Given:
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Question Details:
Is there anybody can explain me how to fin the Thevenin equivalent circuit see by the 1k resistor (R TH and V TH). I know that v0 = 3V but I want to resolve it by Thevenin. I also know that when I open the circuit by the 1K resistor, I don't have any current in the 3 mA branch. Thanx for your help.
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The 4k ohm , the 2k ohm resistors , and the dependant voltage source are irelevant as far as the Thevenin circuit the
1k ohm resistor sees . Instead of a 1k ohm resistor , let the load resistor value vary . The ideal 3mA current source
delivers 3mA to this load resistor no matter what its value is . For the circuit drawn , you said
and you are
right ---
.What if the load resistor is doubled to 2k ohms ? We then have :
. What if we let the load resistor go to infinity ? This is the Thevenin voltage :
. The Thevenin resistance is :
These values are not very useful either . The more useful equivalent circuit that the 1k ohm resistor sees is the Norton
This means we can replace the circuit with an ideal
current source in series with the 1k ohm resistor as far as
the 1k ohm resstor is concerned.