123-ME · Thevenin, Norton, Superposition & the Wheatstone Bridge

Thevenin, Norton, Superposition & the Wheatstone Bridge

Slides 2.6.1 · 2.6.2 · 2.7 · 2.8 — Exercise 6, Tutorial 5

Node and mesh analysis will solve any resistive circuit, but they solve it all at once: change one load resistor and you re-solve the whole system. This chapter buys you something better. Every technique here — superposition, Thevenin, Norton, the bridge — is a consequence of one fact: a resistive circuit is a linear system. Linearity lets you take a circuit apart source-by-source, and it lets you collapse an arbitrarily ugly network into one source and one resistor that a load cannot tell apart from the original. On the exam these are the questions where the "obvious" method takes twenty minutes and the right method takes three.

Without Thevenin: same network, three loads, three full re-derivations — every source and every other resistor gets re-accounted for each time.

With Thevenin: reduce the network to two numbers once. Every load after that is one divider calculation on a two-element circuit — the whole reason this chapter exists.

Where this is going — pacing budget

Roughly 25 min on linearity + superposition (including the power trap, which is worth a mark on almost every exam that asks it), 35 min on Thevenin/Norton (the workhorse), 15 min on the Wheatstone bridge, then the practice set. If you are short on time: learn the three ways to find $R_{Th}$ and when each one is legal. That single decision is where most lost marks live.

1. Linearity — the license for everything in this chapter

Write node-voltage analysis for any resistive circuit with $N$ unknown node voltages. Every equation is a KCL sum of terms $\,(v_j - v_k)/R\,$ and known source currents. Collect them and you always get the same shape:

\mathbf{G}\,\mathbf{v} \;=\; \mathbf{i}_s

\mathbf{G}

= conductance matrix (resistors only).

\mathbf{i}_s

= the sources.

\mathbf{v}

= node voltages.

The resistors live entirely in $\mathbf{G}$ ; the sources live entirely on the right-hand side. So the solution is

\mathbf{v} \;=\; \mathbf{G}^{-1}\,\mathbf{i}_s

which is a linear function of the source vector. Two consequences fall straight out, and they are the whole chapter:

Additivity → superposition. If $\mathbf{i}_s = \mathbf{i}_A + \mathbf{i}_B$ , then $\mathbf{v} = \mathbf{G}^{-1}\mathbf{i}_A + \mathbf{G}^{-1}\mathbf{i}_B$ . The response to all sources is the sum of the responses to each source alone.
Affine terminal law → Thevenin/Norton. Pull one pair of terminals out of the network and the relation between the terminal voltage $v$ and terminal current $i$ can only be a straight line, $v = V_{Th} - R_{Th}\,i$ . A straight line has exactly two parameters — which is why exactly two measurements pin down any network.

What "linear" excludes

Resistors, independent sources, and linear dependent sources (like $3i_x$ or $2v$ ) are all fine. Diodes, transistors, and anything with a squared term are not. And note carefully: power is not a linear function of the sources — $p = i^2R$ is quadratic. That single fact is the most-tested trap in this chapter (§2.3).

2. Superposition

The recipe follows directly from additivity: solve the circuit once per independent source, with all the other independent sources deactivated, then add the answers.

x \;=\; x' + x'' + x''' + \cdots

where

x

is any voltage or current, and each term is that quantity with exactly one independent source left active.

Deactivating a source means setting its value to zero, and what a zero-valued source is follows from its definition:

A voltage source with $v = 0$ forces zero volts across itself no matter the current — that is a short circuit (a plain wire).
A current source with $i = 0$ forces zero current no matter the voltage — that is an open circuit (a gap).

Do not memorise this backwards. Derive it in three seconds from "what does a zero-valued version of this element do?" and you will never flip it under exam pressure.

2.1 The deactivation sequence, drawn

This is Exercise 6 P2.94's circuit: a 30 V source feeding a 10 Ω resistor into a node, a 5 Ω resistor from that node to ground carrying the current $i$ we want, and a 3 A source pushing current up into the same node.

Both sources active. Target:

i

down through 5 Ω.

3 A source → open. Divider: 30·5/15 = 10 V.

30 V source → short. Divider: 3·10/15 = 2 A.

Worked example — Exercise 6, P2.94 (superposition)

Voltage source alone. Open the 3 A source. The 10 Ω and 5 Ω are now a simple series pair across 30 V, so a voltage divider gives the 5 Ω voltage $v' = 30\cdot\frac{5}{10+5} = 10$ V.

i' = v'/5 = 2

A, downward.

Current source alone. Short the 30 V source. Now the 3 A splits between 10 Ω (through the short, back to ground) and 5 Ω. A current divider sends the larger share to the smaller resistor:

i'' = 3\cdot\frac{10}{10+5} = 2\ \text{A, downward}

Add. Both contributions point down through the 5 Ω, so they add with the same sign: $i = i' + i'' = 4$ A.

✓

Cross-check by node analysis (one equation, both sources live):

\frac{v-30}{10}+\frac{v}{5}=3 \;\Rightarrow\; 3v = 60 \;\Rightarrow\; v = 20\ \text{V}

i = 20/5 = 4

A. Agrees.

i = 4

A downward through the 5 Ω resistor.

