Question

Assume that for a certain circuit, you have temporarily removed the load and observed an open-circuit voltage voc​=0.44 V and short-circuit current isc​=67.8 mA.

a. Draw and label BOTH the Thevenin and Norton equivalent circuits

Answer 1

The Thevenin and Norton equivalent circuits are based on the provided open-circuit voltage and short-circuit current. The Thevenin equivalent has a voltage of 0.44 V with a resistor of 6.49Ω, and the Norton equivalent has a current source of 67.8 mA with the same resistance.

Step-by-step explanation:

Based on the question, we are asked to analyze a circuit and provide the Thevenin and Norton equivalent circuits. From the given information, the open-circuit voltage VOC is 0.44 V and the short-circuit current ISC is 67.8 mA.

The Thevenin equivalent consists of a voltage source VTh in series with a resistor RTh, while the Norton equivalent consists of a current source INo in parallel with a resistor RNo. Since we have the open-circuit voltage and short-circuit current, we can identify VTh as 0.44 V and INo as 67.8 mA. The resistance in both Thevenin and Norton circuits is the same and can be calculated using Ohm's law as RTh = RNo = VOC / ISC which is 0.44 V / 67.8 mA = 6.49Ω.