Use Norton equivalent circuits: Exercises — Hardware deep dive

Before we start, one shared picture of the vocabulary. A Norton equivalent is a current source $I_{N}$ (an arrow that pushes a fixed number of amps into a node no matter what) sitting next to — in parallel with — a single resistor $R_{N}$ . "Parallel" means both components connect the same two dots. The two dots are the terminals: the only place the outside world (a load) may touch the circuit.

Figure s01 — anatomy of the Norton box (read this before the exercises). The picture below shows the two-component model. On the left, the yellow circle with an up-arrow is the current source $I_{N}$ : it always injects the same amps upward into the top rail. In the middle, the blue zig-zag is the resistor $R_{N}$ , wired between the same top and bottom rails as the source — that shared wiring is what "parallel" means (both components touch the same two dots). On the right, the two red dots are the terminals; the outside voltage $V$ appears across them. Trace the loop with your eye: current leaves the source up the left wire, and at the top node it may either leak down through $R_{N}$ or flow out to the right toward the terminals. That fork is the whole story of Norton.

$I_{N}$ = the current through a short (a plain wire, $0 Ω$ ) laid across the terminals. A short forces the terminal voltage $V = 0$ .
$R_{N}$ = the resistance you see looking into the terminals after you deactivate every independent source (a voltage source becomes a wire; a current source becomes a gap/open).

Where the terminal law comes from (build it, don't memorise it)

We keep using $I_{L} = I_{N} - V / R_{N}$ . Here is why it is true, from two circuit laws only.

Figure s02 — the single node where everything splits. Look at the top node in the picture below. Three wires meet there: the source arrow bringing $I_{N}$ in, the resistor $R_{N}$ carrying a leak current down, and the terminal wire carrying $I_{L}$ out to the load.

Kirchhoff's Current Law (KCL) says: current into a node = current out of it. Nothing piles up. So $I_{N} = I_{leak} + I_{L}$ .
Ohm's Law gives the leak: the resistor $R_{N}$ has the terminal voltage $V$ across it (same two dots), so the current down through it is $I_{leak} = V / R_{N}$ .

Substitute the second into the first and solve for the load current: $I_{N} = \frac{V}{R _{N}} + I_{L} ⟹ I_{L} = I_{N} - \frac{V}{R _{N}} .$

That is the terminal law — no memorisation, just KCL at one node plus Ohm's Law for one resistor. Every check in this page below rests on it.

Level 1 — Recognition

Exercise 1.1 (L1)

A Norton equivalent is drawn as a current source of $I_{N} = 5 A$ in parallel with $R_{N} = 8 Ω$ . The terminals are left open (nothing attached). What is the open-circuit terminal voltage $V_{oc}$ ?

Recall Solution 1.1

WHAT: "Open" means no load, so no current can leave through the terminals: $I_{L} = 0$ . WHY: By the terminal law $I_{L} = I_{N} - V / R_{N}$ , setting $I_{L} = 0$ forces all of $I_{N}$ through $R_{N}$ . Compute: from $0 = I_{N} - V_{oc} / R_{N}$ we get $V_{oc} = I_{N} R_{N} = 5 \times 8 = 40 V$ . This $V_{oc}$ is exactly the Thévenin voltage $V_{T h}$ (see Thévenin equivalent circuits).

Exercise 1.2 (L1)

Same source ( $I_{N} = 5 A$ , $R_{N} = 8 Ω$ ). Now place a short circuit across the terminals. What current flows through the short?

Recall Solution 1.2

WHAT: A short is a $0 Ω$ wire, so $V = 0$ across the terminals. WHY: In the terminal law the resistor's leak is $V / R_{N} = 0/8 = 0$ . Nothing leaks, so everything the source pushes goes into the short. Compute: $I_{L} = I_{N} - V / R_{N} = 5 - 0 = 5 A$ . The short-circuit current is $I_{N}$ — that is the whole definition.

Exercise 1.3 (L1)

A Thévenin equivalent is given: $V_{T h} = 24 V$ , $R_{T h} = 6 Ω$ . State its Norton twin $(I_{N}, R_{N})$ .