2.2 Dependent sources are never deactivated

Classic lost mark

Superposition splits the independent source vector. A dependent source ( $3i_x$ , $2v$ , …) is not part of $\mathbf{i}_s$ — it is part of the network's own linear behaviour, i.e. it belongs in $\mathbf{G}$ . So in every sub-circuit the dependent source stays active, with its controlling variable re-evaluated for that sub-circuit. Zeroing it is not "one more source deactivated", it is deleting part of the circuit, and it gives a wrong answer. (Tutorial 5 P1 is exactly this circuit — a $3v$ dependent source that must remain live through both passes.)

2.3 Why superposition does not apply to power

Suppose source A alone drives $i'$ through a resistor $R$ , and source B alone drives $i''$ . With both active, $i = i' + i''$ — that part is legal. Now compute the power:

p = i^2R = (i'+i'')^2R = \underbrace{i'^2R}_{p'} + \underbrace{i''^2R}_{p''} + \underbrace{2\,i'i''R}_{\text{cross term}}

The cross term $2i'i''R$ is the whole story. It is not zero in general, so $p \neq p' + p''$ . Power is a quadratic function of the source vector, and only linear functions superpose.

The correct procedure

Superpose the current (or voltage), then square it once at the end: find $i = i'+i''$ first, then $p = i^2R$ . Never add sub-powers. Tutorial 5 P4 does exactly this: superpose to $i = 2.5 + 0.375 - 1 = 1.875$ A, then $P_{3\Omega} = (1.875)^2(3) \approx 10.55$ W.

Self-check 1

3. Thevenin equivalent circuits

Take any linear resistive network, no matter how many sources and resistors, and bring out two terminals $a$ – $b$ . From linearity, the terminal voltage must be an affine function of the terminal current:

v \;=\; V_{Th} \;-\; R_{Th}\, i

A straight line — so a source in series with a resistor reproduces it exactly.

That equation is a voltage source $V_{Th}$ in series with a resistance $R_{Th}$ (walk KVL around it and you get the same line). So no load can distinguish the original network from that two-element equivalent. Thevenin's theorem is not a trick; it is the statement that a straight line has two parameters.

Original network driving a load.

Thevenin equivalent — identical

v

–

i

a

–

b

3.1 The two intercepts: why $R_{Th} = v_{oc}/i_{sc}$

You know the terminal law is the line $v = V_{Th} - R_{Th}i$ . To identify its two parameters, take the two easiest points on it — the intercepts.

Open the terminals. No load means $i = 0$ , so $v = V_{Th}$ . Therefore

V_{Th} = v_{oc}

The open-circuit voltage is the Thevenin voltage, by definition of the line's

v

-intercept.

Short the terminals. A short forces $v = 0$ , so $0 = V_{Th} - R_{Th}i_{sc}$ :

i_{sc} = \frac{V_{Th}}{R_{Th}} \qquad\Longrightarrow\qquad R_{Th} = \frac{v_{oc}}{i_{sc}}

The slope of the line is

-R_{Th}

, and slope = (rise)/(run) between the two intercepts

= -v_{oc}/i_{sc}

Ideal voltmeter = infinite resistance ⇒

i = 0

⇒ reads

V_{Th}

Ideal ammeter = zero resistance ⇒

v = 0

⇒ reads

i_{sc}

Two intercepts fix the line:

v_{oc}

i_{sc}

, and

R_{Th} = v_{oc}/i_{sc}

3.2 Three ways to get $R_{Th}$ — and when each is legal

Method	How	Legal when
Source deactivation	Short every independent voltage source, open every independent current source, then reduce the resulting resistor network seen from $a$ – $b$ by series/parallel.	No dependent sources. Fastest by far — usually one line of arithmetic.
$v_{oc}/i_{sc}$	Solve the circuit twice (open, then shorted), divide.	Always — unless $v_{oc} = i_{sc} = 0$ (no independent sources at all), which makes the ratio $0/0$ .
Test source	Kill only the independent sources, apply a test $v_T$ (or $i_T$ ) at $a$ – $b$ , solve for the resulting $i_T$ , and take $R_{Th} = v_T/i_T$ .	Always, and the only method for a network whose sources are all dependent. This is what Tutorial 5 P2's key does to get $R_{Th} = 1\ \Omega$ .

The single most common error in this chapter

Using source deactivation on a circuit that contains a dependent source. You cannot "turn off" a $3i_x$ source — its value is set by the circuit, not by you, so setting it to zero changes the network. When you see a diamond-shaped source symbol, switch to $v_{oc}/i_{sc}$ or a test source.

Worked example — Exercise 6, P2.80 (Thevenin and Norton)

A 10 V source feeds a 10 Ω resistor into node $A$ . At $A$ : a 5 Ω resistor to ground, a 1 A source pushing current up into $A$ , and the open output terminals.

Open terminals

a

–

b

a

ties straight to node

A

b

to the return rail. No load connected while solving for

v_{oc}

i_{sc}

, and

R_{Th}

$v_{oc}$ by node analysis (terminals open, so no current leaves at $A$ ):

\frac{v_A - 10}{10} + \frac{v_A}{5} = 1 \;\Rightarrow\; 3v_A = 20 \;\Rightarrow\; v_{oc} = 6.67\ \text{V}

$i_{sc}$ . Short $a$ – $b$ , so $v_A = 0$ . The 5 Ω now has zero volts across it and carries nothing; everything goes into the short:

i_{sc} = \frac{10}{10} + 1 = 2\ \text{A}

$R_{Th} = v_{oc}/i_{sc} = 6.67/2 = 3.33\ \Omega$ .

