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Anonymous Tue 29 Sep 22:53:59 2020 No.12176509 View Reply Original Report

Quoted By: >>12176551 >>12176578 >>12176947

Can someone help me understand this please.

If I have a voltage source modelled as an ideal voltage source outputting A volts in series with a resistance (output impedance) of B ohms, then using V=IR on the resistor shows that I is A/B, that is that output impedance lowers output current (as I would expect). But, shouldn't the output voltage also be lowered by the output impedance, or does impedance only affect current? And, if voltage is lowered, why does this happen?

Anonymous

Anonymous Tue 29 Sep 2020 23:03:39 No.12176536 Report

Quoted By:

If you're such a brainlet that you can't understand basic ohms' law then you have no business touching electronics or even posting on /sci/
Also do your own damn homework kid we're not a tutoring service.

Anonymous

Anonymous Tue 29 Sep 2020 23:04:05 No.12176537 Report

Quoted By: >>12176562

ideal voltage source works like this: it fixes a constant voltage difference between + and - side, no matter what happens outside
ideal current source(*) works like this: it fixes a constant current flowing through the current source, no matter what happens outside

(*) not included in your image, but worth to mention at the same time you are introduced to voltage sources. It's represented by an arrow in the circle, similar to voltage source

Anonymous

Anonymous Tue 29 Sep 2020 23:09:07 No.12176551 Report

Quoted By: >>12176562

>>12176509
>But, shouldn't the output voltage also be lowered by the output impedance, or does impedance only affect current?
the voltage won't change because that's how ideal voltage sources work: they force/fix a constant voltage between their leads
but something similar would happen if it was a current source instead. That is, if a current source pumped constant current I, then the voltage would depend on the impedance - opposite to the previous scenario, the voltage would rise if the impedance would rise

Anonymous

Anonymous Tue 29 Sep 2020 23:11:14 No.12176562 Report

Quoted By: >>12176596

>>12176551
>>12176537
i think i have worded myself poorly. when i talk about the output voltage, i am not referring to the output of the ideal voltage source, i am referring to the output of the real voltage source which is modelled as an ideal voltage source connected in series with a resistor. The diagram I attached is also misleading because it doesn't show the terminals of the real voltage source it is supposed to be modelling so sorry about that

Anonymous

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Anonymous Tue 29 Sep 2020 23:17:10 No.12176578 Report

Quoted By:

>>12176509
>But, shouldn't the output voltage also be lowered no nigga
>output of the real voltage source
Are you asking: "When I load a 12 V battery, does that 12 volt batter no longer supply 12 volts?" The short answer is no.

Anonymous

Anonymous Tue 29 Sep 2020 23:23:46 No.12176596 Report

Quoted By:

>>12176562
Ah, now I see.
In this case if we increase the load resistance then sure, there is less current going through the circuit, but you also have to remember that we need less current to make the same voltage drop on a bigger resistor.
Consider two resistors, one with small resistance and second with bigger resistance. But this time we control te current instead of voltage. To produce the same voltage drop on both resistor, you need more current going through the small resistor, and less current going through the big resistor.

also, somewhat relevant - see the Jacobi's Law:
https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

Anonymous

Anonymous Wed 30 Sep 2020 03:45:30 No.12176947 Report

Quoted By:

>>12176509
Is R on the diagram the internal resistance of your voltage source? Then it is a short circuit. Or is it a load resistor? Then it is in series with the internal resistance not shown in your picture. They form a voltage divider but the internal resistance is usually very small compared to the load so the output voltage should remain constant. Unless the current is high enough to cause a significant voltage drop on the internal resistance. You can see that effect by connecting a small resistor across a battery.

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