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posted by  carlos1978 on 11/6/2009 5:16:54 PM  |  status: Closed  |  Earned Karma: 25

forget to say that I will rate Life saver!!!

Course Textbook Chapter Problem Needs by
Analog Circuits Fundamentals of Electric Circuits (3rd) by Alexander, Sadiku 4 29 11/6/2009 at 9:00:00 PM
Question Details:
Is there anybody can explain me how to fin the Thevenin equivalent circuit see by the 1k resistor (RTH and VTH).
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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posted by Tachyon12 on 11/7/2009 2:25:02 AM  |  status: Live
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Response Details:
Thevenin equivalent of this circuit is not meaningful. When you open the circuit by removing the load resistance then, you will get Vth to be infinite. This is because, the ideal current source will always pump in 3mA of current. Since the circuit is not closed, the electron have no where to go. So the positive terminal will be at +∞Volt.

But you can find the Norton Equivalent of this circuit.
1. First short the load resistance. The short circuit current(Isc) will be 3mA

2. Next, remove the load resistance. The open circuit voltage will be +∞ V ( due to reasons already mentioned)

So you get Rth= ∞/3 = ∞Ω

So the norton equivalent circuit is


Note: The Voltage Controlled Voltage Source(VCVS) will have no effect in the equivalent circuit, because the current by this source will only flow through 4k and 2k resistors, not though Rload.
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