Thevenin and Norton Equivalent Circuits

Simplification for circuits with a port (two nodes) that is connected to a load

Thevenin

Simplified down to one voltage source ($V_{Th}$) in series with one resistor ($R_{Th}$)

Finding

• In practice
  1. Disconnect the load
  2. measure the open circuit voltage, $V_{oc}=V_{Th}$
  3. measure the short-circuit current, $i_{sc}$
  4. solve for $R_{Th}=\frac{V_{Th}}{i_{sc}}$
• When only independent
  1. Disconnect the load
  2. Calculate the open circuit voltage, $V_{oc}=V_{Th}$
  3. measure the short-circuit current, $i_{sc}$
  4. Either
     • Solve for $R_{Th}$
     • Deactivate all the independent sources and solve for $R_{eq}$ through port AB, $R_{eq}$ = $R_{Th}$
• When only dependent
  1. Disconnect the load
  2. $V_{oc}$ = $V_{Th}=0$
  3. Connect a known test source (current, $I_{test}$ or voltage, $V_{test}$) — use 1 for those test values
  4. Solve for the corresponding voltage, $V_{test}^{‘}$ or current, $I_{tset}^{‘}$ at port $R_{Th}=V_{test}/I_{test}^{‘}=V_{test}^{‘}/I_{test}$
• Independent and Dependent Sources
  1. Disconnect the load
  2. Calculate the open circuit voltage, $V_{oc}=V_{Th}$
  3. Calculate the short-circuit current, $i_{sc}$
  4. Optionally: Deactivate independent sources and connect a known test source at the load terminals and solve for the corresponding voltage
  5. Solve for $R_{th}$

Maximum Power Transferred to Load

$$p=i^2{R}_L={\left(\frac{V_{Th}}{R_{Th}+R_L}\right)}^2{R}_L$$

Power is maximized when $R_L=R_{Th}$ so

$$p_{max}=\frac{V_{Th}^2}{4R_{Th}}$$

Norton

Simplified down to one current source in parallel with one resistor

From Thevenin, use a Source Transformation