Hi everyone!
I have a simple question, but I have the impression I'm missing some important concepts of circuits. (especially open en short circuits)

this is the given circuit:

Asked is to give the norton equivalent of this circuit from A to B with the help of superposition.

turning off all the sources gives Ri=1kΩ

What I did further:
turning off the current source:

2 resistances in serie => Ik,1=6V/2kΩ = 3mA

turning off the voltage source:

the voltage over the 1kΩ is 3mA*2kΩ=6V
the currents are in parallel => Ik,2=3mA+(6V/1kΩ)=9mA

=> Ik=Ik,1+Ik,2=12mA

But when I don't think about superposition I can clearly see Ik=6mA due to 3mA+(6V/2kΩ)

What did I do wrong? I think there's a concept about short/open circuits I'm missing.

PS: sorry about the embarassing Paint-skills

 

Hi everyone!
I have a simple question, but I have the impression I'm missing some important concepts of circuits. (especially open en short circuits)

this is the given circuit:

Asked is to give the norton equivalent of this circuit from A to B with the help of superposition.

turning off all the sources gives Ri=1kΩ

Okay, that looks good.

What I did further:
turning off the current source:

2 resistances in serie => Ik,1=6V/2kΩ = 3mA

No, I don't see two resistances in series. I see a 6 V source in parallel with a 2 kΩ resistor and a 1 kΩ resistor. You're looking for the current through the 1 kΩ resistor. It might help if you sketched in the actual short circuit path; it's the current through the short circuit path that you want to find.

turning off the voltage source:

the voltage over the 1kΩ is 3mA*2kΩ=6V
the currents are in parallel => Ik,2=3mA+(6V/1kΩ)=9mA

Ah, no. The short circuit guarantees that the voltage across the 1 kΩ is zero. All current will pass through the short and none will take the 1 kΩ path. Again, it might help if you were to actually sketch in the short circuit.

 

Thanks for your reply,

so:

gives a 3mA current through the 2kΩ. But how can you determine the current through the 1kΩ? Can I say that also 3mA travels through the 6V-voltage which results in Ik,1=6mA? And if so, why can I say that?

I understand that if you turn off the 6V-voltage point a en b are at the same potential, but why would the short circuit guarantee that theres no voltage between b and c? I'm afraid that's a concept i don't understand.

 

First try to find VAB voltage for circuit in fig.1b.
Next find VAB voltage for circuit in fig.1c
And next for circuit from fig.1d find norton equivalent.

 

Thanks for your reply,

so:
<picture snipped>

gives a 3mA current through the 2kΩ. But how can you determine the current through the 1kΩ? Can I say that also 3mA travels through the 6V-voltage which results in Ik,1=6mA? And if so, why can I say that?

Let's rearrange the circuit diagram a bit. The "broken" branch with the suppressed current source is omitted because no current can flow through an open circuit so that branch plays no part:

Note that the short circuit between the terminals A and B has been indicated. What's the current through the 1k resistor and the short?

I understand that if you turn off the 6V-voltage point a en b are at the same potential, but why would the short circuit guarantee that theres no voltage between b and c? I'm afraid that's a concept i don't understand.

Again, redraw the circuit including the short between A and B:

The suppressed 6 V source has "taken out" its parallel 2k resistor. Note that the short between A and B parallels the 1k resistor. A short is equivalent to a zero ohm resistor... so that means the parallel combination of the short and the 1k is...?