Norton equivalent problem

vaggabond · Feb 19, 2014

Hi Guys,

I'm trying to find R_NO and solve the problem (b) below, however I'm not getting the right answer. According to the book , the correct answer is R_NO=15ohms, I_NO=2A.

If I short all voltage supplies and open all current sources then the R_NO should be 10ohm || 30ohm = 7.5ohms. Am I doing something wrong?

Thanks

DTFT · Feb 19, 2014

If you have a dependent source in the circuit, which (b) does, you can't just simplify the resistors.

You may find page three of thishelpful.

vaggabond · Feb 19, 2014

DTFT said:
If you have a dependent source in the circuit, which (b) does, you can't just simplify the resistors.

You may find page three of thishelpful.

Thank you, I'll check it.

vaggabond · Feb 20, 2014

DTFT said:
If you have a dependent source in the circuit, which (b) does, you can't just simplify the resistors.

You may find page three of thishelpful.

Hey DTFT, I believe I solved it. Do you mind checking if its correct? Thanks again

godfreyl · Feb 20, 2014

vaggabond said:
I'm trying to find R_NO and solve the problem (b) below

The current through the voltage source on the left is the same as the current through the 10 Ohm resistor, and we are told that the voltage across it is double the voltage across the 10 Ohm resistor. Thus the voltage source can be replaced by a 20 Ohm resistor.

The rest is easy:
a) 20Ω in series with 10Ω = 30Ω
b) 30Ω in parallel with 30Ω = 15Ω

Norton equivalent problem

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