Sorry, I got a little carried away. I'll let the formulas stay, but take away the numbers.Paw1, this forum is to help the OP with homework, not do the homework for them.
And its confusing to them when you do it incorrectly.
The thing to note about the circuit is, what is the impedance of a voltage source?
You cannot add resistance and reactance like that. They are not in phase. You have to add the complex impedance.Impedance: Z = XC + R = 1016 Ohm
Thanks! Should I remove the text in my post so it won't cause any confusion?Your mistake is here:
You cannot add resistance and reactance like that. They are not in phase. You have to add the complex impedance.
Z = Zr + Zc
Zr = R
Zc = jXc
Z = R + jXc
I don't think you can ignore the output impedance of a voltage source..................................
This looks like a typical homework assignment, I don't think output impedance is to be considered in this case.
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Remember I had tasks like this in school, and output impedance was always omitted. You are right though, in a real world situation (and theoretically of course) it will matter.I don't think you can ignore the output impedance of a voltage source.
Assuming it is infinity does not correspond to a standard voltage source and any real world voltage source does not act that way (e.g. a lab power supply).
If the homework solver is supposed to assume that without any instruction to that effect then the homework is incorrect.
The voltage across the resistor will be 7.072 V a couple of times during one cycle. I think the task wants you to find the voltage across the capacitor at these moments.There is something wrong with this circuit. How can voltage across 1kΩ resistor be equal to 7.072V for Vin = 10Vrms, F = 10kHz and C = 1μF ??
Are you kidding me ? How can voltage be equal 7.072V a couple of times during ONE cycle ?The voltage across the resistor will be 7.072 V a couple of times during one cycle. I think the task wants you to find the voltage across the capacitor at these moments.
V1 is surely a sine wave. It will be 7.072 V on the way up to the positive peak and on the way down from the positive peak. A small amount of voltage will stay across the capacitor, so the source signal will have a slightly higher potential than 7.072 V in these moments. That's what I meant.Are you kidding me ? How can voltage be equal 7.072V a couple of times during ONE cycle ?
So this 7.072V is not a RMS value? And V1 is not a sine wave ?
Also do you know the II Kirchhoff's law ??
You can't just discard one of them. Which one would you discard? Why that one?You're correct. I didn't notice the extra voltages. There are two many numbers in that problem making it inconsistent. You have to discard one of the numbers.
Why would you think that someone is assuming the output impedance of that source to be infinity? I didn't see that suggested, only that, as a homework problem if nothing else and with the information given, that the expectation might be that the source impedance not be taken into account.I don't think you can ignore the output impedance of a voltage source.
Assuming it is infinity does not correspond to a standard voltage source and any real world voltage source does not act that way (e.g. a lab power supply).
If the homework solver is supposed to assume that without any instruction to that effect then the homework is incorrect.
If you want to include R1 in the calculation then you need to ignore the voltage source impedance (which makes it infinity).Why would you think that someone is assuming the output impedance of that source to be infinity? I didn't see that suggested, only that, as a homework problem if nothing else and with the information given, that the expectation might be that the source impedance not be taken into account.
We ignore source impedances all the time. ....................................
Huh? This isn't making any sense.If you want to include R1 in the calculation then you need to ignore the voltage source impedance (which makes it infinity).
Of course we ignore voltage source impedance but only when a low source impedance has little effect on the measurement. Here a low source impedance has a big effect on the circuit. An ideal zero source impedance removes R1 from the circuit.
Thus I don't see how you can ignore it. Certainly it's very bad engineering practice to do so and will lead a student to ignore the source impedance when he shouldn't.
I can see we are talking in circles.........................................
How does ignoring the voltage source impedance make it infinity? It makes it ZERO! If you assume it's infinite then you have no current at all. Remember, the source impedance for a voltage source is in series with the source and load.
Why do you need to ignore the voltage source impedance if you want to include R1 in the calculation?
How does an ideal zero source impedance remove R1 from the circuit?
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Could you please sketch a schematic, because I can't picture what you have in mind. Now you are talking about two sources. Where is this second source coming from? Are you thinking that the meter across R1 is a voltage source?I can see we are talking in circles.
I'm not talking about the voltage sources impedance effect on the voltage from the source. I'm taking about its effect on the effective impedance of R1 as seen by V1.
If both source impedances are zero then there is an effective AC short across R1 so all V1 sees is the impedance of C1. This means the AC voltage across C1 is 10VAC with an average value of -7.072VDC.
That's what both a Spice simulation and a real world circuit would show.