✓

Cross-check by deactivation (legal here — both sources are independent): short the 10 V, open the 1 A, and look in from $a$ – $b$ :

R_{Th} = 10 \parallel 5 = \frac{10\cdot 5}{15} = 3.33\ \Omega \;\;\checkmark

Thevenin: 6.67 V in series with 3.33 Ω. Norton: 2 A in parallel with 3.33 Ω.

Worked example — Thevenin/Norton with a dependent source

A 12 V source feeds a 2 Ω resistor into node $A$ ; call the current flowing rightward through it $i_x$ . At $A$ , a dependent current source worth $3i_x$ pulls current from $A$ down to ground. A 4 Ω resistor runs from $A$ out to terminal $a$ ; terminal $b$ is ground. This is exactly the §3.2 warning circuit: a diamond source, so only two of the three $R_{Th}$ methods are legal here.

Diamond source symbol ⇒ deactivation is off the table (§3.2). Only

v_{oc}/i_{sc}

and the test source remain.

$v_{oc}$ . Terminals open, so no current flows in the 4 Ω, and the dependent source is the only path out of $A$ besides the source branch: $i_x = 3i_x \Rightarrow i_x = 0$ . With zero drop across the 2 Ω,

v_{oc} = v_A = 12\ \text{V}

$i_{sc}$ . A dependent source can never be deactivated — its value is set by the circuit, not by you (§3.2's red note) — so the only legal route to $R_{Th}$ is $v_{oc}/i_{sc}$ or a test source. Short $a$ – $b$ : the 4 Ω now runs from $A$ straight to ground, in parallel with the dependent source's path. With $i_x = (12-v_A)/2$ , KCL at $A$ (current in = current out):

i_x = 3i_x + \frac{v_A}{4} \;\Longrightarrow\; \frac{12-v_A}{2} = 3\cdot\frac{12-v_A}{2} + \frac{v_A}{4}

Clearing fractions ( $\times 4$ ): $2(12-v_A) = 6(12-v_A) + v_A \Rightarrow 3v_A = 48 \Rightarrow v_A = 16$ V, so

i_{sc} = \frac{v_A}{4} = 4\ \text{A} \qquad\Longrightarrow\qquad R_{Th} = \frac{v_{oc}}{i_{sc}} = \frac{12}{4} = 3\ \Omega

✓

Cross-check by test source. Kill only the 12 V source (short it — that's legal, it's independent), keep the dependent source live, and inject a 1 A test current $i_T$ into terminal $a$ . With the source shorted, the 2 Ω now runs from $A$ straight to ground, so $i_x = -v_A/2$ (defined into $A$ , as before). All of $i_T$ enters $A$ through the 4 Ω. KCL at $A$ :

i_x + i_T = 3i_x \;\Longrightarrow\; i_T = 2i_x \;\Longrightarrow\; i_x = \tfrac{1}{2}\ \text{A} \;\Longrightarrow\; v_A = -2i_x = -1\ \text{V}

The test current drops $i_T R_2 = (1)(4) = 4$ V across the 4 Ω on its way in, so $v_T = v_A + 4 = 3$ V, giving

R_{Th} = \frac{v_T}{i_T} = \frac{3}{1} = 3\ \Omega \;\;\checkmark

Thevenin: 12 V in series with 3 Ω. Norton: 4 A in parallel with 3 Ω (

I_N = V_{Th}/R_{Th} = 12/3 = 4

A ✓). Deactivation was never on the table — both legal methods agree.

Open circuit: no current, no drop across the hidden

R_{Th}

, so the meter reads

V_{Th}

directly.

Loaded: current now flows through the hidden

R_{Th}

too, and its drop is what "sag" is.

The battery interpretation (Exercise 6, P2.81)

A real battery is a Thevenin equivalent: an ideal source behind an internal resistance — not a metaphor, a literally accurate circuit model, which is why the two diagrams above draw an ordinary battery with a dashed box labeled "hidden" around its own Thevenin parts. Measure 9 V open-circuit, then hang a 100 Ω load and watch it sag to 6 V. The load and $R_{Th}$ form a divider: $6 = 9\cdot\frac{100}{100+R_{Th}}$ , so $R_{Th} = 50\ \Omega$ . Two voltmeter readings, no access to the inside of the battery — that is Thevenin's theorem earning its keep.

4. Norton equivalents and source transformation

Rearrange the same terminal law for $i$ instead of $v$ :

i \;=\; \frac{V_{Th}}{R_{Th}} - \frac{v}{R_{Th}} \;=\; I_N - \frac{v}{R_N}

Read as KCL at the terminal: a current source

I_N

in parallel with

R_N

I_N = i_{sc} = \frac{V_{Th}}{R_{Th}}, \qquad R_N = R_{Th} = \frac{v_{oc}}{i_{sc}}

It is literally the same line, algebraically re-solved. So Thevenin and Norton are two readings of one fact, and you can swap between them at will — source transformation, which is often the fastest way to collapse a ladder of alternating sources and resistors.

Source transformation. Same two terminals, same straight line, two spellings.

Why this speeds up ladders

A voltage source in series with $R$ becomes a current source in parallel with $R$ — and parallel resistors combine, while series ones did not (they were separated by the source). Alternate transformations and reductions and a five-element ladder collapses in a few strokes, with no simultaneous equations at all.

Worked example — find the Norton equivalent directly, three ways

A 12 V source feeds a 4 Ω resistor into node $A$ ; a 6 Ω resistor runs from $A$ to ground; terminals $a$ – $b$ sit open right at $A$ and ground. No current source this time — the point is to go straight for $I_N$ and $R_N$ instead of finding Thevenin first and converting.

