It was a typo. I meant to say volts not amps. The solution says 1.2V by voltage division: 2V * 6/(4+6) =1,2V. It doesn't make sense to me either.Do you want the voltage (which has units of volts) or the current (which has units of amps)?
Assuming you mean 0.8 V (and not 0.8 A), then that is correct. Where did you get the supposedly correct answer of 1.2 V (again, I'm assuming that it's supposed to be V and not A)?
Note that the 6V source and the left 6 Ω resistor has no influence on anything to the right of the 2 V source -- the 2 V source isolates the left side from the ride side by clamping the voltage at the top-center node. The means that the right side is nothing more than a simple voltage divider and the voltage across the 4 Ω resistor has to be less than the voltage across the 6 Ω resistor (since they have the same current flowing in them) and since the sum of the two has to be 2 V, the voltage drop across the 4 Ω resistor has to be less than half of 2 V. So an answer of 1.2 V would make no sense.
Whoever did the solution made several mistakes. The didn't track their units properly (sadly that's way too common), they misapplied a common formula (we all make mistakes) and they didn't bother to ask if their answer made sense (which is a pretty major one, in my view).It was a typo. I meant to say volts not amps. The solution says 1.2V by voltage division: 2V * 6/(4+6) =1,2V. It doesn't make sense to me either.
That would be my university professor, the university is closed so I cant ask him about the solution. But do you agree that the thevenin equivalent circuit would consist of a 0.8 voltage source and a 2.4 ohms resistor, connected in series?Whoever did the solution made several mistakes. The didn't track their units properly (sadly that's way too common), they misapplied a common formula (we all make mistakes) and they didn't bother to ask if their answer made sense (which is a pretty major one, in my view).
As I said, we all make mistakes, so it's not fair to judge the entire body of work based on one botched solution, but it would give me cause to suspect (and expect) a certain level of sloppiness in the future from that source's material.
Yes, that's correct.That would be my university professor, the university is closed so I cant ask him about the solution. But do you agree that the thevenin equivalent circuit would consist of a 0.8 voltage source and a 2.4 ohms resistor, connected in series?
That looks like the node equation for the node Vx. The 3Ω resistor is not relevant because the current through that branch is not affected by the actual value of that resistance - it depends only on the current source. Not exactly sure what the " |•12Ω " is supposed to mean.It still doesnt make sense. Why did he skip the 3 ohm resistor then?
by Jake Hertz
by Luke James
by Luke James
by Jake Hertz