ECE 211 · Lecture 24: The Examination Review, Worked Through

§1 · Superposition

One source at a time, then add

The partial answers are obtained with every independent source but one turned off: a voltage source is shorted, a current source is opened, and a dependent source is never touched. The full statement of the procedure is on the reference page.

The problem

The node voltage $V_A$ is sought in a circuit driven by a 12 V source behind a 4 Ω resistor and by a 2 A source injected at the same node, with a 6 Ω resistor to the reference.

A single nodal equation would also answer the question. Superposition is applied here because the same circuit is reused in Section 2, where the two partial answers make the equivalent easy to check.

The voltage source acting alone

The current source is opened. The 4 Ω and 6 Ω resistors are then in series across the 12 V source, and voltage division gives

V_{A1} = \tm{sp.vs}{12}\cdot \frac{\tm{sp.r6}{6}}{\tm{sp.r4}{4} + \tm{sp.r6}{6}} = 7.2\ \text{V}

The current source acting alone

The voltage source is shorted. The 4 Ω and 6 Ω resistors are then in parallel across the 2 A source, so

V_{A2} = \tm{sp.is}{2}\cdot \left(\tm{sp.r4}{4} \parallel \tm{sp.r6}{6}\right) = 2(2.4) = 4.8\ \text{V}

Add the partial answers

V_A = V_{A1} + V_{A2} = 7.2 + 4.8 = \boxed{12\ \text{V}}

Both partial answers are positive because both sources drive current down through the 6 Ω resistor in the same direction. A partial answer of the opposite sign is perfectly normal and is simply added with its sign.

Check by a single nodal equation

With the reference at the bottom rail, KCL at node A reads

\frac{V_A - 12}{4} + \frac{V_A}{6} = 2 \;\Longrightarrow\; 3(V_A - 12) + 2V_A = 24 \;\Longrightarrow\; 5V_A = 60 \;\Longrightarrow\; V_A = 12\ \text{V} \ \checkmark

Agreement between a one-shot method and superposition is the expected outcome; disagreement almost always traces to a source that was turned off incorrectly. Linearity and superposition were developed in Lecture 4.

§2 · The Thevenin and Norton equivalents

The same circuit, seen from the load terminals

The load is taken to be the 6 Ω resistor of Section 1, so the port is opened at its two terminals and the three experiments are run on what remains. The three routes to the resistance, and when each is permitted, are tabulated on the reference page.

Choose the port

The 6 Ω resistor is removed and the terminals it occupied are named a and b. Everything left of the port is the network to be replaced; the load itself is never part of the equivalent.

The open-circuit voltage

With the port open, no current leaves terminal a, so the 2 A delivered by the current source must return through the 4 Ω resistor into the 12 V source. Walking from the reference up through the source and then across the resistor against the current gives

\tm{tv.vt}{V_T} = 12 + (2)(4) = \boxed{20\ \text{V}}

The short-circuit current

With the port shorted, node A sits at zero volts. The 12 V source then drives $12/4 = 3$ A through the 4 Ω resistor, and the 2 A source adds its own current, so

\tm{tv.isc}{I_N} = 3 + 2 = \boxed{5\ \text{A}}

The resistance, and the check

Both independent sources are turned off: the voltage source is shorted and the current source is opened. Looking in from the port, only the 4 Ω resistor stands between the two terminals, so $R_T = 4\ \Omega$ . The ratio route agrees:

R_T = \frac{V_{oc}}{I_{sc}} = \frac{20}{5} = \boxed{4\ \Omega}

Reattaching the load reproduces the Section 1 answer

The equivalent is a 20 V source in series with 4 Ω. With the 6 Ω load restored,

V_A = 20\cdot\frac{6}{4 + 6} = 12\ \text{V} \ \checkmark

which is the number superposition produced. Note also that the two partial answers of Section 1 are visible inside the equivalent: 12 V of the 20 V is contributed by the voltage source and 8 V by the current source, and voltage division scales both by $6/10$ to give 7.2 V and 4.8 V. Thevenin and Norton equivalents were developed in Lectures 4 and 5.

§3 · Maximum power transfer

The board example: 55 V behind 4 Ω

The load that draws the most power is the one that matches the Thevenin resistance of the network driving it:

\boxed{\ R_L = R_T\ } \qquad \boxed{\ P_{L,\max} = \frac{V_T^2}{4R_T}\ }

For a 55 V source behind 4 Ω the matched load is therefore $R_L = 4\ \Omega$ , at which the load voltage is half of the source voltage:

V_L = \frac{55}{2} = 27.5\ \text{V}, \qquad P_{L,\max} = \frac{V_L^2}{R_L} = \frac{(55/2)^2}{4} = 189.06\ \text{W}

The equivalent driving the load. The load resistance is set by the slider; the current and the load voltage follow, and the animated dots run faster as the current rises.

