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back to what we were talking about, R in series is adding more R
Just working on KISS's calculation in post #411, I'm seeing of you had equal R & V they would cancel each other out so you would get no I, so I'm trying to work out what you are trying to show me
So if I had 10V & 10ohm but the 10ohm were 2x 5ohm R in parallel, I would get less R even though values are the same because the R is halved by division, hence, allowing power to flow
On the frequency front, I can see where I was misunderstanding, I was thinking if 1A = 1Hz how do they get 1MHz, where do they get all that power from, but it works backwards doesn't it, as R increases so does frequency, so we don't need more power for a higher frequency, we actually need less but at the same time as we lift frequency we are also increasing R against I making it take longer, so this has something to do with charge times, where 't' comes into play
Sort of, but we need to clarify the math.
Let's start again with a basic two resistor parallel circuit:
View attachment 62417
To solve for the equivalent resistance for this circuit, we use the following formula (LateX isn't working):
R1 X R2 divided by R1 + R2. This can also be written: (R1*R2)/(R1+R2).
So, in this case: (10*40)/(10+40), or 400/50, which equals 8.
So in words, the equivalent resistance of a parallel, 2 resistor circuit consisting of a 10Ω and 40Ω resistors is 8Ω.
It isn't so important to know why the formula is used as it is to realize that this is how it's done.
You might ask: "Why not use an 8Ω resistor, instead"? Well, when you can't find a certain value resistor (and an 8Ω resistor IS NOT a normal value [and neither is a 40, for that matter]), it's OK to create an "equivalent" value "network" circuit of multiple resistors to achieve that.
And since we've been investigating RC timing circuits, for a very specific RC time constant we might have to do just that.
One last example. Let's say you wanted an 18Ω resistor (also an odd value, as is 40Ω, but I wanted to keep the math simple). This is how to get there:
View attachment 62418
In other words, an equivalent 8 ohm parallel circuit (composed of 10 and 40 ohm resistors) in series with an 10 ohm resistor.
400/50, which equals 8.
1/Rp= 1/R1+1/R2+1/R3+...1/Rn (Try it!)
So if there are two, 10 ohm resistors.
1/Rp = 1/10+1/10
1/Rp = 2/10
Rp = 10/2 or 5
I solved it using fractions, you can solve it using decimals too.
1/Rp = 0.1+0.1
1/Rp = 0.2
Rp = 1/0.2
Rp =5
The formula would work for 100 resistors in parallel and 1,000,000 resistors in parallel. It doesn't matter. It will even work for 1.
Using 10 ohms: 1/Rp = 1/10; Rp = 10
Each resistor can have a different value.
You learned a while ago that R*C has the units of ohms * farads (Usually written as (ohms- Farads) and that reduced to the units of seconds (s). I mistakenly used t. Units that are multiplied together show up as what looks like subtraction. (ohm-cm) is one such unit.
The length of one cycle is called the period and it's measured in time units, like 20 mS. 1/(period) is the frequency. So 1/20 mS is 50 Hz. See if you can get that answer on your calculator?
Capacitors change their behavior based on frequency. When capacitors are part of a timing circuit, the quantity R*C is usually involved. That quantity is called the TIME CONSTANT.
So ignoring V for a moment, I can see the time taken by R & cap is equal, so these are setting for the charge time & nothing else... so I was right in thinking R was the control of the cap, I just couldn't relate it to time
I think I'm starting to get the picture now, a cap resists instant change, therefore it takes time to build the charge, so there are lots of different uses
Muttley said:found it, when you put this in calculator you get 0.05?
You need to be able to keep this rule firmly set in your mind: under normal circumstances, neither Voltage (V), nor Current (I) can or will alter resistance (R). NEVER.
But the reverse IS true: resistance WILL alter both current AND Voltage.