Way 1 — source transformation, by inspection. Transform the 12 V–in–series–with–4 Ω branch into its Norton form: $I = 12/4 = 3$ A in parallel with 4 Ω. That parallel source now sits directly between $A$ and ground alongside the 6 Ω — the circuit is already the Norton equivalent, no equations needed:

I_N = 3\ \text{A}, \qquad R_N = 4\parallel 6 = \frac{24}{10} = 2.4\ \Omega

Way 2 — $i_{sc}$ directly on the original circuit. Short $a$ – $b$ ( $v_A = 0$ ): the 6 Ω carries nothing, so all the source current takes the short:

i_{sc} = \frac{12}{4} = 3\ \text{A} = I_N\;\;\checkmark

Way 3 — deactivation for $R_N$ , plus $v_{oc}$ as a cross-check. Short the 12 V and look in from $a$ – $b$ : $4\parallel 6 = 2.4\ \Omega = R_N$ , same as way 1. For $v_{oc}$ , terminals open, one node equation:

\frac{v_A-12}{4}+\frac{v_A}{6}=0 \;\xrightarrow{\times 12}\; 3(v_A-12)+2v_A=0 \;\Rightarrow\; 5v_A=36 \;\Rightarrow\; v_{oc}=7.2\ \text{V}

Consistency check across all three: $V_{Th} = I_N R_N = 3\times 2.4 = 7.2$ V $\,=\, v_{oc}$ ✓

Norton: 3 A in parallel with 2.4 Ω — found by inspection via source transformation, then confirmed by direct

i_{sc}

and by deactivation.

5. Widget: the black-box probe

Below is a sealed network. You cannot see inside it. All you can do is what a lab bench lets you do: open the terminals and read a voltmeter, short them and read an ammeter, or hang a load and read both. Take measurements, plot them, and see if you can name $V_{Th}$ and $R_{Th}$ before you hit Reveal — every point you record must land on one straight line, and that line is the answer.

Black-box Thevenin probe

Load resistance 8.0 Ω

terminal V —

terminal I —

load P —

Take an Open reading and a Short reading — that is already enough.

Read the plot, not the meter

At any one load the ratio $v/i$ you measure is just $R_L$ — it tells you nothing directly about $R_{Th}$ . What identifies the network is the slope of the line through your points: $R_{Th} = -\Delta v/\Delta i$ , equivalently the $v$ -intercept over the $i$ -intercept, $v_{oc}/i_{sc}$ . One measurement is never enough; two of them anywhere on the line are always enough.

Self-check 2

6. The Wheatstone bridge

Two voltage dividers hung across the same source, compared against each other. Drawn as a diamond it looks exotic; drawn flat it is obviously just two dividers, and that is the way to think about it.

Classic diamond.

G

= galvanometer (or any load) across

c

–

d

Same circuit, flat: two dividers compared at

c

and

d

6.1 The balance condition, derived

Suppose the bridge is balanced: zero current through the detector arm. That single assumption does all the work, because if no current crosses the bridge then $R_1$ and $R_2$ carry the same current $i_L$ as each other (nothing is siphoned off at $c$ ), and likewise $R_3$ and $R_4$ share a common current $i_R$ . Each side is then a clean, undisturbed voltage divider:

v_c = v_s\,\frac{R_2}{R_1+R_2}, \qquad v_d = v_s\,\frac{R_4}{R_3+R_4}

Zero current through the detector also means zero volts across it (Ohm's law on the detector's own resistance), so $v_c = v_d$ :

\frac{R_2}{R_1+R_2} = \frac{R_4}{R_3+R_4}

\Rightarrow\; R_2R_3 + R_2R_4 = R_1R_4 + R_2R_4 \;\Rightarrow\; \boxed{R_1R_4 = R_2R_3}

Equivalently

R_1/R_2 = R_3/R_4

— the two dividers have the same ratio.

Why anyone cares

The balance condition contains no $v_s$ . So a bridge measurement is immune to supply drift, and it is a null measurement — you are not asking a meter "how many volts?", only "is this zero?", which instruments do far more accurately. Adjust a known decade resistor until the galvanometer reads nothing, and $R_{\text{unknown}} = R_{\text{known}}\cdot\frac{R_2}{R_1}$ , to whatever accuracy your ratio resistors have. This is how strain gauges, load cells, and RTD thermometers are read to this day.

6.2 The unbalanced bridge: Thevenin to the rescue

Off balance, the detector does draw current, so the two dividers are no longer independent and the divider formulas above stop being exact. Rather than write two node equations every time, Thevenin the bridge at the detector terminals — with the detector removed, the two dividers are undisturbed again, so:

v_{oc} = v_s\!\left(\frac{R_4}{R_3+R_4} - \frac{R_2}{R_1+R_2}\right), \qquad R_{Th} = (R_1\!\parallel\! R_2) + (R_3\!\parallel\! R_4)

R_{Th}

: short the source and the two divider tops collapse onto the two bottoms — each side becomes a parallel pair, and the two pairs are in series along the detector path.

Then the detector current for any galvanometer resistance is one division: $i_G = v_{oc}/(R_{Th}+R_G)$ . Practice problem 3 below runs this end to end.