Recall Solution 1.3

Rule (Source transformation): $R_{N} = R_{T h}$ and $I_{N} = V_{T h} / R_{T h}$ .

$R_{N} = 6 Ω$ .
$I_{N} = 24/6 = 4 A$ . Why: Both describe the same $I$ – $V$ line. The short-circuit current of the Thévenin box is $V_{T h} / R_{T h}$ , and that is by definition $I_{N}$ .

Level 2 — Application

Exercise 2.1 (L2)

A $18 V$ source connects through $R_{1} = 3 Ω$ (in series) to a node A. From node A, $R_{2} = 6 Ω$ drops to the bottom rail. The terminals are taken across $R_{2}$ . Find the Norton equivalent $(I_{N}, R_{N})$ .

Figure s03 — the Ex 2.1 network. The yellow battery on the left is the $18 V$ source; the blue zig-zag along the top is $R_{1}$ in series; the green vertical zig-zag is $R_{2}$ ; the red dots are the output terminals taken across $R_{2}$ .

Recall Solution 2.1

Step 1 — $I_{N}$ (short the terminals). Shorting the terminals joins both ends of $R_{2}$ , so $R_{2}$ sees $0 V$ and carries no current — it is invisible. The $18 V$ source then drives only $R_{1}$ : $I_{N} = \frac{18}{3} = 6 A .$ Step 2 — $R_{N}$ (kill the source). Replace the $18 V$ source with a wire. Now $R_{1}$ and $R_{2}$ both bridge the same two nodes (top rail joined by the wire, bottom rail) — they are in parallel: $R_{N} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2 Ω.$ Result: $I_{N} = 6 A$ , $R_{N} = 2 Ω$ .

Exercise 2.2 (L2)

Attach a load $R_{L} = 2 Ω$ to the Norton source from 2.1. Find the load current $I_{L}$ and load voltage $V$ .

Recall Solution 2.2

Where the current-divider comes from (derive, don't quote). The load $R_{L}$ and the Norton resistor $R_{N}$ hang across the same two terminals, so they share the same voltage $V$ . By Ohm's Law each carries $I_{L} = V / R_{L}$ and $I_{l e ak} = V / R_{N}$ . By KCL the source current splits between them: $I_{N} = I_{L} + I_{l e ak} = V / R_{L} + V / R_{N} = V (\frac{1}{R _{L}} + \frac{1}{R _{N}})$ . Solve for $V$ : $V = \frac{I _{N}}{\frac{1}{R _{L}} + \frac{1}{R _{N}}} = I_{N} \cdot \frac{R _{N} R _{L}}{R _{N} + R _{L}} .$ Now the load current is $I_{L} = V / R_{L}$ , and the $R_{L}$ cancels: $I_{L} = \frac{V}{R _{L}} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}} .$ That is the current divider — it is just KCL + Ohm's Law, not a magic formula. Notice the load gets the other resistor's value $R_{N}$ on top (more resistance elsewhere ⇒ more current steered your way). Plug in: $I_{L} = 6 \cdot \frac{2}{2 + 2} = 6 \cdot \frac{1}{2} = 3 A .$ Forecast check: equal resistors ⇒ current splits in half ⇒ $3 A$ . ✓ $V = I_{L} R_{L} = 3 \times 2 = 6 V .$

Exercise 2.3 (L2)

For the Norton source $I_{N} = 6 A$ , $R_{N} = 2 Ω$ , use the terminal law to find $I_{L}$ when the load holds the terminals at $V = 4 V$ .

Recall Solution 2.3

Direct substitution into $I_{L} = I_{N} - V / R_{N}$ (derived above from KCL): $I_{L} = 6 - \frac{4}{2} = 6 - 2 = 4 A .$ Meaning: at $4 V$ , the resistor leaks $2 A$ back, leaving $4 A$ for the load.

Level 3 — Analysis

Exercise 3.1 (L3)

A current source of $2 A$ is in parallel with $R_{1} = 5 Ω$ ; that combination connects through $R_{2} = 5 Ω$ (in series) out to the terminals. Find the Norton equivalent seen at the terminals.

Figure s04 — the Ex 3.1 network. The yellow current source pushes $2 A$ up; blue $R_{1}$ sits in parallel with it; green $R_{2}$ carries the current sideways to the red output terminals.

