OpenSpecEngineering

Converts a Thevenin equivalent (a voltage source in series with a resistance) into the Norton equivalent (a current source in parallel with the same resistance), and back. Use it to simplify a circuit step by step: sources in series add in Thevenin form, and sources in parallel add in Norton form.

Equations

I_n = V_th/R_th     V_th = I_n·R_n     R_n = R_th

V_th is the Thevenin voltage, the voltage at the two terminals with nothing connected (open circuit). I_n is the Norton current, the current that flows when the terminals are shorted together. R_th = R_n is the resistance seen looking back into the terminals with every source switched off [1]. Both forms deliver the same voltage and current to any load, which is why one can replace the other.

Inputs and limits

• Choose the direction under Convert. Only the fields for that direction show.
• The resistance must be positive. An ideal voltage source (zero resistance) has no Norton form, and an ideal current source (infinite resistance) has no Thevenin form.
• This page converts a known equivalent. Finding V_th and R_th for a whole network is a separate step: find the open-circuit voltage, then either the short-circuit current or the resistance with the sources off.

Common mistakes

• Putting the Norton resistance in series with the current source. It goes in parallel.
• Treating the equivalent as identical inside. It only matches at the terminals: the power lost inside the Thevenin and Norton forms is not the same.
• Switching off sources wrongly when finding R_th. A voltage source becomes a short circuit; a current source becomes an open circuit.

Worked example

Two DC sources run in parallel to feed one load. Source A has an open-circuit voltage of 7.5 V and an internal resistance of 0.6 Ω. Source B has 7.2 V and 0.4 Ω. What single source do they make together?

1. Choose Convert Thevenin → Norton, and enter Thevenin voltage (V_th) 7.5 V and Thevenin/Norton resistance (Rth = Rn) 0.6 Ω. Press Solve to get Norton current (I_n): 12.5 A.
2. Repeat with 7.2 V and 0.4 Ω to get Norton current (I_n): 18 A.
3. In parallel the Norton currents add, 12.5 + 18 = 30.5 A, and the resistances combine in parallel, 0.6 × 0.4/1.0 = 0.24 Ω.
4. Choose Convert Norton → Thevenin, enter Norton current (I_n) 30.5 A and Norton/Thevenin resistance (Rn = Rth) 0.24 Ω, and press Solve. The Solution shows: Thevenin voltage (V_th): 7.32 V
   Thevenin resistance (R_th): 0.24 Ω

Check: the combined voltage must lie between the two source voltages, closer to the stiffer source B, and 7.32 V does. With no load, a circulating current of (7.5 − 7.2)/1.0 = 0.3 A flows from A into B, and 7.5 − 0.3 × 0.6 = 7.32 V agrees.