Norton equivalents

YOU WILL BE ABLE TO

State Norton's theorem and find I_N and R_N.
Convert between Thevenin and Norton forms (source transformation).
Use repeated source transformations to collapse a network.
Choose the form that fits the problem.

Norton's theorem is Thevenin's mirror image: any linear two-terminal network is also exactly equivalent to one ideal CURRENT source I_N in parallel with one resistor R_N. Same claim, dual form — and the two equivalents describe the same network, so they are locked together by Ohm's law.

The Norton form: a current source in parallel with R_N. Same network, same terminals, same behaviour as the Thevenin form.

R_N = R_{Th}

I_N = \frac{V_{Th}}{R_{Th}}

V_{Th} = I_N R_N

source transformation

I_N has its own direct meaning: it is the short-circuit current at the terminals — short a–b and all of I_N is forced through the short (none takes the R_N path, which has the full 0 V across it). So the two "experiments" from Lesson 2 ARE the two equivalents: the open-circuit experiment measures Thevenin's voltage, the short-circuit experiment measures Norton's current, and their ratio is the shared resistance.

WORKED EXAMPLE

Norton of the Lesson 2 network

Same network as Lesson 2's worked example: 12 V source, R_1 = 4 kΩ, R_2 = 12 kΩ, terminals a–b.
Directly: short a–b. The short removes R_2 from the picture, so i_sc = 12/4 = 3 mA = I_N.
R_N = R_Th = 3 kΩ (same kill-the-sources computation).
Consistency: V_Th = I_N R_N = 3 mA × 3 kΩ = 9 V — matching Lesson 2 exactly, as it must.

I_N = 3 mA, R_N = 3 kΩ

WORKED EXAMPLE

Source transformation as a solving technique

Transformations work in both directions and can be chained: any (source + series R) may be redrawn as (source ∥ R), and vice versa, without changing anything outside.
Chain: 9 V in series with 3 kΩ → transform → 3 mA ∥ 3 kΩ. Suppose a 6 kΩ resistor hangs in parallel at the terminals: combine 3 ∥ 6 = 2 kΩ, giving 3 mA ∥ 2 kΩ.
Transform back: 3 mA × 2 kΩ = 6 V in series with 2 kΩ. The network shrank by one element with no equations solved.
Repeat until the circuit is trivial — this is a legitimate complete solving method for ladder-shaped networks.

Alternate transforms and combinations until one loop remains.

PRACTICE

Convert a Thevenin equivalent of 10 V in series with 2 kΩ to its Norton form.

For Lesson 2's practice network (15 V in series with 6 kΩ, terminals across a 3 kΩ to ground): find the Norton equivalent directly via the short-circuit current, and confirm it matches the Thevenin answer (5 V, 2 kΩ).

Collapse by source transformations: 8 V in series with 4 kΩ, whose terminals also have a 4 kΩ resistor in parallel. Give the final Thevenin form.

A Norton equivalent is 6 mA ∥ 1 kΩ. What current flows into a short across its terminals, and what voltage appears across an open?