by

\boxed{\ R_L = n^2R_T \quad\text{for a turns ratio } 1 : n\ }

Developed in Lectures 17 and 18.

The frequency response

The last block of the course replaces a single frequency by all of them. The circuit is redrawn in the $s$ domain with $Z_L = sL$ and $Z_C = 1/sC$ , the transfer function $H(s)$ is formed as the ratio of the output phasor to the input phasor, and $H(j\omega)$ is then plotted against frequency.

Step	What is produced	Lecture
Redraw in the s domain	impedances sL and 1/sC; every DC technique still applies	19
Form H(s)	a ratio of polynomials, whose roots are the zeros and poles	19
Factor and plot	the Bode magnitude and phase, as the sum of the factor asymptotes	20, 21
Classify	low-pass, high-pass, band-pass, or band-stop, from the two extreme gains	22
Cut off	ω_c from \|H(jω_c)\|² = \|H\|²_max / 2	23

§10 · Where every topic was developed

The index, and where to find a worked example

Each row names the lecture that develops the topic in full and, where one exists, the section of Lecture 24 that works a numbered example of it.

Topic	Developed in	Worked example
KCL and the branch current method	2	Lecture 2
Series, parallel, and the two dividers	2	Lecture 2
Y-Δ and the bridge	2	Lecture 2
Nodal analysis and the supernode	3	Lecture 3
Linearity and superposition	4	Lecture 24 §1
Thevenin and Norton equivalents	4, 5	Lecture 24 §2
Maximum power transfer	5, 13	Lecture 24 §3
Operational amplifiers	6	Lecture 24 §4
Energy storage and the instant of switching	7	Lecture 24 §5
First-order switched circuits	8, 9	Lecture 24 §5
Second-order switched circuits	10, 11	Lecture 24 §6
Sinusoids, phasors, and impedance	12	Lecture 24 §7
AC power and the power triangle	13, 14	Lecture 24 §7
Balanced three-phase circuits	15, 16	Lecture 24 §8
Magnetically coupled circuits	17	Lecture 17
Transformers and impedance matching	18	Lecture 24 §9
Transfer functions and the frequency response	19	Lecture 19
Bode plots	20, 21	Lecture 21
Filters, first and second order	22	Lecture 22
Cutoff frequencies of second-order filters	23	Lecture 23

The practice problem that closes Lecture 24 is cumulative: it combines a dependent source, both routes to a Thevenin equivalent, and maximum power transfer in a single circuit.