Intuition The one-line ideaAny messy linear two-terminal network — no matter how many resistors and sources — behaves, as seen from its two output terminals, like a single current source in parallel with a single resistor . That pair is the Norton equivalent . Replace the mess, keep the behavior.
Intuition Why it must be trueA linear network's terminal behavior is fully described by the relationship between the terminal voltage V V V I I I straight line : I = a V + b I = a\,V + b I = a V + b two numbers (slope and intercept). So any linear network needs only two components to reproduce that line:
a current source (the intercept → current when terminals shorted),
a resistor (the slope → how current falls as voltage rises).
That's exactly a current source in parallel with a resistor. There is nothing left to describe.
Definition Norton equivalentA linear two-terminal network is equivalent to a current source I N I_N I N R N R_N R N
I N I_N I N Norton current = the current that flows through a short circuit placed across the terminals.R N R_N R N Norton resistance = the resistance seen looking into the terminals with all independent sources deactivated (voltage sources → short, current sources → open).
Worked example The 3-step methodStep 1 — Find I N I_N I N short the two output terminals; compute the current through that short.
Why this step? A short forces V = 0 V=0 V = 0 I L = I N − 0 = I N I_L = I_N - 0 = I_N I L = I N − 0 = I N reads off the intercept directly.
Step 2 — Find R N R_N R N turn off all independent sources (V-source → wire, I-source → gap) and find the equivalent resistance between the terminals.
Why this step? The slope of the I I I V V V
Step 3 — Draw I N ∥ R N I_N \parallel R_N I N ∥ R N
Worked example Two resistors and a sourceA 12 V 12\text{ V} 12 V R 1 = 4 Ω R_1 = 4\,\Omega R 1 = 4 Ω R 2 = 6 Ω R_2 = 6\,\Omega R 2 = 6 Ω R 2 R_2 R 2
Step 1 (I N I_N I N Short the output terminals. This shorts out R 2 R_2 R 2 I N = 12 4 = 3 A . I_N = \frac{12}{4} = 3\text{ A}. I N = 4 12 = 3 A . Why? With the short, R 2 R_2 R 2 R 1 R_1 R 1
Step 2 (R N R_N R N Kill the source (12 V → wire). Now R 1 R_1 R 1 R 2 R_2 R 2 parallel :
R N = 4 ⋅ 6 4 + 6 = 2.4 Ω . R_N = \frac{4\cdot 6}{4+6} = 2.4\,\Omega. R N = 4 + 6 4 ⋅ 6 = 2.4 Ω. Why parallel? Deactivating the source ties the tops of R 1 R_1 R 1 R 2 R_2 R 2 R 2 R_2 R 2
Result: I N = 3 A ∥ R N = 2.4 Ω I_N = 3\text{ A} \parallel R_N = 2.4\,\Omega I N = 3 A ∥ R N = 2.4 Ω
Worked example Add a load and forecastAttach a R L = 2.4 Ω R_L = 2.4\,\Omega R L = 2.4 Ω R N R_N R N R L R_L R L I L = I N ⋅ R N R N + R L = 3 ⋅ 2.4 4.8 = 1.5 A . I_L = I_N\cdot\frac{R_N}{R_N+R_L} = 3\cdot\frac{2.4}{4.8} = 1.5\text{ A}. I L = I N ⋅ R N + R L R N = 3 ⋅ 4.8 2.4 = 1.5 A . Why current-divider? Two equal resistances share the 3 A 3\text{ A} 3 A Forecast: equal resistors ⇒ half each ⇒ 1.5 A. ✓ Verified.
Load voltage: V = I L R L = 1.5 × 2.4 = 3.6 V V = I_L R_L = 1.5\times2.4 = 3.6\text{ V} V = I L R L = 1.5 × 2.4 = 3.6 V
Common mistake Leaving sources ON when findingR N R_N R N Why it feels right: "The sources are part of the circuit, so keep them." Why it's wrong: R N R_N R N slope , purely resistive; live sources add a fixed offset that corrupts a simple resistance reading. Fix: deactivate independent sources (V→short, I→open) before computing R N R_N R N dependent sources — use a test source method for those.)
