- 2068 Chaitra · 6 marks
Norton theorem steps and comparison
Short answer
Norton’s theorem: any linear two-terminal network can be replaced by a current source IN (short-circuit current) in parallel with RN (resistance seen from terminals with sources zeroed). Compared to Thevenin: Norton uses a current source in parallel, Thevenin a voltage source in series; RN = Rth; conversion by IN = Vth/Rth.
State Norton's description theorem and list the steps for Nortonizing a circuit. Compare the Norton's equivalent circuit to the Thevenin's equivalent circuit.
Norton's theorem
Any linear two-terminal network containing sources and resistors can be replaced, as seen from its terminals, by a current source I_N in parallel with a resistance R_N. I_N is the short-circuit current between the terminals and R_N is the resistance seen from the terminals with all independent sources set to zero.
Steps for Nortonizing a circuit
- Remove the load and mark the terminals a and b.
- Short a-b and calculate the short-circuit current I_N.
- Set all independent sources to zero (voltage sources shorted, current sources opened) and find R_N between a and b.
- Draw I_N in parallel with R_N and reconnect the load: I_L = I_N × R_N/(R_N + R_L).
Norton equivalent
o----+--------+------o a
(I_N) [R_N] [R_L]
o----+--------+------o b
Norton equivalent versus Thevenin equivalent
| Point | Norton | Thevenin |
|---|---|---|
| Source | current source I_N | voltage source V_th |
| Resistance | R_N in parallel | R_th in series |
| Key quantity | short-circuit current | open-circuit voltage |
| Resistance value | R_N = R_th | R_th |
| Conversion | I_N = V_th/R_th | V_th = I_N R_N |
| Load current | I_L = I_N R_N/(R_N + R_L) | I_L = V_th/(R_th + R_L) |
| Convenient for | parallel networks (nodal analysis) | series networks (mesh analysis) |
They are equivalent at the load terminals (source transformation), but not inside: with an open load, the Norton circuit dissipates power in R_N while the Thevenin circuit does not.
✓ Answer checked by Claude Opus 5.5 (high effort)
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