Understand
Treat everything to the left of the load terminals as a one-port. Ask what voltage/current law the load sees.
E-002Foundational25–40 min
Thévenin and Norton equivalence
Replace a loaded resistor network by two different source models and prove that the load cannot tell the difference.
01
Engineering problem
A 12 V source feeds R1 = 3 kΩ in series with a node where R2 = 6 kΩ goes to ground. A load RL = 4.7 kΩ will later be connected from VOUT to ground. Before connecting the load, determine the Thévenin and Norton equivalents seen at VOUT. Then use each equivalent to predict the loaded voltage and current.
ValuesVS = 12 VR1 = 3 kΩR2 = 6 kΩRL = 4.7 kΩ
Original loaded network
The load is connected here only for the final equivalence check. Remove RL conceptually when extracting VTH and RTH.
02
What this exercise should prove
01
Interpret Thévenin/Norton as equivalent terminal behavior rather than algebraic tricks.
02
Find VTH from the open-circuit voltage.
03
Find RTH by deactivating the independent voltage source.
04
Convert exactly between Thévenin and Norton forms.
05
Verify that the original, Thévenin and Norton circuits predict the same loaded operating point.
03
Derivation checkpoint
ThéveninVTH = VOC = VS · R2/(R1 + R2)RTH = R1 || R2VTH = 8.000 VRTH = 2.000 kΩ
NortonIN = VTH/RTHRN = RTHIN = 4.000 mARN = 2.000 kΩ
04
Two drawings, one terminal law
Thévenin form
Same port behavior represented by an ideal voltage source in series with RTH.
Norton form
Same port behavior represented by an ideal current source in parallel with RN = RTH.
05
Understand → Calculate → Transform → Verify → Explain
Calculate
Remove RL, compute VOC = VTH, then zero the 12 V source and look back into the port to find RTH.
Transform
Compute IN = VTH/RTH and draw the Norton source in parallel with RN = RTH.
Verify
Reconnect RL and compute VL and IL from both equivalents. The results must match.
Explain
Explain why internal circuits with different drawings can still be indistinguishable from the load.
06
Questions
- What is the open-circuit voltage at VOUT?
- After deactivating the 12 V ideal source, what resistance is seen looking into VOUT?
- What Norton current corresponds to the Thévenin equivalent?
- What voltage appears across RL = 4.7 kΩ?
- What load current flows?
- Why do the Thévenin and Norton drawings look different but predict the same terminal behavior?
Hints
Progressive hints
Reveal hint 1
With RL removed, R1 and R2 form an ordinary divider.
Reveal hint 2
When an ideal voltage source is set to 0 V, replace it by a short circuit.
Reveal hint 3
After that short, R1 and R2 are both connected from VOUT to ground.
Reveal hint 4
Use IN = VTH/RTH for the Norton form.
Go further
Go further
The detailed course derives the terminal law, covers dependent sources and test-source methods, solves multi-node resistor networks, then applies Thévenin to real source–load design and measurement loading.
07
Common misconceptions
01
Leaving RL connected while calculating VTH.
02
Opening the ideal voltage source instead of shorting it when deactivating it.
03
Using R1 + R2 for RTH even though the zeroed source makes them parallel.
04
Treating Norton as an approximation instead of an exactly equivalent terminal model.
05
Comparing internal branch currents between equivalent models; only terminal behavior is guaranteed to match.
Reference
Reference result
Reveal only after completing your own derivation.
Reveal reference result
VTH = 8.000 VRTH = 2.000 kΩIN = 4.000 mAVL = 5.612 VIL = 1.194 mA
08
Evidence earned
The useful evidence is not that you saw the theorem; it is that you can derive, transform and verify the port model.
Derivation
Correct VTH, RTH and IN with stated source-deactivation reasoning.
Equivalence
Same VL and IL obtained from the original, Thévenin and Norton forms.
Explanation
Clear statement of what terminal equivalence does and does not imply.
Simulation
Optional MAJAL DC verification of all three circuits.