Self-check 3

Worked example — balancing a bridge (find the unknown leg)

Left leg: $R_1 = 800\ \Omega$ on top, $R_2 = 200\ \Omega$ on the bottom. Right leg: $R_3 = 1200\ \Omega$ on top, unknown $R_4$ on the bottom. Supply $v_s = 10$ V. Find $R_4$ that zeroes the galvanometer, then confirm it two ways.

Apply the balance condition from §6.1 directly:

R_1R_4 = R_2R_3 \;\Longrightarrow\; R_4 = \frac{R_2R_3}{R_1} = \frac{200\cdot 1200}{800} = 300\ \Omega

Check via the ratio form. $R_1/R_2 = 800/200 = 4$ and $R_3/R_4 = 1200/300 = 4$ — equal, so the two dividers really do split $v_s$ identically.

✓

Check by computing both node voltages directly.

v_c = v_s\cdot\frac{R_2}{R_1+R_2} = 10\cdot\frac{200}{1000} = 2\ \text{V}, \qquad v_d = v_s\cdot\frac{R_4}{R_3+R_4} = 10\cdot\frac{300}{1500} = 2\ \text{V}

$v_c = v_d$ , so $v_{oc}$ across the galvanometer is zero regardless of $R_G$ — the bridge is balanced.

R_4 = 300\ \Omega

balances the bridge;

v_c = v_d = 2

V confirms it.

Worked example — unbalanced bridge, Thevenin vs. direct node analysis

Left leg: $R_1 = 2$ kΩ on top, $R_2 = 2$ kΩ on the bottom. Right leg: $R_3 = 1$ kΩ on top, $R_4 = 3$ kΩ on the bottom. Supply $v_s = 9$ V, galvanometer $R_G = 500\ \Omega$ . Find $i_G$ .

Check balance first. $R_1R_4 = (2\text{k})(3\text{k}) = 6\times10^6$ but $R_2R_3 = (2\text{k})(1\text{k}) = 2\times10^6$ — unequal, so current will flow.

Thevenin at the galvanometer. Remove it; each leg is an undisturbed divider:

v_c = 9\cdot\frac{2}{2+2} = 4.5\ \text{V}, \qquad v_d = 9\cdot\frac{3}{1+3} = 6.75\ \text{V}

v_{oc} = v_d - v_c = 2.25\ \text{V} \quad (d\text{ positive w.r.t. }c)

Deactivate the source for $R_{Th}$ — each leg collapses to a parallel pair, in series along the $c$ -to- $d$ path:

R_{Th} = (R_1\parallel R_2)+(R_3\parallel R_4) = 1\text{k} + 750 = 1750\ \Omega

i_G = \frac{v_{oc}}{R_{Th}+R_G} = \frac{2.25}{1750+500} = \frac{2.25}{2250} = 1\ \text{mA}

✓

Cross-check by direct node analysis with the galvanometer in place (top rail at 9 V, bottom grounded, unknowns $v_c$ , $v_d$ ):

\text{node }c:\ \frac{v_c-9}{2000}+\frac{v_c}{2000}+\frac{v_c-v_d}{500}=0 \;\xrightarrow{\times 2000}\; 6v_c-4v_d=9

\text{node }d:\ \frac{v_d-9}{1000}+\frac{v_d}{3000}+\frac{v_d-v_c}{500}=0 \;\xrightarrow{\times 3000}\; 10v_d-6v_c=27

Solving: $v_d = 6$ V, $v_c = 5.5$ V, so

i_G = \frac{v_d-v_c}{500} = \frac{0.5}{500} = 1\ \text{mA}\;\;\checkmark

Unbalanced.

v_{oc} = 2.25

R_{Th} = 1750\ \Omega

i_G = 1

mA from

d

c

7. Maximum power transfer — and the trap in it

Hang $R_L$ on a Thevenin source. The load power is

p_L = i^2R_L = \frac{V_{Th}^2\,R_L}{(R_{Th}+R_L)^2}

Differentiate with respect to $R_L$ and set to zero and you get the familiar matching condition $R_L = R_{Th}$ , giving $p_{max} = V_{Th}^2/(4R_{Th})$ . Note the sanity check at the extremes: at $R_L = 0$ there is current but no load voltage; as $R_L \to \infty$ there is voltage but no current; the maximum sits between them.

Exercise 6, P2.91 — read which resistor is the variable

" $V_{Th} = 20$ V, load $R_L = 5\ \Omega$ fixed — choose $R_{Th}$ for maximum load power." The matching condition does not apply: it was derived by varying $R_L$ with $R_{Th}$ fixed. Here it is the other way round, and $p_L = \frac{400 R_L}{(R_{Th}+5)^2}$ is monotonically decreasing in $R_{Th}$ — every ohm of source resistance is wasted heat. The best you can do is $R_{Th} \to 0$ , giving $p_L = 20^2/5 = 80$ W. Matching is not "always make them equal"; it is "the best a fixed source can do for a load you get to choose."

8. Practice problems

Every solution below was worked twice, by two independent methods, and shipped only where both agreed. Try them cold before opening the solutions.

P1 — Thevenin, Norton, and a load~10 min

(a) Find the Thevenin equivalent at $a$ – $b$ . (b) Write the Norton equivalent. (c) An $8\ \Omega$ load is connected across $a$ – $b$ : find its current and power.

