Norton's Theorem

Basic Electrical & Electronics Engineering — Network Theorems, NEC licence examination syllabus (Nepal Engineering Council).

Norton's Theorem

The same idea as Thevenin, asked the other way round — and one line converts between them.

The idea: the same socket, described differently

Thevenin asks: what voltage does this circuit offer, and what resistance sits behind it?

Norton asks a different question about the same circuit: if I short the terminals with a thick wire, how much current pours out?

Both descriptions are complete. It is like describing a car by its top speed or by its engine power — different numbers, same car, and either can be worked out from the other.

The conversion

I_N = V_th / R_th the short-circuit current R_N = R_th the resistance is UNCHANGED and back: V_th = I_N × R_N

That is genuinely all of it. If you can do Thevenin, you can do Norton — find the Thevenin form and divide.

🔀 The structural difference

TheveninVoltage source in series with R.

NortonCurrent source in parallel with the same R.

Worked example — the same circuit as Thevenin

A 24 V source with 4 Ω, a 12 Ω to ground, and an 8 Ω load. Thevenin gave 1.636 A. Let us watch Norton reach the same number by a different route.

Step 1 — R_N

Identical to R_th. Same circuit, same sources killed, same resistance looking in:

R_N = (4 × 12)/(4 + 12) = 3 Ω

Step 2 — I_N, the short-circuit current

Put a wire across a–b. That wire has zero resistance, so it is a far easier path than the 12 Ω beside it — essentially all the current takes the wire, and the 12 Ω is bypassed entirely.

Only the 4 Ω is left in the path: I_N = 24/4 = 6 A

💡 A short circuit across a resistor removes it from the calculation. Current prefers the zero-resistance path, so anything parallel with a short may as well not exist. Spotting that turns this step into one division.

The conversion formula agrees, as it must:

I_N = V_th/R_th = 18/3 = 6 A ✓

Step 3 — reconnect the load

Norton is a current source with R_N in parallel, so the 6 A splits between R_N and the load. We want the load's share, so the other branch (R_N) goes on top:

I_L = 6 × 3/(3 + 8) = 6 × 3/11 = 1.636 A

Both methods, one answer

Thevenin: 18/(3+8) = 1.636 A Norton: 6 × 3/(3+8) = 1.636 A

Which to use

⚖️ Pick the one that means less work

TheveninWhen the sources are voltages in series, and when the question asks for a voltage.

NortonWhen there are current sources or many parallel branches, and when the question asks for a current.

💡 If you get stuck halfway through one, converting to the other costs a single division. That is a genuinely useful escape route under exam pressure.

Exam notes

📝 What is asked

Source transformation"Convert this to its Norton equivalent" — one division and a redraw. Nearly free marks.

Same-circuit-both-waysSolve by Thevenin and by Norton, and show they agree. Say explicitly that R is unchanged.

Common slipForgetting the current divider at the end. Norton's R is in parallel, so the load never gets all of I_N.

Syllabus points

In, Rn; equivalent circuit (numerical)

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