Use Norton equivalent circuits: Worked examples — Hardware deep dive

Intuition What this page is for

The parent note taught the recipe: short for the current, kill for the resistance. Here we grind that recipe against every kind of circuit it can meet — one source, two sources, a current source, a dependent source, degenerate zero-cases, a negative-current case, an active negative-resistance case, and a word problem. If a case exists, you will have seen it worked before an exam shows it to you.

Every symbol used here was defined in the parent note. As a one-line refresher:

$I_{N}$ = current through a wire placed across the two output terminals.
$R_{N}$ = resistance seen at the terminals with sources switched off (voltage source → wire, current source → gap).
Terminal law: $I_{L} = I_{N} - \frac{V}{R _{N}}$ , where $V$ is the voltage across the terminals and $I_{L}$ the current handed to the load. We will actually apply this law in Ex 6 and Ex 9, not just quote it.

The scenario matrix

Before working anything, let's list every case class a Norton problem can belong to. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	Single voltage source, series–parallel resistors	baseline recipe	Ex 1
B	Single current source	source stays as current, deactivate = open	Ex 2
C	Two independent sources	use Superposition or nodal for $I_{N}$	Ex 3
D	Dependent source present	can't kill it; test-source for $R_{N}$	Ex 4
E	Degenerate: short-circuit branch ( $R \to 0$ )	$R_{N} = 0$ , model becomes ideal source	Ex 5
F	Degenerate: open branch / no source	$I_{N} = 0$ or $R_{N} \to \infty$	Ex 6
G	Sign case: negative $I_{N}$ / reversed-polarity load	apply the terminal law with signs	Ex 7
H	Active negative $R_{N}$ (strong dependent source)	test source gives $R_{N} < 0$	Ex 8
I	Word problem + Maximum power transfer ( $P_{max}$ )	translate words → circuit, compute $P_{max}$	Ex 9
J	Exam twist: find $R_{N}$ by the $V_{oc} / I_{sc}$ method	two-measurement shortcut	Ex 10

The signs to watch: a current source can push current the "wrong" way through the short (negative $I_{N}$ , Ex 7), and a strong dependent source can make the terminal slope run backwards (negative $R_{N}$ , Ex 8). Both are worked below.

Cell A — baseline: one voltage source

Short a–b, find $I_{N}$ . Placing a wire across a–b joins both ends of $R_{2}$ , so $R_{2}$ carries $0 V$ and is invisible. Only $R_{1}$ limits the source: $I_{N} = \frac{24}{8} = 3 A .$ Why this step? A short forces terminal voltage $V = 0$ , and the terminal law gives $I_{L} = I_{N} - 0 = I_{N}$ . The short reads off $I_{N}$ directly.
Kill the source, find $R_{N}$ . Replace $24 V$ with a wire. Now the tops of $R_{1}$ and $R_{2}$ are tied together and both hang between the same two nodes a–b — they are in parallel: $R_{N} = \frac{8 \cdot 12}{8 + 12} = \frac{96}{20} = 4.8 Ω.$ Why this step? $R_{N}$ is the slope of the $I$ – $V$ line, set purely by resistors (Ohm's Law); sources only shift the line, so we switch them off to isolate it.
Draw $I_{N} = 3 A ∥ R_{N} = 4.8 Ω$ .

Verify: cross-check with Thévenin equivalent circuits. Open-circuit voltage should be $V_{T h} = I_{N} R_{N} = 3 \times 4.8 = 14.4 V$ . Independently: with a–b open, the terminal draws nothing, so $R_{2}$ carries the full source current $24/ (8 + 12) = 1.2 A$ and $V_{oc} = 1.2 \times 12 = 14.4 V$ . ✓ Match.

So $I_{N} = 3 A$ exactly as the forecast's "hold that thought" hinted — $R_{2}$ vanished under the short.

Cell B — a current source instead

Short a–b, find $I_{N}$ . A wire across a–b gives a $0 Ω$ path in parallel with both $R_{1}$ and $R_{2}$ . All $2 A$ prefers the wire (zero resistance), so: $I_{N} = 2 A .$ Why this step? Any resistor in parallel with a short is bypassed — the entire source current flows through the short. The current source hands its whole $2 A$ to the short.
Deactivate the source, find $R_{N}$ . For a current source, "off" means open (a gap) — a current source that pushes zero current is an open circuit. With that gap, $R_{1}$ and $R_{2}$ both bridge a–b in parallel: $R_{N} = \frac{5 \cdot 20}{5 + 20} = \frac{100}{25} = 4 Ω.$ Why open, not short? Shorting a current source would create a second $0 Ω$ loop and misread the resistance. Opening it removes its influence while leaving the resistor network intact.
Draw $I_{N} = 2 A ∥ R_{N} = 4 Ω$ .