(a) $v_{oc}$ . With $a$ – $b$ open, no current flows in $R_3$ , so there is no drop across it and $v_{oc} = v_A$ . One node equation at $A$ :

\frac{v_A - 24}{6} + \frac{v_A}{12} = 3 \;\;\xrightarrow{\times 12}\;\; 2v_A - 48 + v_A = 36 \;\Rightarrow\; 3v_A = 84 \;\Rightarrow\; v_A = 28\ \text{V}

So $V_{Th} = v_{oc} = 28$ V.

$R_{Th}$ — method 1, deactivation (legal: both sources independent). Short the 24 V, open the 3 A. Looking in from $a$ – $b$ : $R_3$ in series with the parallel pair $R_1 \parallel R_2$ :

R_{Th} = 4 + (6\parallel 12) = 4 + \frac{6\cdot 12}{18} = 4 + 4 = 8\ \Omega

$R_{Th}$ — method 2, $v_{oc}/i_{sc}$ (independent cross-check). Short $a$ – $b$ , so $R_3$ now runs from $A$ to ground:

\frac{v_A - 24}{6} + \frac{v_A}{12} + \frac{v_A}{4} = 3 \;\xrightarrow{\times 12}\; 2v_A - 48 + v_A + 3v_A = 36 \;\Rightarrow\; v_A = 14\ \text{V}

i_{sc} = \frac{14}{4} = 3.5\ \text{A} \qquad R_{Th} = \frac{28}{3.5} = 8\ \Omega\;\checkmark

(b) Norton: $I_N = i_{sc} = 3.5$ A in parallel with $R_N = R_{Th} = 8\ \Omega$ . Consistency: $I_N = V_{Th}/R_{Th} = 28/8 = 3.5$ A ✓

(c) Via the equivalent:

i_L = \frac{28}{8+8} = 1.75\ \text{A}, \qquad p_L = i_L^2R_L = (1.75)^2(8) = 24.5\ \text{W}

Cross-check by direct node analysis on the full original circuit with the load attached (two nodes: $v_A$ and $v_a$ ):

\text{node }a:\ \frac{v_a - v_A}{4} + \frac{v_a}{8} = 0 \;\xrightarrow{\times 8}\; 3v_a = 2v_A

\text{node }A:\ \frac{v_A-24}{6} + \frac{v_A}{12} + \frac{v_A - v_a}{4} = 3 \;\xrightarrow{\times 12}\; 6v_A - 3v_a = 84

Substituting $3v_a = 2v_A$ : $\,6v_A - 2v_A = 84 \Rightarrow v_A = 21$ V, $v_a = 14$ V. Then $i_L = 14/8 = 1.75$ A and $p_L = 14^2/8 = 24.5$ W — both methods agree.

V_{Th} = 28

R_{Th} = 8\ \Omega

; Norton: 3.5 A ∥ 8 Ω; with an 8 Ω load,

i_L = 1.75

A and

p_L = 24.5

P2 — Superposition, and the power trap~10 min

Both sources drive current toward $v$ (same polarity sense, $+$ on top). Branch 3 carries $i$ .

(a) Find $i$ by superposition. (b) Find the power in the 2 Ω resistor. (c) Compute the power each source would produce in it acting alone, add those, and explain the discrepancy quantitatively.

(a) 12 V alone (short the 6 V source, which puts its $3\ \Omega$ from $v$ straight to ground, in parallel with the $2\ \Omega$ ):

3\parallel 2 = \frac{6}{5} = 1.2\ \Omega, \qquad v' = 12\cdot\frac{1.2}{6+1.2} = 12\cdot\frac{1.2}{7.2} = 2\ \text{V}

so $i' = v'/2 = 1$ A.

6 V alone (short the 12 V source, putting its $6\ \Omega$ in parallel with the $2\ \Omega$ ):

6\parallel 2 = 1.5\ \Omega, \qquad v'' = 6\cdot\frac{1.5}{3+1.5} = 6\cdot\frac{1.5}{4.5} = 2\ \text{V}

so $i'' = 1$ A. Both push current downward through the $2\ \Omega$ , so

i = i' + i'' = 2\ \text{A}, \qquad v = 4\ \text{V}

Cross-check 1 — node analysis with both sources live:

\frac{v-12}{6} + \frac{v-6}{3} + \frac{v}{2} = 0 \;\xrightarrow{\times 6}\; (v-12) + 2(v-6) + 3v = 0 \;\Rightarrow\; 6v = 24 \;\Rightarrow\; v = 4\ \text{V}\;\checkmark

Cross-check 2 — mesh analysis. Clockwise mesh currents $i_a$ (left loop: 12 V, 6 Ω, shared 2 Ω) and $i_b$ (right loop: shared 2 Ω, 3 Ω, 6 V), with the shared-branch current downward equal to $i_a - i_b$ :

-12 + 6i_a + 2(i_a - i_b) = 0 \;\Rightarrow\; 8i_a - 2i_b = 12

2(i_b - i_a) + 3i_b + 6 = 0 \;\Rightarrow\; -2i_a + 5i_b = -6

Solving: $i_a = 4/3$ A, $i_b = -2/3$ A, so $i = i_a - i_b = 2$ A ✓ Three methods, one answer.

(b) $p = i^2R = (2)^2(2) = 8$ W.

(c) Sub-powers: $p' = (1)^2(2) = 2$ W and $p'' = (1)^2(2) = 2$ W, summing to 4 W — only half the true 8 W. The missing piece is exactly the cross term:

2\,i'i''R = 2(1)(1)(2) = 4\ \text{W}, \qquad 2 + 2 + 4 = 8\ \text{W}\;\checkmark

Power superposes only when the cross term vanishes (e.g. one contribution is zero), which is why the rule is: superpose currents, then square once.

i = 2

p_{2\Omega} = 8

W. Adding sub-powers gives 4 W and is wrong — it drops the

2i'i''R = 4

W cross term.

