Use Norton equivalent circuits: Question bank — Hardware deep dive

Before we start, one term used throughout: a two-terminal network means the whole circuit box seen only through its two output wires (its terminals). Everything we say is about what those two wires do, never the internal wiring.

First, see the picture behind every question

Every trap on this page lives on one object: the straight line that relates terminal voltage $V$ and terminal current $I$ . Before defining a single symbol, look at it.

Read the figure left to right:

The horizontal axis is $V$ = the voltage across the two terminals (how hard we push on the outlet). The vertical axis is $I$ = the current out of the top terminal (how much flows).
The line has an equation $I = aV + b$ . Here $a$ is the slope (how steeply current falls as we raise the voltage) and $b$ is the vertical intercept (the current when $V = 0$ ). These letters are just names for "slope" and "intercept" — nothing more.
The red dot where the line crosses the $I$ -axis ( $V = 0$ ) is $I_{N}$ : that is $b$ itself, the short-circuit current.
The point where the line crosses the $V$ -axis ( $I = 0$ ) is the open-circuit voltage $V_{oc} = V_{T h}$ .
The steepness is fixed by resistance: the slope is $a = - 1/ R_{N}$ . The minus sign says "more voltage ⇒ less current out," and the size $1/ R_{N}$ says a small $R_{N}$ makes a steep line.

So the whole Norton method is: find the two red features of this line — its $I$ -intercept ( $I_{N}$ ) and its slope ( $- 1/ R_{N}$ ). Every question below pokes at one of those two.

True or false — justify

True or false: Any two-terminal network can be replaced by a Norton equivalent.

False. Only linear networks (resistors, linear dependent sources, independent sources) give a straight

I

–

V

line; a diode or transistor bends the line, so no single

I_{N} ∥ R_{N}

pair reproduces it.

True or false: The Norton current

I_{N}

is the current the network delivers to any load you attach.

False.

I_{N}

is only the current into a short circuit (zero-ohm load). For any real load

R_{L}

the delivered current is smaller:

I_{L} = I_{N} \cdot R_{N} / (R_{N} + R_{L})

True or false:

R_{N}

and

R_{T h}

are two different resistances you must compute separately.

False. They are the same number,

R_{N} = R_{T h}

— both are the slope

- 1/ R_{N}

of the same

I

–

V

line, found the same way (deactivate independent sources, look in).

True or false: To find

R_{N}

you deactivate every source in the circuit.

False. You deactivate only independent sources (V→short, I→open). Dependent sources stay live because they scale with circuit variables — they belong to the slope and must be probed with a test source.

True or false: A voltage source is deactivated by removing it (leaving a gap).

False. A dead voltage source forces

0 V

across itself, which is a short (a wire). Removing it (open) is how you deactivate a current source, not a voltage source.

True or false: The Norton and Thévenin models of the same network give different terminal behavior.

False. They describe the same straight line, just parameterised differently — pick whichever is convenient; a load sees no difference.

True or false: If a network has no independent sources, its Norton current is zero.

True. With no independent sources the short-circuit current is

I_{N} = 0

(the line passes through the origin), and the network reduces to a pure resistance

R_{N}

with no source.

True or false: Doubling every resistance in the network doubles

I_{N}

False. Doubling all resistances halves the short-circuit current (more opposition), so

I_{N}

halves, while

R_{N}

doubles.

True or false: The Norton resistance

R_{N}

is always a positive number.

False. With dependent sources inside, the equivalent resistance can be negative (

R_{N} < 0

) — the network can push more current as you raise the voltage, tilting the

I

–

V

line the "wrong" way. See the edge-case section below.

Spot the error

A student finds

I_{N}

but leaves the load resistor

R_{L}

connected across the terminals while computing it. What's wrong?

The Norton equivalent is the network minus the load. With

R_{L}

still in place you'd measure a mix of network-and-load current, not the true short-circuit current — remove the load first.

A student finds

R_{N}

with the

12 V

source still switched on and reads a resistance off the terminals. What's wrong?

R_{N}

is a pure slope (resistance only); a live independent source adds a fixed current offset that corrupts a simple ohmmeter-style reading. Deactivate independent sources first, then measure resistance.

A student computes

I_{N}

and includes the resistor

R_{2}

that sits directly across the shorted terminals. What's wrong?

The short is a

0 Ω

path in parallel with

R_{2}

, so

R_{2}

has

0 V

and carries no current — it is invisible for

I_{N}

and must be deleted from that calculation.

A student converts Thévenin to Norton using

I_{N} = V_{T h} \cdot R_{T h}

. What's wrong?

The correct relation divides:

I_{N} = V_{T h} / R_{T h}

. Multiplying gives the wrong units and wrong value; think "short-circuit current = voltage ÷ resistance," which is Ohm's law.

A student says "

R_{N} = 0

means the source is broken." What's wrong?

R_{N} = 0

means an ideal voltage source (a stiff, perfect source) — a real and useful case, not a fault. Its terminal voltage stays fixed no matter the load current.