Verify: $V_{oc} = I_{N} R_{N} = 2 \times 4 = 8 V$ . Independently with a–b open, the $2 A$ splits through $R_{1} ∥ R_{2} = 4 Ω$ , giving $V_{a} = 2 \times 4 = 8 V$ . ✓ Units: A·Ω = V. ✓

Forecast answered: barely any work — a current source is the answer once you route it through the short.

Cell C — two sources at once

Short a–b, find $I_{N}$ by Superposition. The short holds node a at $0 V$ , so $R_{3}$ (also across a–b) carries no current — ignore it. Each source drives its own branch straight into the short: $I_{N} = \frac{10}{2} + \frac{6}{3} = 5 + 2 = 7 A .$ Why superposition? In a linear network the short-circuit current is the sum of what each source alone would push through the short (Kirchhoff's Current Law at the shorted node). Both sources point current into a and out the short, so they add.
Kill both sources, find $R_{N}$ . Replace both voltage sources with wires. Now $R_{1}$ , $R_{2}$ , $R_{3}$ all connect a to b (their far ends are all grounded through the wires) — three resistors in parallel: $\frac{1}{R _{N}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1 \Rightarrow R_{N} = 1 Ω.$ Why this step? Deactivating both sources ties every resistor's far end to the same rail, so all three bridge a–b.
Draw $I_{N} = 7 A ∥ R_{N} = 1 Ω$ .

Verify: $V_{oc} = 7 \times 1 = 7 V$ . Independent nodal check with a–b open, at node a: $\frac{V - 10}{2} + \frac{V - 6}{3} + \frac{V}{6} = 0 \Rightarrow 3 (V - 10) + 2 (V - 6) + V = 0 \Rightarrow 6 V = 42 \Rightarrow V = 7 V .$ ✓

Forecast confirmed: the two short-circuit currents added to $7 A$ .

Cell D — a dependent source

Short a–b, find $I_{N}$ . Short forces $V_{a} = 0$ , so $V_{x} = 0$ , so the dependent source $g V_{x} = 0$ contributes nothing. Only the $12 V$ source drives the short through $R_{1}$ : $I_{N} = \frac{12}{4} = 3 A .$ Why this step? The dependent source is controlled by $V_{x}$ ; the short zeroes $V_{x}$ , silencing it. $R_{2}$ is shorted out too, so $R_{1}$ alone sets the short-circuit current.
Find $R_{N}$ with a test source — do NOT deactivate the dependent source. Why this step? A dependent source has no independent "on/off" — its output tracks $V_{x}$ , which we are about to change. You cannot switch off something that only reacts. So we kill only the independent $12 V$ source (→ wire), then apply a known test voltage $V_{t} = 1 V$ at a–b and measure the current $I_{t}$ it draws; their ratio $R_{N} = V_{t} / I_{t}$ is the resistance the dependent source helps define. With the independent source off, $V_{a} = 1$ , $V_{x} = 1$ , and KCL at node a (current out through resistors minus current injected by the dependent source $= I_{t}$ ): $I_{t} = 0.25 \frac{1}{R _{1}} + 0.25 \frac{1}{R _{2}} - 0.25 \cdot 1 g V_{x} = 0.25 + 0.25 - 0.25 = 0.25 A .$ $R_{N} = \frac{V _{t}}{I _{t}} = \frac{1}{0.25} = 4 Ω.$ The dependent source's minus sign here reduces $I_{t}$ , raising $R_{N}$ above the plain $R_{1} ∥ R_{2} = 2 Ω$ .
Draw $I_{N} = 3 A ∥ R_{N} = 4 Ω$ .