R_L (Ω) 4

I_L = 6.875 A · V_L = 27.50 V · R_L = 4.0 Ω · P_L = 189.1 W · 100.0% of the maximum

The load power against the load resistance. The curve is flat near its peak, so a load within a factor of two of the match still collects more than eighty per cent of the available power; far from the match the penalty grows quickly in both directions.

The same answer read from the Norton form

The board works the example a second time in Norton form as a check. The Norton source is $I_N = V_T/R_T = 55/4 = 13.75$ A in parallel with 4 Ω, and with the matched 4 Ω load attached the current divider splits that source in half.

The Norton form of the same problem. Two equal resistances in parallel receive equal shares of the source current, so the load current is half of 13.75 A whichever route is taken.

I_L = \frac{\tm{mn.in}{13.75}}{2} = 6.875\ \text{A}, \qquad P_L = I_L^2 \tm{mn.rl}{R_L} = (6.875)^2(4) = 189.06\ \text{W} \ \checkmark

The two routes must agree, because the two equivalents describe the same network. The board rounds the answer to 189 W; the unrounded value is 189.0625 W. Note that at the match the source resistance dissipates exactly as much power as the load, so the efficiency is fifty per cent, which is why matching is a signal-circuit goal and not a power-delivery goal.

§4 · An inverting amplifier

Two rules, then a nodal equation

The ideal model supplies zero input current and equal input voltages, which together reduce the circuit to one KCL equation. The rules, and the two places where KCL must not be written, are stated on the reference page.

The circuit

A 1 V source drives a 2 kΩ input resistor into the inverting input; a 10 kΩ resistor closes the feedback path from the output back to the same node. The non-inverting input is grounded, and a 1 kΩ load hangs on the output.

Apply the two rules

The non-inverting input is tied to the reference, so $V_p = 0$ . The second rule then forces $V_n = 0$ as well: the inverting node is a virtual ground, held at zero volts without being connected to the reference. The first rule forces the current into that input to be zero.

The word "virtual" matters. The node is at zero volts but carries no current to ground, so it must not be treated as a short to the reference.

KCL at the inverting node

Only two branches carry current into the virtual ground, the input resistor and the feedback resistor, because the op-amp input takes none:

\frac{\tm{ox.vin}{1} - 0}{\tm{ox.r1}{2\ \text{k}\Omega}} = \frac{0 - V_o}{\tm{ox.rf}{10\ \text{k}\Omega}} \;\Longrightarrow\; i_1 = 0.5\ \text{mA}

The output voltage

V_o = -\frac{R_f}{R_1}\,V_{in} = -\frac{10}{2}(1) = \boxed{-5\ \text{V}}

The inverting gain depends only on the ratio of the two resistors, and the load does not appear in it: an ideal op-amp holds its output voltage regardless of what is drawn from it.

The output current, which needs the forbidden equation

Because $i_{\text{out}}$ is now the quantity asked for, KCL at the output node is permitted. Taking $i_{\text{out}}$ as the current leaving the amplifier, $i_1 = 0.5$ mA arriving through the feedback resistor, and $V_o/R_L = -5$ mA leaving through the load,

i_{\text{out}} = \frac{V_o}{R_L} - i_1 = -5 - 0.5 = \boxed{-5.5\ \text{mA}}

The negative sign says that 5.5 mA flows into the output terminal: with a negative output voltage the amplifier sinks current rather than sourcing it. The magnitude must be checked against the rated output current of a real device, which is the physical reason this quantity is ever asked for.

Operational amplifiers were developed in Lecture 6, including the test-source route to a Thevenin resistance when an amplifier is present.

§5 · A first-order switched circuit

Three numbers fill in one formula

Every voltage and every current in a first-order DC switched circuit follows

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}

so the work is entirely the computation of $f(0^+)$ , $f(\infty)$ , and $\tau$ . The element laws and the continuity statements they rest on are tabulated on the reference page.

The problem

The switch has been closed for a long time and opens at $t = 0$ . The capacitor voltage is sought for $t > 0$ .

