Discussion in 'Homework Help' started by Zortz, May 8, 2014.

1. Zortz Thread Starter New Member

May 2, 2014
16
0
Hey!
Problem understanding this:

Situation is: frequency is infinity, meaning that capacitators are fully loaded and act as short circuit!?

So without the capacitators this is quite easy voltage divider:
R2/R1+R2 = 2/4 = 0.5

This however is not the correct answer because capacitators are doing some magic that i cannot understand here. So what am i missing?

Thanks !

2. WBahn Moderator

Mar 31, 2012
18,261
4,941
What does it mean for a capacitor to be "fully loaded"?

If the capacitors are acting as a short circuit, then that means that the voltmeter is shorted out by C2, right? But it also means that it is connected across the R1/R2 junction by C1, right?

Forget about the frequency being infinite for now and get the transfer function for the circuit as a function of frequency. Then let evaluate that function in the limit that the frequency goes to infinity.

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3. Zortz Thread Starter New Member

May 2, 2014
16
0
I understand that if capacitator is charged i will measure same voltage as input - so transfer would be 1.

Since C2 is shorted i can ignore R2. So from C2 i measure same as input but because there is C1 that is also fully loaded i measure another input.
So could transfer be 2? Feels wrong.

I have no skills to construct the transfer function, i will try to look this up.

//edit

I simulated this circuit and found the answer to be 0. I dont understand why it is like that.
I guess my problem is that i dont understand capacitators at AC, so i will explore that subject.

Last edited: May 8, 2014
4. WBahn Moderator

Mar 31, 2012
18,261
4,941
You are mixing concepts pretty badly.

What does it mean for a capacitor to be "charged". And what is the voltage that you are "measuring" and what is the "input" that it is supposedly the same as? Remember that for a capacitor undergoing an AC voltage/current, the capacitor charge is constantly changing and goes from positive, to zero, to negative and back through zero to positive again. Hence, what does it mean to say that this capacitor is "charged"?

At high frequency a capacitor looks more and more like a short circuit in terms of the ratio of the magnitude of the voltage across it to the magnitude of the current through it. A better way of saying this is that its "impedance" (the AC equivalent of resistance) goes to zero. But this does not say anything about the phase relationship between the voltage and the current.

Even more to the point, let's forget about C1 and C2 altogether for the moment and just look at the resistive voltage divider formed by R1 and R2. What would that transfer function be if we let R1 and R2 both get smaller and smaller? Let's say we divide both by one million. Now R2 is essentially zero and so we would expect the meter to read 0V. But now R1 is also essentially zero and so we would expect the meter to read whatever the supply voltage is (let's call it Vin). Both can't be true. The problem is that we have, on the surface, an indeterminate situation.

$
V_{out} \; = \; V_{in} $$\frac{R2}{R1+R2}$$
\
A_v \; = \; \frac{V_{out}}{V_{in}} \; = \; \frac{R2}{R1+R2}
$

If we now divide R1 and R2 by some large factor K (which we let go to infinity) we have

$
A_v \; = \; \frac{\frac{R2}{K}}{\frac{R1}{K}+\frac{R2}{K}}
\
A_v \; = \; \frac{\frac{R2}{\infty}}{\frac{R1}{\infty}+\frac{R2}{\infty}}
\
A_v \; = \; \frac{0}{0+0} \; = \; ????
$

But we can easily resolve this by dividing top and bottom by R2 to get

$
A_v \; = \; \frac{1}{1+\frac{R2}{R1}}
$

Now when we divide both resistances by the same value of K, the Ks cancel out no matter how large they are, leaving us with a fixed output ratio.

A similar thing will happen with your capacitance values.

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5. Zortz Thread Starter New Member

May 2, 2014
16
0
Thank you for all the help WBahn.
I will deal with DC for now because yes i am mixing things up badly.
I think it makes sense to understand DC before moving to AC.
Ill be back!