[SOLVED] How to isolate ''t'' in: Vx= Vs(e(-t/RC))

[Gauthier]

Hi,

Let me give you the problem that bring me to that question,

First, I have 24 Vcc source that aliments R1=62KΩ in series with C1= 1000µF, very basic circuit...

The first question asks me to calculate the voltage drop at C1 and curent in the circuit at intervals of 30 secondes and to trace the exponential curves. The capacitor is charging from 0V, with the time constant of 62 and this formula Vc=Vs(1-e(-t/RC)) for Voltage at C1 and this one Ix = Iinitial(e(-t/RC)). The way I have calculate the initial current is 24V/62kΩ supposing that at the begining C1 is like a short.

Now the second question asks me to trace both curve of the same circuit, the only parameter that changes is when the charging begin C1 is already at 5V of Voltage drop.
I'm ok for the voltage curve: VC1= 24V+(5V-24V)e(-30/62), so when the switch is turned on, the VC1=5V and at 30 sec VC1=12.2V compared to when C1 start charging from 0V, when the switch is turned on VC1 =0V and at 30 secondes VC1=9.2V

My problem is with the the current curve: I think I have to find the current when the VC1 is at 5V, first I used the voltage divider formula...
If there's a 5V drop at C1 there's a 19V drop at R1... so 19V=(62kΩ/Rtot)24V Rtot= 78.3kΩ

Initial current= 24V/78.3KΩ
Initial current= 306µA

After I can use this formula: 306µA(e(-30/62))...

Do I treat the resistance of capacitor alimented with C.C. voltage source juste like a real resistance or it's capacitive reactance like with a sinusoidal alimentation.... I'm a little confuse.

And I would like to know, example if I have this: 5V=24V( (1-e(-t/62)), How do I isolate the ''t'' ?

Best regards

thanks

Alex

 

[ljcox]

Hi,

Let me give you the problem that bring me to that question,

First, I have 24 Vcc source that aliments R1=62KΩ in series with C1= 1000µF, very basic circuit...

The first question asks me to calculate the voltage drop at C1 and curent in the circuit at intervals of 30 secondes and to trace the exponential curves. The capacitor is charging from 0V, with the time constant of 62 and this formula Vc=Vs(1-e(-t/RC)) for Voltage at C1 and this one Ix = Iinitial(e(-t/RC)). The way I have calculate the initial current is 24V/62kΩ supposing that at the begining C1 is like a short.

Now the second question asks me to trace both curve of the same circuit, the only parameter that changes is when the charging begin C1 is already at 5V of Voltage drop.
I'm ok for the voltage curve: VC1= 24V+(5V-24V)e(-30/62), so when the switch is turned on, the VC1=5V and at 30 sec VC1=12.2V compared to when C1 start charging from 0V, when the switch is turned on VC1 =0V and at 30 secondes VC1=9.2V

My problem is with the the current curve: I think I have to find the current when the VC1 is at 5V, first I used the voltage divider formula...
If there's a 5V drop at C1 there's a 19V drop at R1... so 19V=(62kΩ/Rtot)24V Rtot= 78.3kΩ

Initial current= 24V/78.3KΩ
Initial current= 306µA

After I can use this formula: 306µA(e(-30/62))... No, you have to do it by integration as in my attachment.

Do I treat the resistance of capacitor alimented with C.C. voltage source juste like a real resistance or it's capacitive reactance like with a sinusoidal alimentation.... I'm a little confuse.

You can't treat a capacitor as a resistance & you can only use reactance in steady state AC circuits, ie. circuits driven by sineusoidal sources.

And I would like to know, example if I have this: 5V=24V( (1-e(-t/62)), How do I isolate the ''t'' ? See below.

Best regards

thanks

Alex

Alex,
5V=24V( (1-e(-t/62)) is transposed as follows

5V/24V = 1-e(-t/62)

-e(-t/62) = 5/24 - 1

e(-t/62) = 1 - 5/24 = (24 - 5)/24 = 19/24

Now take the natural log of both sides.

-t/62 = ln (19/24)

So t = 62 ln(24/19)

Len

 

[Onigece]

Is this what you're looking for?
t = -RC ln(Vx/Vs)

 

[ljcox]

It would be less confusing for the OP to do it this way.

Vx= Vs(e(-t/RC))

So Vx/Vs = e(-t/RC)

Thus e(t/RC) = Vs/Vx

So t = RC ln (Vs/Vx)

 

[Gauthier]

OK I watched a coples vids about logarithms on youtube...

so,

Vx= Vs(e(-t/RC))

Vx/Vs = e(-t/RC)

ln(Vx/Vs) = ln (e(-t/RC))

ln = log base e of x

loge^(Vx/Vs) = loge^(e^(-t/RC))

Using the cancellation laws I can keep only the last exposent at the right side.

loge^(Vx/Vs) = -t/RC

RC ln (Vx/Vs) = -t

t = (RC ln (Vs/Vx))/-1

t = -1 RC ln (Vx/Vs)

t = RC ln (Vx/Vs)^-1

Answer: t = RC ln (Vs/Vx)

Is that correct???

Alex

 

[ljcox]

OK I watched a coples vids about logarithms on youtube...

so,

Vx= Vs(e(-t/RC)) Ideally, it should be Vx= Vs(e^(-t/RC))

Vx/Vs = e(-t/RC)

ln(Vx/Vs) = ln (e(-t/RC))

ln = log base e of x

loge^(Vx/Vs) = loge^(e^(-t/RC)) You don't need the exponent sign ^ I would write it as log_e(Vx/Vs) = log_e(e^(-t/RC))

Using the cancellation laws I can keep only the last exposent at the right side. ??

loge^(Vx/Vs) = -t/RC There is no exponent in this expression.

RC ln (Vx/Vs) = -t

t = (RC ln (Vs/Vx))/-1

t = -1 RC ln (Vx/Vs)

t = RC ln (Vx/Vs)^-1

Answer: t = RC ln (Vs/Vx)

Is that correct??? Yes subjuct to my suggestions in red above

Alex

Yes, it is essentially what I did in my previous post, except that I took short cuts.

 

[Gauthier]

yeah the thing about the cancellation law was for the string above: loge(Vx/Vs) = loge(e^(-t/RC))

si I puted exposents every where it's my bad lol, I didn't find the real base e on my keyboard..

but tanks for giving you time Len, I learned coples of very interesting things.

 

[ljcox]

You're most welcome.

Len