Step 1: the circuit at t = 0⁻

Long-closed means DC steady state, so the capacitor is an open circuit and the 24 V source sees the 2 kΩ and 6 kΩ resistors in series. Voltage division gives

v_C(0^-) = 24\cdot\frac{6}{2 + 6} = 18\ \text{V}

Step 2: the circuit at t = 0⁺

Continuity carries that value across the switching instant: $v_C(0^+) = v_C(0^-) = 18$ V. This is the initial condition. The current at that instant is not continuous and is computed separately if it is wanted: $i_C(0^+) = (24 - 18)/2\ \text{k}\Omega = 3$ mA.

Step 3: the circuit at t = ∞

With the switch open, the 6 kΩ branch is disconnected entirely. At the new steady state the capacitor is again an open circuit, so no current flows anywhere and no voltage is dropped across the 2 kΩ resistor:

v_C(\infty) = 24\ \text{V}

Step 4: the time constant, and the answer

The capacitor is removed and the 24 V source is shorted. Looking in at the two terminals left behind, only the 2 kΩ resistor appears, because the 6 kΩ branch is open. Therefore

\tau = R_T C = (2000)(25\times10^{-6}) = 0.05\ \text{s} = 50\ \text{ms}

\boxed{\ v_C(t) = (18 - 24)e^{-t/0.05} + 24 = 24 - 6e^{-20t}\ \text{V}, \quad t \ge 0\ }

The response climbs from its initial value toward its final value, covering 63.2 per cent of the gap in one time constant and settling within one per cent after five. First- order switched circuits were developed in Lectures 7 through 9.

§6 · A second-order switched circuit

Two initial conditions, one characteristic equation

A circuit containing two independent storage elements obeys $y'' + Ay' + By = K$ , whose characteristic equation $s^2 + As + B = 0$ classifies the response and whose two constants are fitted to the complete response, final value included. The four-step process is stated on the reference page.

The problem

A 12 V source is connected to a series combination of R = 5 Ω, L = 1 H, and C = 1/6 F when the switch closes at $t = 0$ . No energy is stored beforehand. The capacitor voltage is sought.

Initial conditions

Nothing is stored at $t = 0^-$ , so $v_C(0^-) = 0$ and $i_L(0^-) = 0$ , and continuity carries both across the instant. The derivative condition follows from the capacitor law, with the series current equal to the inductor current:

v_C'(0^+) = \frac{i_C(0^+)}{C} = \frac{i_L(0^+)}{C} = 0

Final condition

At $t = \infty$ the inductor is a short and the capacitor is an open circuit, so no current flows and no voltage is dropped across the resistor or the inductor:

v_C(\infty) = 12\ \text{V}

The differential equation and its roots

KVL around the loop with $i = C\,v_C'$ substituted gives the standard series form:

v_C'' + \frac{\tm{rl.r}{R}}{\tm{rl.l}{L}}v_C' + \frac{1}{\tm{rl.l}{L}\,\tm{rl.c}{C}}v_C = \frac{V_s}{LC} \;\Longrightarrow\; v_C'' + 5v_C' + 6v_C = 72

s^2 + 5s + 6 = 0 \;\Longrightarrow\; s = -2,\ -3

The roots are real and distinct, so the response is overdamped. The final value read from the equation agrees with the circuit: $K/B = 72/6 = 12$ V ✓.

Fit the constants and write the answer

v_C(t) = 12 + A_1e^{-2t} + A_2e^{-3t}

v_C(0^+) = 0:\quad 12 + A_1 + A_2 = 0 \qquad v_C'(0^+) = 0:\quad -2A_1 - 3A_2 = 0

The second equation gives $A_1 = -\tfrac32 A_2$ ; substituting into the first gives $A_2 = 24$ and $A_1 = -36$ :

\boxed{\ v_C(t) = 12 - 36e^{-2t} + 24e^{-3t}\ \text{V}, \quad t \ge 0\ }

Check the answer at both ends

At $t = 0$ the expression gives $12 - 36 + 24 = 0$ ✓, and its derivative $72e^{-2t} - 72e^{-3t}$ gives $0$ at $t = 0$ ✓. As $t \to \infty$ both exponentials vanish and 12 V remains ✓. The inductor current follows as $i_L = C\,v_C' = 12\left(e^{-2t} - e^{-3t}\right)$ A, which peaks at $t = \ln 1.5 = 0.405$ s with a value of $48/27 = 1.778$ A.

The same circuit at every damping

Only the resistance is moved below; $L = 1$ H and $C = 1/6$ F are held, so $\omega_0 = \sqrt{1/LC} = \sqrt6 = 2.449$ rad/s stays fixed while $\alpha = R/2L$ sweeps past it. The boundary sits at $R = 2\sqrt6 = 4.899\ \Omega$ .

