# RL to RC circuit

## Homework Statement

Design a RC circuit so it has the same V vs time characteristics
*See attached diagram*

## The Attempt at a Solution

I really have no idea how to approach this, but I do know,

RC circuit:

Vc(t)=Vb(1-e^-(t/RC))

RL circuit:
i(t)=(Vb/R)(1-e^-(tR/L))

I have no idea what to do with the current equation... can I use ohm's law? Any help would be great!!

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gneill
Mentor
Can you sketch the vout versus time curve for the given circuit? (I'm assuming that the input will be some DC voltage applied at time t = 0).

Well I believe that the Vout for the RL would be an exponential decay starting at Vb and decreasing down to zero, whereas the Vout for the RC would be an exponential growth starting at zero and increasing to Vb

gneill
Mentor
That's right. Can you give the time constant for that exponential curve?

I believe for RC it is RC and for RL it is L/R

So assuming that i would want the time constants the same i would set

RC=L/R

So that would give me that when C=L/R^2

gneill
Mentor
Okay! Now comes the tricky bit. You've got your two components, R and C, and you want the output to be a decaying voltage.

The input to the circuit is a step; the switch is closed and the input voltage goes immediately from zero to some value Vin. What is the initial voltage drop across an uncharged capacitor?

I believe that the initial voltage drop across an uncharged capacitor is 0, and as time goes on, it approaches Vin exponentially

gneill
Mentor
That's right. So initially none of the voltage will be dropped by the C, so all of it must be dropped by the R. Further, you want all of that voltage presented at the output of the circuit. So what arrangement of the R and C would have the initial output voltage equal to the initial input voltage?

maybe put the capacitor first then the resistor?

gneill
Mentor
How can you confirm that that would work? What (and why) are the initial and final output voltages for that arrangement?

Well cause then initially the first voltage drop across the capacitor (which would be zero) and then the second voltage drop would come across the resistor (which we would have Vin=Vout). As time goes on, the voltage drop across the capacitor would increase exponentially, so the Vout (voltage drop across the resistor) would therefore have to decrease exponentially cause if we subtract the voltage drop across the capacitor from Vin, it would give us exponential decay.

One more question: Would the value of C really matter? Should it be equal to L/R^2 as I said before?

gneill
Mentor
Well cause then initially the first voltage drop across the capacitor (which would be zero) and then the second voltage drop would come across the resistor (which we would have Vin=Vout). As time goes on, the voltage drop across the capacitor would increase exponentially, so the Vout (voltage drop across the resistor) would therefore have to decrease exponentially cause if we subtract the voltage drop across the capacitor from Vin, it would give us exponential decay.

One more question: Would the value of C really matter? Should it be equal to L/R^2 as I said before?
Excellent.

Yes, if you want to have the same exponential response curves then the time constants of the two circuits should be the same.