P3 — Unbalanced Wheatstone bridge~12 min

Left leg: $R_1 = 3$ kΩ on top, $R_2 = 1$ kΩ on the bottom, meeting at node $c$ . Right leg: $R_3 = 1$ kΩ on top, $R_4 = 1$ kΩ on the bottom, meeting at node $d$ . A galvanometer of $R_G = 750\ \Omega$ bridges $c$ – $d$ .

(a) Is the bridge balanced? (b) Find the Thevenin equivalent seen by the galvanometer and hence the galvanometer current. (c) What value of $R_1$ would balance the bridge?

(a) $R_1R_4 = (3\text{k})(1\text{k}) = 3\times10^6$ but $R_2R_3 = (1\text{k})(1\text{k}) = 1\times10^6$ . Not equal ⇒ unbalanced, so current will flow through the galvanometer.

(b) Remove the galvanometer. With it gone each leg is an undisturbed divider (bottom node grounded, top node at 12 V):

v_c = 12\cdot\frac{R_2}{R_1+R_2} = 12\cdot\frac{1}{4} = 3\ \text{V}, \qquad v_d = 12\cdot\frac{R_4}{R_3+R_4} = 12\cdot\frac{1}{2} = 6\ \text{V}

v_{oc} = v_d - v_c = 3\ \text{V} \quad (\text{

For $R_{Th}$ , deactivate the source: shorting it ties the two leg-tops to the leg-bottoms, so each leg collapses to a parallel pair, and the path $c \to d$ passes through both in series:

R_{Th} = (R_1\parallel R_2) + (R_3\parallel R_4) = \frac{3000\cdot1000}{4000} + \frac{1000\cdot1000}{2000} = 750 + 500 = 1250\ \Omega

i_G = \frac{v_{oc}}{R_{Th}+R_G} = \frac{3}{1250+750} = \frac{3}{2000} = 1.5\ \text{mA}\ \ (d \to c)

Cross-check by direct node analysis on the full bridge with the galvanometer in place (ground the bottom rail, top rail at 12 V, unknowns $v_c$ and $v_d$ ):

\text{node }c:\ \frac{v_c-12}{3000}+\frac{v_c}{1000}+\frac{v_c-v_d}{750}=0 \;\xrightarrow{\times 3000}\; 8v_c - 4v_d = 12 \;\Rightarrow\; 2v_c - v_d = 3

\text{node }d:\ \frac{v_d-12}{1000}+\frac{v_d}{1000}+\frac{v_d-v_c}{750}=0 \;\xrightarrow{\times 3000}\; 10v_d - 4v_c = 36 \;\Rightarrow\; 5v_d - 2v_c = 18

Adding the two reduced equations: $4v_d = 21 \Rightarrow v_d = 5.25$ V, then $v_c = (3+5.25)/2 = 4.125$ V. So

i_G = \frac{v_d - v_c}{750} = \frac{1.125}{750} = 1.5\ \text{mA}\;\checkmark

Note that with the galvanometer loading the bridge, $v_c$ and $v_d$ are not 3 V and 6 V any more — but their difference drives the same current the Thevenin model predicted, which is the whole point of the equivalent.

(c) Balance requires $R_1R_4 = R_2R_3$ , so $R_1 = R_2R_3/R_4 = (1\text{k})(1\text{k})/(1\text{k}) = 1$ kΩ. Check: then $v_c = 12\cdot\frac{1}{2} = 6$ V $= v_d$ , so $v_{oc} = 0$ and no current flows regardless of $R_G$ ✓

Unbalanced.

v_{oc} = 3

R_{Th} = 1250\ \Omega

i_G = 1.5

mA from

d

c

. Balanced at

R_1 = 1

kΩ.

P4 — Thevenin with a dependent source~12 min

$i_x$ is the current flowing rightward through the 4 Ω resistor into node $A$ . The dependent current source is worth $2i_x$ , arrow pointing from $A$ down to ground. Terminal $b$ is ground.

(a) Find the Thevenin equivalent at $a$ – $b$ — remember, a dependent source is never deactivated. (b) Write the Norton equivalent. (c) A $3\ \Omega$ load is attached at $a$ – $b$ : find its current and power.

(a) $v_{oc}$ . With $a$ – $b$ open, no current flows in the 6 Ω, so $v_{oc} = v_A$ . All of $i_x$ must leave through the dependent source (the only other path), so $i_x = 2i_x \Rightarrow i_x = 0$ . With no drop across the 4 Ω, $v_A = 20$ V.

v_{oc} = 20

$R_{Th}$ via $i_{sc}$ (mandatory here — the source is dependent, so deactivation is illegal). Short $a$ – $b$ : the 6 Ω now runs from $A$ straight to ground, in parallel with the dependent source's path. With $i_x = (20-v_A)/4$ :

i_x = \frac{v_A}{6} + 2i_x \;\Longrightarrow\; -i_x = \frac{v_A}{6} \;\Longrightarrow\; -\frac{20-v_A}{4} = \frac{v_A}{6}

Cross-multiplying: $-6(20-v_A) = 4v_A \Rightarrow -120+6v_A=4v_A \Rightarrow v_A=60$ V, so $i_{sc} = v_A/6 = 10$ A and

R_{Th} = \frac{v_{oc}}{i_{sc}} = \frac{20}{10} = 2\ \Omega

Cross-check by test source. Kill the 20 V source (short it — that's legal, it's independent), keep the dependent source, and inject a 1 A test current into terminal $a$ . Now the 4 Ω runs from $A$ to ground directly, so $i_x = -v_A/4$ (current from ground into $A$ ). Nodal at $a$ : $1 = (v_a - v_A)/6 \Rightarrow v_a = v_A + 6$ . Nodal at $A$ (currents leaving): $v_A/4 - 1 - v_A/2 = 0 \Rightarrow -v_A/4 = 1 \Rightarrow v_A = -4$ V, so $v_a = 2$ V and $R_{Th} = v_a/1 = 2\ \Omega$ ✓ — matches exactly.