Recall Solution 3.1

Step 1 — $I_{N}$ (short the terminals). With the terminals shorted, the far end of $R_{2}$ is tied to the bottom rail. Now $R_{2}$ is in parallel with $R_{1}$ across the current source. The source's $2 A$ splits between $R_{1}$ and $R_{2}$ ; the part through $R_{2}$ reaches the short (current divider, derived in 2.2): $I_{N} = 2 \cdot \frac{R _{1}}{R _{1} + R _{2}} = 2 \cdot \frac{5}{10} = 1 A .$ Step 2 — $R_{N}$ (kill the source). A current source deactivates to an open (a gap). With the source opened, the only route between the two terminals runs through $R_{2}$ to node A, then through $R_{1}$ down to the bottom rail — a single file line: $R_{N} = R_{2} + R_{1} = 5 + 5 = 10 Ω.$ Why series here (not parallel)? With the current source opened, there is exactly one path from terminal to terminal, threading both resistors end-to-end. That is the definition of series. Result: $I_{N} = 1 A$ , $R_{N} = 10 Ω$ .

Exercise 3.2 (L3)

Verify Exercise 3.1 by computing the open-circuit voltage two ways: (a) directly on the original circuit, (b) as $I_{N} R_{N}$ .

Recall Solution 3.2

Sign convention first. Walk from the bottom rail (call it $0 V$ ) up to the "+" terminal, and add a voltage rise for anything that pushes potential up, subtract a drop for anything you climb against current. We choose the top terminal as "+". (a) Original circuit, terminals open. No current can leave the open terminals, so no current flows through $R_{2}$ — it is a dead-end. By Ohm's Law a resistor with zero current has zero voltage: $I_{R_{2}} R_{2} = 0$ . So the terminal voltage equals the node-A voltage. All $2 A$ from the source flows through $R_{1}$ , building $V_{A} = 2 \times 5 = 10 V .$ Then $V_{oc} = V_{A} - I_{R_{2}} R_{2} = 10 - 0 = 10 V$ . The subtraction is there in principle (we climb across $R_{2}$ to reach the terminal), but since $R_{2}$ carries no current its drop is exactly zero, so nothing is actually lost. (b) From Norton: $V_{oc} = I_{N} R_{N} = 1 \times 10 = 10 V$ . ✓ They match, confirming 3.1.

Exercise 3.3 (L3)

A network gives short-circuit current $I_{sc} = 4 A$ and open-circuit voltage $V_{oc} = 20 V$ . Find $I_{N}$ and $R_{N}$ without knowing the internal layout.

Recall Solution 3.3

Key idea: $I_{N}$ is by definition the short-circuit current, and $R_{N} = V_{oc} / I_{sc}$ (the slope of the terminal line = intercept-voltage ÷ intercept-current).

$I_{N} = I_{sc} = 4 A$ .
$R_{N} = \frac{V _{oc}}{I _{sc}} = \frac{20}{4} = 5 Ω$ . Why this works: the straight line $I_{L} = I_{N} - V / R_{N}$ hits $I_{L} = I_{N}$ when $V = 0$ (short) and $I_{L} = 0$ when $V = V_{oc}$ . Setting $0 = I_{N} - V_{oc} / R_{N}$ gives $R_{N} = V_{oc} / I_{N}$ .

Level 4 — Synthesis

Exercise 4.1 (L4)

Two sources. A $12 V$ source through $R_{1} = 4 Ω$ meets a node; a $6 V$ source through $R_{2} = 4 Ω$ meets the same node; from that node $R_{3} = 8 Ω$ goes to the bottom rail. Terminals are across $R_{3}$ . Find $I_{N}$ using Source transformation.

Figure s05 — the Ex 4.1 two-source network. Two yellow battery-plus-resistor branches feed one shared node A; green $R_{3}$ drops from A to the bottom rail; red terminals sit across $R_{3}$ .

Recall Solution 4.1

Strategy: convert each voltage-plus-series-resistor branch into a Norton (current source $∥$ resistor), then merge.