The capacitor voltage for a 12 V step, always ending at the same final value. Below the boundary the response overshoots and rings; above it the response is monotonic and grows slower as the resistance grows, because one of the two roots moves toward the origin.

R (Ω) 5

α = R/2L = 2.50 s⁻¹ · ω₀ = 2.449 rad/s · s = −2.00, −3.00 · overdamped

Second-order switched circuits were developed in Lectures 10 and 11.

§7 · An alternating-current example

Transform, solve, transform back

A source at a single frequency is replaced by its phasor, every element by its impedance, and the circuit is then solved by any technique of the first half of the course. The phasor conventions and the impedance list are on the reference page.

The example circuit in the phasor domain. A source of $100\angle0^\circ$ V at $\omega = 400$ rad/s drives a 6 Ω resistance in series with a 20 mH inductor, whose impedance at that frequency is $j\omega L = j8$ Ω.

The two impedances are in series, so they add, and one division gives the current:

\tm{az.z}{Z} = \tm{az.zr}{6} + \tm{az.zl}{j8} = 10\angle53.13^\circ\ \Omega, \qquad \tm{az.i}{\tilde{I}} = \frac{100\angle0^\circ}{10\angle53.13^\circ} = 10\angle-53.13^\circ\ \text{A}

i(t) = 10\cos(400t - 53.13^\circ)\ \text{A}

The power delivered

Complex power is formed from rms phasors, with the conjugate taken on the current. With $V_{\text{rms}} = 100/\sqrt2 = 70.71$ V at $0^\circ$ and $I_{\text{rms}} = 10/\sqrt2 = 7.071$ A at $-53.13^\circ$ ,

S = \tilde{V}_{\text{rms}}\tilde{I}_{\text{rms}}^{\,*} = (70.71\angle0^\circ)(7.071\angle+53.13^\circ) = 500\angle53.13^\circ = \tm{at.p}{300} + j\,\tm{at.q}{400}\ \text{VA}

The power triangle of the example. The real power is 300 W, the reactive power is 400 VAR, and the apparent power is 500 VA. The load is inductive, so $Q$ is positive, the current lags the voltage, and the power factor is described as lagging.

Two independent checks on the same numbers

The element route confirms both parts without forming $S$ at all:

P = I_{\text{rms}}^2R = (50)(6) = 300\ \text{W}, \qquad Q = I_{\text{rms}}^2X = (50)(8) = 400\ \text{VAR} \ \checkmark

and the magnitude follows as $|S| = I_{\text{rms}}^2|Z| = (50)(10) = 500$ VA ✓. The power factor is $\cos 53.13^\circ = 0.6$ lagging. Phasors and impedance were developed in Lecture 12, and alternating-current power in Lectures 13 and 14.

§8 · A balanced three-phase load

Solve one phase, then multiply by three

The neutral of a balanced system carries no current, so the whole system reduces to the per-phase circuit below. The line and phase relations for both connections are tabulated on the reference page.

The per-phase equivalent, with a source of $120\angle0^\circ$ V rms and a load of $6 + j8$ Ω. The line current follows by one division, and the neutral connection between N and n carries no current.

\tm{tp.ia}{\tilde{I}_a} = \frac{120\angle0^\circ}{\tm{tp.zy}{10\angle53.13^\circ}} = 12\angle-53.13^\circ\ \text{A}

S_{1\phi} = \tilde{V}_\phi\tilde{I}_\phi^{\,*} = (120)(12)\angle53.13^\circ = 1440\angle53.13^\circ = 864 + j1152\ \text{VA}

\boxed{\ S_{3\phi} = 3S_{1\phi} = 4320\angle53.13^\circ = 2592 + j3456\ \text{VA}\ }

Two independent checks, and the Δ equivalent

The element route gives $P_{3\phi} = 3|I_a|^2R = 3(144)(6) = 2592$ W and $Q_{3\phi} = 3(144)(8) = 3456$ VAR ✓. The line-quantity route gives

|S_{3\phi}| = \sqrt3\,V_LI_L = \sqrt3\,(207.85)(12) = 4320\ \text{VA} \ \checkmark

where the line voltage is $V_L = \sqrt3\,(120) = 207.85$ V, leading $\tilde{V}_{an}$ by $30^\circ$ . The same load connected in Δ would carry $Z_\Delta = 3Z_Y = 18 + j24$ Ω, and converting it back to Y is the first move in any Δ problem. Three-phase circuits were developed in Lectures 15 and 16; the examination covers the balanced case only.