(b) Norton: $I_N = i_{sc} = 10$ A in parallel with $R_N = 2\ \Omega$ . Consistency: $I_N = V_{Th}/R_{Th} = 20/2 = 10$ A ✓

(c) Via the equivalent:

i_L = \frac{20}{2+3} = 4\ \text{A}, \qquad p_L = i_L^2R_L = 4^2(3) = 48\ \text{W}

Cross-check by direct nodal analysis on the full original circuit with the 3 Ω load attached (unknowns $v_A$ , $v_a$ ; $i_x=(20-v_A)/4$ ):

\text{node }A:\ \frac{v_A-20}{4} + \frac{v_A-v_a}{6} + 2i_x = 0 \;\xrightarrow{\times 12}\; -v_A - 2v_a + 60 = 0

\text{node }a:\ \frac{v_a-v_A}{6} + \frac{v_a}{3} = 0 \;\Longrightarrow\; v_A = 3v_a

Substituting: $-3v_a - 2v_a + 60 = 0 \Rightarrow v_a = 12$ V, so $i_L = v_a/3 = 4$ A and $p_L = 4^2(3) = 48$ W — both methods agree.

V_{Th} = 20

R_{Th} = 2\ \Omega

; Norton: 10 A ∥ 2 Ω; with a 3 Ω load,

i_L = 4

A and

p_L = 48

P5 — sizing a load for maximum power~8 min

(a) Find the Thevenin equivalent at $a$ – $b$ . (b) What value of $R_L$ delivers maximum power to the load, and what is that power? (c) Show that $R_L = 10\ \Omega$ delivers less, to confirm the match isn't just a formula — it's a real optimum.

(a) $v_{oc}$ . Open at $a$ – $b$ , no drop across the right 3 Ω, so $v_{oc}=v_A$ :

\frac{v_A-15}{3}+\frac{v_A}{6}=0 \;\xrightarrow{\times 6}\; 2(v_A-15)+v_A=0 \;\Longrightarrow\; v_A = 10\ \text{V}

$R_{Th}$ (independent source, deactivation is legal): short the 15 V, and from $a$ – $b$ you see the right 3 Ω in series with $(3\parallel 6)$ :

R_{Th} = 3 + (3\parallel 6) = 3 + 2 = 5\ \Omega

(b) Maximum power transfer: $R_L = R_{Th} = 5\ \Omega$ , and

P_{max} = \frac{v_{oc}^2}{4R_{Th}} = \frac{100}{20} = 5\ \text{W}

Cross-check by direct nodal analysis with $R_L=5\ \Omega$ attached (unknowns $v_A$ , $v_a$ ):

\text{node }A:\ \frac{v_A-15}{3}+\frac{v_A}{6}+\frac{v_A-v_a}{3}=0 \;\Longrightarrow\; 5v_A-2v_a=30

\text{node }a:\ \frac{v_a-v_A}{3}+\frac{v_a}{5}=0 \;\Longrightarrow\; v_A = \tfrac{8}{5}v_a

Substituting: $8v_a - 2v_a = 30 \Rightarrow v_a = 5$ V, so $i_L = v_a/R_L = 1$ A and $p_L = i_L^2R_L = 1^2(5) = 5$ W — matches $P_{max}$ exactly, as it should when $R_L=R_{Th}$ .

(c) With $R_L = 10\ \Omega$ :

i_L = \frac{v_{oc}}{R_{Th}+R_L} = \frac{10}{15} = \tfrac{2}{3}\ \text{A}, \qquad p_L = \left(\tfrac{2}{3}\right)^2(10) = \tfrac{40}{9} \approx 4.44\ \text{W}

Less than 5 W, even though $R_L$ is larger and would carry a "safer" (smaller) current. Bigger $R_L$ isn't better or worse on its own — only matching $R_{Th}$ is optimal.

V_{Th} = 10

R_{Th} = 5\ \Omega

; max power at

R_L=5\ \Omega

is 5 W. At

R_L=10\ \Omega

, only 4.44 W.

Exam checklist for this chapter

1. "Find the current in one particular resistor while everything else is fixed" → Thevenin the rest of the circuit at that resistor's terminals. 2. "This one resistor takes several different values" → Thevenin once, then divide, repeatedly. 3. "Several sources, one output" → superposition, especially if one source makes a whole branch trivial. 4. Any dependent source → never deactivate it; get $R_{Th}$ from $v_{oc}/i_{sc}$ or a test source. 5. Any power question → get the total current first, then square. 6. A four-resistor diamond → check $R_1R_4 = R_2R_3$ before doing any work; if it balances, the bridge arm carries nothing and the circuit falls apart into two dividers.