Common mistake ForgettingR 2 R_2 R 2 I N I_N I N Why it feels right: R 2 R_2 R 2 Why it's wrong: a short across the terminals places a 0 Ω 0\,\Omega 0 Ω R 2 R_2 R 2 Fix: redraw with the short; any element parallel to a short can be deleted.
Common mistake Using the load resistor while computingI N I_N I N R N R_N R N Fix: the Norton equivalent is the network minus the load. Remove the load first, then find I N I_N I N R N R_N R N
Recall Feynman: explain to a 12-year-oldImagine a water tank feeding pipes to your sink. Instead of drawing every pipe, I hand you one card: "this tap pushes 3 cups per second, and there's a leaky pipe of 'thickness 2.4' that steals some of it." Now you can predict how much water reaches any cup you attach — without ever looking at the tank again. Norton = the smallest honest summary of a circuit's outlet.
"Short for the current, kill for the resistance."
I N I_N I N short the terminals. R N R_N R N kill (deactivate) the sources. Then draw N orton = current source N ext to (parallel) a resistor.
What is the Norton equivalent of a linear two-terminal network? A current source
I N I_N I N in parallel with a resistance
R N R_N R N .
How do you find the Norton current I N I_N I N Short the output terminals and compute the current through the short.
How do you find the Norton resistance R N R_N R N Deactivate all independent sources (V→short, I→open) and find the resistance between the terminals.
What is the Norton terminal equation? I L = I N − V / R N I_L = I_N - V/R_N I L = I N − V / R N .
How does R N R_N R N R T h R_{Th} R T h They are equal:
R N = R T h R_N = R_{Th} R N = R T h .
Convert Thévenin to Norton. I N = V T h / R T h I_N = V_{Th}/R_{Th} I N = V T h / R T h , and
R N = R T h R_N = R_{Th} R N = R T h .
Why is the terminal I–V relationship a straight line for a linear network? Superposition/linearity makes
I I I an affine function of
V V V :
I = a V + b I=a V+b I = aV + b , needing only two parameters.
When finding I N I_N I N A short in parallel with it carries all the current, so it has 0 V and is invisible.
Current-divider for a Norton source into load R L R_L R L I L = I N ⋅ R N / ( R N + R L ) I_L = I_N \cdot R_N/(R_N+R_L) I L = I N ⋅ R N / ( R N + R L ) .
Thévenin equivalent circuits — the voltage-source twin; convert freely.Source transformation — the bridge V T h = I N R N V_{Th}=I_N R_N V T h = I N R N Ohm's Law — supplies the V / R N V/R_N V / R N Kirchhoff's Current Law — justifies I L = I N − V / R N I_L = I_N - V/R_N I L = I N − V / R N Maximum power transfer — uses R L = R N R_L=R_N R L = R N Superposition — why terminal behavior is linear/affine.
Linear two-terminal network
Terminal law I equals aV plus b straight line
I_L equals I_N minus V over R_N
Intuition Hinglish mein samjhoDekho, Norton equivalent ka funda simple hai: koi bhi complicated linear circuit — chahe usme 20 resistor aur 5 source ho — agar tum sirf uske do output terminals se dekho, toh woh bas ek current source parallel me ek resistor jaisa behave karta hai. Kyunki linear circuit ka terminal behaviour ek seedhi line hoti hai (I = a V + b I = aV + b I = aV + b
Do number nikalne ka tarika yaad rakho: "Short for current, kill for resistance." I N I_N I N R N R_N R N I N I_N I N R N R_N R N
Ek common galti: log R N R_N R N R N R_N R N I N I_N I N R N R_N R N
Yeh cheez exam aur real design dono me kaam aati hai: ek baar equivalent ban gaya, toh alag-alag load ke liye current (I L = I N ⋅ R N / ( R N + R L ) I_L = I_N \cdot R_N/(R_N+R_L) I L = I N ⋅ R N / ( R N + R L ) R N = R T h R_N = R_{Th} R N = R T h I N = V T h / R T h I_N = V_{Th}/R_{Th} I N = V T h / R T h