Verify: $V_{oc}$ should equal $I_{N} R_{N} = 3 \times 4 = 12 V$ . Independent open-circuit nodal at a (no load current): $\frac{V - 12}{4} + \frac{V}{4} - 0.25 V = 0 \Rightarrow (0.25 + 0.25 - 0.25) V = 3 \Rightarrow 0.25 V = 3 \Rightarrow V = 12 V$ . ✓

Cell E — degenerate: a short across the terminals of a source

Short a–b, find $I_{N}$ . Shorting an ideal voltage source demands $V = 0$ across a device that insists $V = 9 V$ — the model gives infinite current. This is a degenerate case: a pure ideal voltage source has no finite Norton current. Why? $I_{N} = V / R_{N}$ and here $R_{N} = 0$ , so $I_{N} \to \infty$ . The Norton form breaks down.
Kill the source, find $R_{N}$ . Replace $9 V$ with a wire: terminals a–b are joined directly, so $R_{N} = 0 Ω.$
Interpretation: an ideal voltage source is the $R_{N} \to 0$ limit — it has a perfect Thévenin equivalent circuits ( $V_{T h} = 9 V, R_{T h} = 0$ ) but no proper Norton equivalent (you'd need $I_{N} \to \infty$ ). Add any real series resistance $r$ and it recovers: $I_{N} = 9/ r$ , $R_{N} = r$ .

Verify: with $r = 3 Ω$ added, $I_{N} = 9/3 = 3 A$ , $R_{N} = 3 Ω$ , and $V_{oc} = I_{N} R_{N} = 3 \times 3 = 9 V$ . ✓ As $r \to 0$ , $I_{N} \to \infty$ — confirms the degeneracy.

Cell F — degenerate: no source, or an open branch (apply the terminal law)

(F1) Short a–b, find $I_{N}$ . No source ⇒ no drive ⇒ no current: $I_{N} = 0 A$ . Kill "sources" (there are none), find $R_{N} = R_{1} + R_{2} = 10 + 15 = 25 Ω$ . Result: $I_{N} = 0 ∥ R_{N} = 25 Ω$ — this is just a $25 Ω$ resistor (a dead network is a pure resistor). ✓ sanity: a resistor's $I$ – $V$ line passes through the origin, so intercept $I_{N} = 0$ .
Apply the terminal law $I_{L} = I_{N} - V / R_{N}$ . Attach a $5 V$ source externally across a–b (so $V = 5 V$ is forced) and predict the current the network absorbs: $I_{L} = I_{N} - \frac{V}{R _{N}} = 0 - \frac{5}{25} = - 0.2 A .$ Why the minus? $I_{L}$ is defined as current delivered by the network. A dead $25 Ω$ resistor delivers nothing; instead it draws $0.2 A$ , which shows up as a negative $I_{L}$ . This is the terminal law doing real work, not decoration.
(F2) Open branch. The switch open isolates the $5 A$ source from a–b. Shorting a–b draws $I_{N} = 0 A$ ; with no resistor path to the terminals, $R_{N} \to \infty$ (an open circuit). The terminal law then gives, for any $V$ : $I_{L} = 0 - \frac{V}{\infty} = 0 A .$ Why $R_{N} \to \infty$ ? An open circuit is the $R_{N} \to \infty$ limit; no current flows no matter the applied voltage — correct for isolated terminals.

Verify: (F1) $V_{oc} = I_{N} R_{N} = 0 \times 25 = 0 V$ — a dead network shows no open-circuit voltage. ✓ Terminal-law prediction: $V = 5 V$ across $25 Ω$ draws $5/25 = 0.2 A$ (magnitude), so $I_{L} = - 0.2 A$ . ✓ (F2) $I_{L} = 0$ for any $V$ . ✓

Cell G — a sign case: negative $I_{N}$ and a reversed-polarity load

Short a–b, find $I_{N}$ . With a wire across a–b, the source's $4 A$ is forced through the short — but it flows into terminal a and out the source, i.e. into the box, which is the negative direction for $I_{N}$ (defined as current leaving a through the short toward the load): $I_{N} = - 4 A .$ Why negative? $I_{N}$ 's sign encodes direction. A source that drains the terminal produces a short-circuit current pointing the opposite way — the minus sign carries that physics. $R$ is shorted out and irrelevant here.
Kill the source (open it), find $R_{N}$ . Only $R$ bridges a–b: $R_{N} = 3 Ω.$
Attach $R_{L} = 3 Ω$ ; apply the terminal law. The Norton source feeds a divider; equivalently, with $R_{L}$ across the model the terminal voltage satisfies $V = I_{L} R_{L}$ and $I_{L} = I_{N} - V / R_{N}$ . Solve together: $I_{L} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}} = - 4 \cdot \frac{3}{6} = - 2 A .$ The negative $I_{L}$ means current actually flows into terminal a from the load side — a reversed-polarity delivery. The load voltage is $V = I_{L} R_{L} = - 2 \times 3 = - 6 V$ : terminal a sits below b. Why carry the sign through? Flipping to "magnitude 2 A" hides that the box is absorbing, not sourcing. The sign is the answer's most important digit.