§9 · An impedance match through a transformer

The transformer is what makes an unmatched load matchable

An ideal transformer reflects its secondary load into the primary divided by $n^2$ , so the matching condition $R_L = n^2R_T$ selects the turns ratio. The coupled-circuit relations are on the reference page.

An 8 Ω load matched to a 200 Ω source through a 5 : 1 transformer, for which $n = 1/5$ and $n^2 = 0.04$ . The load reflected into the primary is $8/0.04 = 200$ Ω, which matches the source resistance exactly.

The turns ratio is chosen from the two resistances rather than the other way round:

n^2 = \frac{\tm{xm.rl}{R_L}}{\tm{xm.rt}{R_T}} = \frac{8}{200} = 0.04 \;\Longrightarrow\; n = 0.2, \quad\text{a turns ratio of } 5 : 1

How much the match is worth here

With the transformer in place the primary sees 200 Ω, so the source is matched and delivers

P = \frac{V_T^2}{4R_T} = \frac{20^2}{4(200)} = 0.5\ \text{W}

Connecting the same 8 Ω load directly to the source instead gives

P = \left(\frac{20}{208}\right)^2(8) = 0.0740\ \text{W}

so the transformer improves the delivered power by a factor of 6.76. Magnetically coupled circuits and transformers were developed in Lectures 17 and 18.

§10 · Your turn

Practice: a port with a dependent source

The circuit below contains one independent source and one current-controlled current source, where $i_x$ is the current in the 6 Ω resistor flowing from left to right. The problem should be attempted on paper before the hints are opened, in order.

The network to be replaced. The dependent source delivers $2i_x$ , and the port is the terminal pair a and b.

Find the Thevenin equivalent at a-b by two independent routes, then find the load resistance that draws the most power and the value of that power.

Hint 1: the open-circuit voltage

With the port open, no current leaves terminal a. Write KCL at the top node of the dependent source, remembering that the current arriving through the 6 Ω resistor is $i_x$ by definition and that the dependent source contributes $2i_x$ to the same node. What does the resulting equation force $i_x$ to be, and what does that imply about the voltage dropped across the 6 Ω resistor?

Hint 2: the short-circuit current

Short a to b. Terminal a is then at the reference potential, so $i_x$ follows immediately from the 12 V source and the 6 Ω resistor. The short-circuit current is the total current arriving at the top node from both branches.

Hint 3: the test source, as a third route

Turn off the 12 V source only, leaving the dependent source active, and drive the port with 1 A into terminal a. Two unknowns appear, the port voltage and $i_x$ , and two equations relate them: KCL at the top node, and Ohm's law across the 6 Ω resistor with the source shorted.

Solution

Open-circuit voltage. With the port open, KCL at the node joining the 6 Ω resistor, the dependent source, and terminal a reads

i_x + 2i_x = 0 \;\Longrightarrow\; 3i_x = 0 \;\Longrightarrow\; i_x = 0

No current in the 6 Ω resistor means no voltage dropped across it, so terminal a sits at the source potential:

V_T = 12\ \text{V}

Short-circuit current. With a shorted to b, the top node is at zero volts, so

i_x = \frac{12 - 0}{6} = 2\ \text{A}, \qquad I_N = i_x + 2i_x = 3i_x = 6\ \text{A}

R_T = \frac{V_{oc}}{I_{sc}} = \frac{12}{6} = 2\ \Omega

Test source, as the independent check. With the 12 V source shorted and 1 A driven into terminal a, let $V$ be the port voltage. The current in the 6 Ω resistor, still measured left to right, is $i_x = (0 - V)/6 = -V/6$ . KCL at the top node, with all three currents taken as entering, gives

1 + i_x + 2i_x = 0 \;\Longrightarrow\; i_x = -\tfrac13\ \text{A} \;\Longrightarrow\; -\frac{V}{6} = -\frac13 \;\Longrightarrow\; V = 2\ \text{V}

R_T = \frac{V}{1\ \text{A}} = 2\ \Omega \ \checkmark

Maximum power transfer.

R_L = R_T = \boxed{2\ \Omega}, \qquad P_{L,\max} = \frac{V_T^2}{4R_T} = \frac{144}{8} = \boxed{18\ \text{W}}

Check. With the 2 Ω load attached, $I_L = 12/4 = 3$ A and $P_L = (3)^2(2) = 18$ W ✓. Note that the dependent source multiplies the current arriving at the node by three, which is exactly why the port resistance is $6/3 = 2\ \Omega$ rather than the 6 Ω that inspection of the resistors alone would have suggested. This is the reason the series-parallel route is forbidden when a dependent source is present.