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- Step 1 of 5
Refer to Figure 1.1 in the textbook for the representation of asignal source in the Thevenin form and the Norton form.
Figure 1.1 (a) is in Thevenin’s form and Figure 1.1 (b) is inNorton’s form.
Consider that the load terminals are left open. Re-draw thecircuit diagram.
- Step 2 of 5
Determine the open-circuit output voltage in Thevenin’s formcircuit.
Note that there is no current flow in the circuit due to openconnection. Therefore, the source voltage appears across the load.Therefore,
Therefore, the open-circuit output voltage in Thevenin’s form is.
Determine the open-circuit output voltage in Norton’s formcircuit.
Use Ohm’s law to calculate .
Therefore, the open-circuit output voltage in Norton’s form is.
- Step 3 of 5
Consider that the load (output) terminals are shorted together.Re-draw the circuit diagram.
- Step 4 of 5
Determine the short-circuit current in Thevenin’s formcircuit.
Use Ohm’s law to calculate .
Therefore, the short-circuit current in Thevenin’s form circuitis .
Determine the short-circuit current in Norton’s formcircuit.
Note that the current always takes the low impedance path.Therefore,
Therefore, the short-circuit current in Norton’s form circuit is.
- Step 5 of 5
Note that the source transformation can be performed to make thetwo representations equal.
Consider the Thevenin’s form circuit and perform sourcetransformation.
Consider the Norton’s form circuit and perform sourcetransformation.
The resistance in both the representations is equal.
Therefore, the relationship between and for the two circuit representations to be equivalent is .
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