Branch 1: $12 V$ / $4 Ω \Rightarrow 3 A ∥ 4 Ω$ .
Branch 2: $6 V$ / $4 Ω \Rightarrow 1.5 A ∥ 4 Ω$ . Both current sources point into the node (both raise its potential), so add: $3 + 1.5 = 4.5 A$ . Their parallel resistors combine: $4 ∥ 4 = 2 Ω$ . And $R_{3} = 8 Ω$ is also in parallel across the terminals. Short the terminals to read $I_{N}$ : a short holds the top node at the same potential as the bottom rail, so every parallel branch has $0 V$ across it — by Ohm's Law each such resistor then carries $0 A$ and drops out. (This is why a short "kills" the parallel resistances: equal potential on both ends ⇒ no voltage ⇒ no current.) With every parallel resistor idle, the whole $4.5 A$ merged source pours straight into the short: $I_{N} = 4.5 A .$

Exercise 4.2 (L4)

For the same circuit, find $R_{N}$ and then $V_{oc}$ .

Recall Solution 4.2

$R_{N}$ — kill both sources (each $\to$ wire). Now $R_{1}$ , $R_{2}$ , and $R_{3}$ all bridge the node to the bottom rail (the source wires tie their far ends to the bottom rail). Three-way parallel: $R_{N} = (\frac{1}{4} + \frac{1}{4} + \frac{1}{8})^{- 1} = (\frac{2}{8} + \frac{2}{8} + \frac{1}{8})^{- 1} = (\frac{5}{8})^{- 1} = \frac{8}{5} = 1.6 Ω.$ $V_{oc}$ : $V_{oc} = I_{N} R_{N} = 4.5 \times 1.6 = 7.2 V$ . Cross-check with Superposition later if you like — the merged model already encodes it.

Exercise 4.3 (L4)

Attach $R_{L} = 1.6 Ω$ to the source of 4.1–4.2 and find the power delivered to the load.

Recall Solution 4.3

Current divider (from 2.2): $R_{N} = R_{L} = 1.6 Ω$ (equal), so current splits in half: $I_{L} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}} = 4.5 \cdot \frac{1.6}{3.2} = 2.25 A .$ Power: $P_{L} = I_{L}^{2} R_{L} = (2.25)^{2} \times 1.6 = 5.0625 \times 1.6 = 8.1 W .$ Note: $R_{L} = R_{N}$ is the Maximum power transfer condition — this is the largest power any load can pull from this source.

Level 5 — Mastery

Exercise 5.1 (L5) — Degenerate case: zero source

A network has no independent sources — just resistors: $R_{1} = 10 Ω$ from terminal to node, then $R_{2} = 15 Ω$ from node to the other terminal, with the node also tied to nothing else. What is the Norton equivalent seen at the terminals?

Recall Solution 5.1

Step 1 — $I_{N}$ : with no sources, nothing drives current. Short the terminals → the short carries $0 A$ : $I_{N} = 0 A .$ Step 2 — $R_{N}$ : the two resistors are in series between the terminals (single-file path): $R_{N} = R_{1} + R_{2} = 10 + 15 = 25 Ω.$ Interpretation: a source-free network is just a pure resistor — the Norton current source vanishes (an open, since $0 A$ into a parallel resistor does nothing) and only $R_{N} = 25 Ω$ remains. This is the limiting case where the $I$ – $V$ line passes through the origin ( $b = 0$ ).

Exercise 5.2 (L5) — Degenerate case: $R_{N} \to \infty$

Suppose a network's short-circuit current is $I_{N} = 3 A$ and its open-circuit voltage is infinite (the terminal voltage keeps climbing without bound as you reduce the load current). What is $R_{N}$ , and what does the equivalent become?

Recall Solution 5.2

From $V_{oc} = I_{N} R_{N}$ : if $V_{oc} \to \infty$ with finite $I_{N}$ , then $R_{N} \to \infty$ . What it becomes: an ideal current source alone. An infinite parallel resistor is an open — it leaks nothing — so $I_{L} = I_{N} - V /\infty = I_{N}$ for any $V$ . The load always gets exactly $3 A$ no matter what. Picture: the $I$ – $V$ line is perfectly flat (horizontal at $I = 3 A$ ); zero slope means infinite $R_{N}$ (slope $= - 1/ R_{N} \to 0$ ).