Verify: plug back into the terminal law: $I_{L} = I_{N} - V / R_{N} = - 4 - (- 6) /3 = - 4 + 2 = - 2 A$ . ✓ Self-consistent, and $V = I_{L} R_{L} = - 2 \times 3 = - 6 V$ . ✓ Units: A·Ω = V. ✓

Forecast confirmed: a draining source gives a negative $I_{N}$ , and the terminal law propagates that sign cleanly into a reversed-polarity load.

Cell H — active network: a negative Norton resistance

Apply a test source. Force $V_{t} = 1 V$ at a, so $V_{x} = 1$ . KCL at node a (current out through $R_{1}$ minus current injected by the dependent source $= I_{t}$ ): $I_{t} = \frac{V _{t}}{R _{1}} - g V_{x} = \frac{1}{6} - 0.5 \cdot 1 = 0.1667 - 0.5 = - 0.3333 A .$ Why this step? No independent source and a dependent source ⇒ the only way to read the terminal slope is to probe with a known $V_{t}$ and see the response $I_{t}$ .
Compute $R_{N}$ . $R_{N} = \frac{V _{t}}{I _{t}} = \frac{1}{- 1/3} = - 3 Ω.$ Why negative? The dependent source pumps more current into node a ( $0.5 A$ ) than $R_{1}$ can drain ( $0.1667 A$ ) at $1 V$ , so raising the terminal voltage makes the network push current back out — the $I$ – $V$ slope runs backwards. A negative $R_{N}$ is the signature of an active network (it can deliver power indefinitely, like an oscillator's negative-resistance stage).
Interpretation. The Norton model still holds formally: $I_{L} = I_{N} - V / R_{N}$ with $R_{N} = - 3 Ω$ . But this network is not passive — connect the wrong load and it may not settle. That is exactly why exam problems flag "check the sign of $R_{N}$ ".

Verify: self-check the KCL: at $V_{t} = 1$ , out through $R_{1}$ is $1/6$ ; injected is $0.5$ ; net drawn $I_{t} = 1/6 - 1/2 = - 1/3 A$ , so $R_{N} = 1/ (- 1/3) = - 3 Ω$ . ✓ Sign is negative ⇒ active network. ✓

Cell I — real-world word problem + maximum power

(a) $I_{N}$ : short the sensor terminals. The battery drives its EMF through $R_{i}$ + cable in series: $I_{N} = \frac{6}{R _{i} + R _{cable}} = \frac{6}{0.5 + 1.5} = \frac{6}{2} = 3 A .$ Why this step? Shorting the terminals ( $V = 0$ ) makes the battery push its full current through the total series resistance.
$R_{N}$ : kill the battery EMF. Replace the $6 V$ EMF with a wire; the internal and cable resistances remain in series: $R_{N} = R_{i} + R_{cable} = 0.5 + 1.5 = 2 Ω.$
(b) Max power transfer. Power in $R_{L}$ is maximised when $R_{L} = R_{N} = 2 Ω$ (Maximum power transfer).

Definition The source-transformation bridge

"Source transformation" (Source transformation) is the pair of conversion rules between the two equivalents: $R_{N} = R_{T h}$ and $V_{T h} = I_{N} R_{N}$ . It lets us jump from the Norton numbers to the open-circuit voltage $V_{T h}$ without redrawing the circuit. Here it gives $V_{T h} = I_{N} R_{N} = 3 \times 2 = 6 V$ (which equals the battery EMF, as it must).