A student draws the Norton current source in series with

R_{N}

. What's wrong?

Norton is a current source in parallel with

R_{N}

; in series it would force the whole

I_{N}

through the load regardless of load, which is not how the real network behaves.

A student kills the dependent source too, then reads

R_{N}

straight off the resistors. What's wrong?

With a live dependent source you can't just "look in" — you must apply a test source

V_{t es t}

, measure the resulting

I_{t es t}

, and take

R_{N} = V_{t es t} / I_{t es t}

with all independent sources off but the dependent source active.

Why questions

Why does a Norton equivalent need exactly two numbers, not three or four?

The terminal law is a straight line

I = aV + b

, and a line is fixed by exactly two parameters — the slope

a = - 1/ R_{N}

and the intercept

b = I_{N}

, i.e. one resistance and one current.

Why is the short-circuit current the "intercept" of the

I

–

V

line?

Shorting forces terminal voltage

V = 0

, and at

V = 0

the line reads

I = b

— the intercept, which is precisely

I_{N}

Why do the independent sources only shift the

I

–

V

line and never tilt it?

By superposition, the independent-source contributions add a constant offset

b

to the terminal current; the tilt (slope

a

) comes only from resistors and dependent sources, which is why deactivating independent sources isolates

R_{N}

Why can we treat any linear box as just two components even if it has hundreds inside?

Terminal behavior is fully captured by the single

I

–

V

relationship, and that relationship is one line — the internal detail is unobservable from the two wires, so two components suffice.

Why does the current-divider formula use

R_{N}

(not

R_{L}

) in the numerator?

Derive it: the source pushes

I_{N}

into a node where

R_{N}

and

R_{L}

sit in parallel, sharing one voltage

V

. Then

V = I_{N} (R_{N} ∥ R_{L})

and

I_{L} = V / R_{L} = I_{N} \cdot \frac{R _{N} R _{L} / ( R _{N} + R _{L} )}{R _{L}} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}}

. The

R_{L}

cancels, leaving the other branch's resistance

R_{N}

on top — so more current reaches the load when

R_{N}

is large. See the figure below.

Why must dependent sources stay active when finding

R_{N}

A dependent source's value tracks a circuit variable, so it changes the slope of the

I

–

V

line; killing it would change the network's actual resistance-behavior and give a wrong

R_{N}

Edge cases

What is the Norton equivalent of a single ideal current source

I_{0}

alone?

I_{N} = I_{0}

with

R_{N} = \infty

(open) — an ideal current source already is the Norton form with an infinite parallel resistance.

What is the Norton equivalent of a single ideal voltage source

V_{0}

alone?

R_{N} = 0

and

I_{N} \to \infty

in the limit — better expressed as its Thévenin twin (

V_{T h} = V_{0}

R_{T h} = 0

); the Norton current source is degenerate here.

Can the Norton resistance be negative, and what does that look like on the

I

–

V

line?

Yes — a network with a dependent source can have

R_{N} < 0

. Find it by the test-source method: apply

V_{t es t}

, measure

I_{t es t}

; if

R_{N} = V_{t es t} / I_{t es t}

comes out negative, the

I

–

V

line tilts the opposite way (current rises with voltage). This is real in active circuits (e.g. amplifiers) and can cause oscillation — it is not an arithmetic mistake.

What happens to delivered current

I_{L}

R_{L} \to 0

(short load)?

All of

I_{N}

reaches the load:

I_{L} \to I_{N}

, since the short bypasses

R_{N}

entirely and the leak term vanishes.

What happens to delivered current

I_{L}

R_{L} \to \infty

(open load)?

I_{L} \to 0

and the terminal voltage rises to the open-circuit voltage

V = I_{N} R_{N} = V_{T h}

; no current can flow into an open.

For maximum power to the load, what should

R_{L}

be, and does that maximise

I_{L}

Set

R_{L} = R_{N}

for maximum power (see Maximum power transfer); this does not maximise current — current is largest at

R_{L} = 0

, but power there is zero.

I_{N} = 0

but

R_{N} \neq = 0

, what does the box do?

It is a passive resistor: no source inside, so it delivers no current on its own but still presents resistance

R_{N}

to whatever you attach.

Recall One-line self-test

Cover every answer and re-derive the two rules from scratch. "Short for the current, kill (deactivate independent sources) for the resistance." ::: $I_{N}$ = short-circuit current; $R_{N}$ = resistance with independent sources off, dependent sources kept (use a test source if any dependent source is present).

Connections

Use Norton equivalent circuits — the parent method these traps drill.
Thévenin equivalent circuits — the voltage-source twin; same line, other parameters.
Source transformation — the $V_{T h} = I_{N} R_{N}$ bridge tested above.
Ohm's Law — behind $I_{N} = V_{T h} / R_{T h}$ and the leak term.
Kirchhoff's Current Law — justifies the terminal split at the parallel node.
Maximum power transfer — the $R_{L} = R_{N}$ edge case.
Superposition — why independent sources shift but never tilt the line.