Exercise 5.3 (L5) — Full multi-source mastery

A $9 V$ source through $R_{1} = 3 Ω$ feeds node A. A $3 A$ current source pushes into node A directly. From node A, $R_{2} = 6 Ω$ goes to the terminals' top; the terminals' bottom is the reference rail. Find the complete Norton equivalent $(I_{N}, R_{N})$ and the load current when $R_{L} = 4 Ω$ is attached.

Figure s06 — the Ex 5.3 mixed-source network. A yellow battery-plus- $R_{1}$ branch and a yellow ideal current source both feed the same node A; green $R_{2}$ carries current out to the red terminals; the bottom rail is the reference.

Recall Solution 5.3

Step 1 — $I_{N}$ (short terminals). The short ties the far end of $R_{2}$ to the reference rail, so $R_{2}$ hangs between node A and the rail. Two drivers reach node A:

The $9 V$ / $3 Ω$ branch → convert to Norton: $3 A ∥ 3 Ω$ .
The ideal $3 A$ current source (already a current source). Merge the two current sources at node A: $3 + 3 = 6 A$ , with the single parallel resistor $R_{1} = 3 Ω$ from the converted branch (the ideal source adds no parallel resistor). This $6 A$ at node A now splits between $R_{1}$ (down to the rail) and $R_{2}$ (out to the short). By the current divider (from 2.2), the branch through $R_{2}$ to the short gets the other resistor's share $R_{1} / (R_{1} + R_{2})$ : $I_{N} = 6 \cdot \frac{R _{1}}{R _{1} + R _{2}} = 6 \cdot \frac{3}{3 + 6} = 6 \cdot \frac{1}{3} = 2 A .$ Step 2 — $R_{N}$ (deactivate both sources: V→short, I→open). With the voltage source shorted, $R_{1} = 3 Ω$ runs from node A to the rail. With the current source opened, its branch disappears entirely. Looking into the terminals: $R_{2} = 6 Ω$ in series to node A, then $R_{1} = 3 Ω$ to the rail — one single-file path: $R_{N} = R_{2} + R_{1} = 6 + 3 = 9 Ω.$ Step 3 — Attach $R_{L} = 4 Ω$ (current divider again): $I_{L} = I_{N} \cdot \frac{R _{N}}{R _{N} + R _{L}} = 2 \cdot \frac{9}{9 + 4} = 2 \cdot \frac{9}{13} = \frac{18}{13} \approx 1.385 A .$ Result: $I_{N} = 2 A$ , $R_{N} = 9 Ω$ , $I_{L} = 18/13 \approx 1.385 A$ .

Answer key (at a glance)

Ex 1.1

V_{oc}

40 V

Ex 1.2 short current?

5 A

Ex 1.3

(I_{N}, R_{N})

4 A, 6 Ω

Ex 2.1

(I_{N}, R_{N})

6 A, 2 Ω

Ex 2.2

(I_{L}, V)

3 A, 6 V

Ex 2.3

I_{L}

4 A

Ex 3.1

(I_{N}, R_{N})

1 A, 10 Ω

Ex 3.3

(I_{N}, R_{N})

4 A, 5 Ω

Ex 4.1

I_{N}

4.5 A

Ex 4.2

(R_{N}, V_{oc})

1.6 Ω, 7.2 V

Ex 4.3

P_{L}

8.1 W

Ex 5.1

(I_{N}, R_{N})

0 A, 25 Ω

Ex 5.3

(I_{N}, R_{N}, I_{L})

2 A, 9 Ω, 18/13 \approx 1.385 A

Connections

Use Norton equivalent circuits — the parent recipe these exercises drill.
Thévenin equivalent circuits — the voltage-source twin used in Ex 1.1, 1.3.
Source transformation — the engine of Ex 1.3, 4.1, 5.3.
Ohm's Law — every $V = I R$ and leak term.
Kirchhoff's Current Law — the node splits in Ex 3.1, 5.3 and the terminal-law derivation.
Maximum power transfer — the $R_{L} = R_{N}$ case in Ex 4.3.
Superposition — why terminal behavior is linear and multi-source merging is legal.