Matched load current and power: $I_{L} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}} = 3 \cdot \frac{2}{4} = 1.5 A, P_{max} = I_{L}^{2} R_{L} = 1. 5^{2} \times 2 = 4.5 W .$ Equivalently, the standard closed form $P_{max} = \frac{V _{T h}^{2}}{4 R _{N}} = \frac{6 ^{2}}{4 \times 2} = \frac{36}{8} = 4.5 W$ agrees. Why match? Too small an $R_{L}$ and voltage collapses; too large and current starves. The sweet spot is $R_{L} = R_{N}$ , giving $P_{max} = 4.5 W$ .

Verify: Power in $R_{N}$ at match: $P_{R_{N}} = 1. 5^{2} \times 2 = 4.5 W$ too, so efficiency $= 4.5/ (4.5 + 4.5) = 50%$ . ✓ Forecast: matching gives exactly half the power to the load, not most of it. ✓ Both formulas give $P_{max} = 4.5 W$ . ✓ Units: A²·Ω = W. ✓

Cell J — exam twist: $R_{N}$ from two measurements

$I_{N}$ is the short-circuit current — read directly. $I_{N} = I_{sc} = 4 A .$ Why this step? $I_{N}$ is defined as the terminal current when shorted. The ammeter measured exactly that.
$R_{N}$ from the ratio of the two readings. Both meters probe the same $I$ – $V$ line. Open circuit gives $V_{oc} = I_{N} R_{N}$ ; solve for $R_{N}$ : $R_{N} = \frac{V _{oc}}{I _{sc}} = \frac{20}{4} = 5 Ω.$ Why this step? The line runs from intercept $(V = 0, I = I_{sc})$ to $(V = V_{oc}, I = 0)$ ; its slope magnitude is $I_{sc} / V_{oc} = 1/ R_{N}$ . This is the professional way to measure $R_{N}$ without opening the box or knowing internal sources.
Draw $I_{N} = 4 A ∥ R_{N} = 5 Ω$ , equivalently Thévenin equivalent circuits $V_{T h} = 20 V, R_{T h} = 5 Ω$ .

Verify: self-consistency: $V_{oc} = I_{N} R_{N} = 4 \times 5 = 20 V$ ✓ and $I_{sc} = V_{oc} / R_{N} = 20/5 = 4 A$ ✓. Attach $R_{L} = 5 Ω$ : $I_{L} = 4 \cdot 5/10 = 2 A$ , load voltage $2 \times 5 = 10 V$ — exactly half of $V_{oc}$ , the matched-load signature. ✓

Recall Self-test across the whole matrix

One current source ( $2 A$ ) alone across $R_{1} ∥ R_{2} = 4 Ω$ : what is $I_{N}$ ? ::: $2 A$ — the whole source current takes the short. Two voltage sources $10 V /2Ω$ and $6 V /3Ω$ into a shorted node: $I_{N}$ ? ::: $5 + 2 = 7 A$ by superposition. A current source that drains terminal a with $4 A$ : $I_{N}$ ? ::: $- 4 A$ — the sign flips because current enters the box. Test source gives $I_{t} = - 1/3 A$ at $V_{t} = 1 V$ : $R_{N}$ ? ::: $- 3 Ω$ — an active (negative-resistance) network. Matched load on $V_{T h} = 6 V, R_{N} = 2 Ω$ : $P_{max}$ ? ::: $V_{T h}^{2} / (4 R_{N}) = 36/8 = 4.5 W$ . An ideal $9 V$ source with no series resistance: does a Norton equivalent exist? ::: No — $R_{N} = 0$ forces $I_{N} \to \infty$ ; add series $r$ to fix. Black box reads $V_{oc} = 20 V$ , $I_{sc} = 4 A$ : $R_{N}$ ? ::: $R_{N} = 20/4 = 5 Ω$ . With a dependent source present, how do you get $R_{N}$ ? ::: Kill only independent sources, apply a test source, $R_{N} = V_{t} / I_{t}$ .

Connections

Use Norton equivalent circuits — the parent recipe these examples exercise.
Superposition — used in Ex 3 to add short-circuit currents.
Kirchhoff's Current Law — the node law behind every $I_{N}$ .
Ohm's Law — every $V / R$ and $R$ -combination.
Thévenin equivalent circuits & Source transformation — the $V_{oc} = I_{N} R_{N}$ bridge and cross-checks.
Maximum power transfer — the matched-load $P_{max}$ result in Ex 9.

Worked examples — Use Norton equivalent circuits