Bytobi2247
Hello, I have the following circuit where I need to do a node voltage analysis:
I'm not sure, but in an ideal circuit, don't all resistors that are connected in parallel to a voltage source and all those that are connected in series to a current source cancel each other out?
In the solution the resistors were not eliminated, but why not?
(1 votes)
Loading...Good morning, I'd like to become an electrical engineer later in life and would like to study at a university of applied sciences. Before that, however, I'd like to complete an apprenticeship, as I need to complete the practical part of my university entrance qualification. I've been considering an apprenticeship as an electronics technician, but…
Does each of these pins of the USB-A have 5V?
Hello, does anyone here know the HP 48GX computer? My batteries leaked a little (not much). Does anyone know the best way to clean the contact plate? Or can I just leave it as is. Works perfectly… Thank you
Hello, I have a question about relativistic energy correction. In a previous assignment, I calculated the final velocity of an electron while neglecting gravity. Now I'm supposed to calculate the electron's actual final velocity using relativistic energy correction. My problem is that I can't find anything about relativistic energy correction. Can someone help me? Best…
What is the name of this component on the circuit board, and what is its function? I'm looking forward to additional information. 😉
If you’re R1 overcrowd and the right-hand resistors 3, 4, 5, 6 take out, then of course nothing will be added to the node potentials I marked yellow.
What doesn’t work?
If in your result the yellow nodes of R1, R3, R4, R5, R6 depend, you did something wrong.
@isohypse If I understood it correctly, you stroke 3,4,5 and 6 and the R1, as it flows through it, remains in the circuit and thus also the node 3?
What do you mean “painting”? I’ve already written: R1 can be replaced by a bridge (short-close), take the others out (replace each with infinite). Also in the approach you don’t need to take into account all these resistances at all.
What? superfluous we can’t know. You change a circuit to better calculate the rest. In reality, the node 3 can have an important function and it also has a well-defined potential. Similarly, current flows through resistors 3,4,5,6… also this current may have a meaning. The removal is only used to make it easier to achieve a result. And as I said, the result for the yellow node potentials does not depend on the resistors. No more and no less…
@isohypse Okay, then I get it, thank you.
Yes, of course you lose a knot purely formal, but the tension is trivial to calculate if you already have the yellow knots.
ah moment, because the R1 can be bridged, the potential at 3 and reference is equal and thus the node 3 is superfluous or?
but if I bridge R1 I don’t lose node 3?
Resistances can’t be smashed out anywhere.
The drawing’s pretty blurred. The first error, which is roughly in the eye: a current measuring device is connected parallel to the resistor R2.
Parallel to R2, there is probably also a power source – like the source I1. It’s not a measuring instrument!
Ups. Doesn’t change any resistance. It can also be seen where U3 is present.
The circuit is idealized
From where to where U3 is created? From where should the resistance be calculated?
the circuit is idealized
The circuit may still be so idealized. If it is not clear to WELCHE Voltage and WELCHEN Resistance, you cannot calculate anything.
No resistance is to be calculated, a node voltage analysis is made to find voltages
First of all, it would be important to know what is even asked!
Only then can one say which resistances play no role in the answer.
“Screeds” can’t resist at all – and if you want to “paint” (because they don’t play a role), then you have to say what “stroke” means: replace with short-circuit or idling?
Your cross doesn’t say anything.
@Lutz28213 My adoption: It is an ideal voltage source with a resistor of 0 ohm, so 100% of the current flows through the source and no current through the resistors and therefore they play no role for the circuit.
Yes – in principle, that is true. I’m just missing the question and the non-technical term “stroke” bothered me without saying “take off” or “short-circuit.”