so you work out frequency on a circuit by simply dividing V = R by cap value
but surely this resistance must be allowed for in the circuit? Or is it so small that it is negligible ?
As KISS was explaining, there's "resistor" resistance (CONSTANT and linear) and capacitive "resistance", more accurately called "reactance" that VARIES and is NOT linear because of its varying response to frequency, again, as demonstrated by the FG examples.
Does that help?
We'll keep at this concept as long as necessary...
I think I understand now, I need to think of 'Reactance' as more of a control of a component rather than resistance even though they sort of mean the same thing, this is just a miniscule amount in real terms hence measuring on graphs so it enables to see what we are on about & set the required frequency
So now I moved my understanding towards the phase (which is the start of the cycle (Hz) but am I now right in thinking that the reactance is the ability to control the time of the complete cycle by using different value caps giving us required frequency
Just had another look at LOG graph, so what your showing me is the fact that the lower the V the higher the frequency, the higher the V the lower the frequency,
it is simply a graph showing what gain (V) (or my words-charge) you would have at a set frequency in the cap over one Hz
So that would make KISS's example of -3db always 707m the charge capacity at a given frequency, does that make Xc a simple way of finding a charge rate for cap over one Hz, then somehow it is going to have to be tied in with the phase - start time of the cycle
(My emphasis).This would mean you cannot choose any old R as it affects the frequency, it is part of the circuit in a bigger way than I ever imagined, they actually a crucial role to play, so if the cap is the cycle monitor, R must be the control room setting
CBB said:As a side note, isolating the current phase from the voltage phase will be easy enough to demo, but a bear to explain, at this stage.
You want to make it simple:
Frequency was the input (control) and a resistance was the output.
This is close enough for now.
Yep, you were up too late last night. The temperatures are starting to get warmer on this side of the pond.
I think it would be useful to put AC aside for a bit
I'll include one of the math solutions.
One 1kΩ resistor circuit:
The math: V=10mA * 1000Ω, Therefore V=10VDC. Now you're on your own.
One 500Ω resistor circuit:
Two 500Ω resistors circuit:
Two 500Ω resistors circuit (again, with a change):
Do the math so you can see the why of the current.
Now here's one that shows only the voltage output of V1 and the current through R1.
What is the value of R1?
And, finally, at what voltage is V1 set to?
Have fun...CBB
V can be altered around circuit but I is spread even throughout circuit
Cowboybob said:The math: V=10mA * 1000Ω, Therefore V=10VDC.
Muttley600 said:3000ohms*0.004mA = 12V
look at that, even decimal point in the right place
Muttley600 said:What is the value of R1?
1.5V / 2.5mA = 600ohms
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