Problem with 'simple' RC circuits

[lpau001]

Hey there! First post, been browsing through the forums for a little and I have found problems like this, but I still am getting the wrong answer when I try and do the things suggested.

Circuit is simple, and in series.
Vs = 23 V
R1 = 87 Ohms
C=6.1 microfarads

with a switch, of course.

How long after the switch is closed does the voltage across the resistor drop to 13V?

I tried getting the current going through the resistor at that moment in time by using Ohm's law and inputting 13V = I / 87 Ohms. Getting approx: .14943A going through R1 at time 't'.

Then using the formula

I=Io(1-e^(-t/RC)) Io is assuming current at time '0' right when the switch closes.

Well after calculations I get 't' to be 4.4203E-4 sec which is wrong.

Any help? Thanks in advance.

 

[cepheid]

Welcome to PF!

I don't think your formula for the current is correct. It results in a current that starts at 0 and then increases toward a steady state value of I₀. But if you think about it, the opposite should happen. The current should be non-zero initially as the capacitor is charging, but should then approach zero as the capacitor gets closer to being fully charged.

 

[lpau001]

Thanks Cepheid!

What do you mean the current starts at 0? I believe the starting current is around .26437 A, and that is Io.

Is the formula wrong? That is what I was given in class..

Did I misunderstand?

I tried doing the opposite, but unfortunate as it may be I cannot take the natural log of a negative number...

edited for politeness!

 

[gneill]

Welcome to Physics Forums.

The nice thing about these circuits with time constants is that everything follows exponential functions with the same time constant. In this case the current is going to start at a maximum value and decay down to zero (when the switch is first closed the capacitor is uncharged and looks like a short circuit, so the current is maximum). It'll have a function like

I(t) = Io e^-t/τ

where τ = RC

The voltage on the resistor will follow a similar curve, starting at 23V and heading towards zero over time, so it'll look like:

VR(t) = 23V e^-t/τ

 

[lpau001]

Thank you Gneil!
That worked!
Now I need an explanation.. haha!
Why did the formula change from

VR(t)=V[1-e^(-t/RC)]

to

VR(t)=V[e^(-t/RC)] ?

I have the first equation in my notes, but apparently that's wrong! Can you show me the derivation? or maybe just hint at it so I could figure it out on my own?
Thanks again, I appreciate the timely reply of you and Cepheid.

 

[lpau001]

Wait I just saw that the equation gneil gave is for discharging capacitors.
Why is that the case in this problem? I thought the problem was a charging problem..

 

[cepheid]

lpau001: Re-read my post. I was not saying that the current should start at 0. I was saying that if you use the equation that YOU wrote down, the current would start at 0 and increase up to a steady state value, and that's how I could tell your equation was wrong.

In any case, gneill gave you the answer, and to answer you latest question: there is no inconsistency here. The voltage of the capacitor as a function of time is given by:

[tex] V_C(t) = V_0(1 - e^{-t/RC}) [/tex]

Where V₀ is the source voltage. The current through the circuit is equal to the voltage across the resistor, divided by its resistance. Assuming that the resistor has one end connected to a battery terminal, and the other end connected to a capacitor terminal, then the voltage across the resistor is given by:

[tex] V_R(t) = V_0 - V_C(t) [/tex]

[tex] = V_0 - V_0(1 - e^{-t/RC}) [/tex]

[tex] = V_0[1 -(1 - e^{-t/RC})] [/tex]

[tex] = V_0[1 - 1 + e^{-t/RC}] [/tex]

[tex]= V_0e^{-t/RC}[/tex]

The current is then:

[tex] I(t) = \frac{V_R(t)}{R} = \frac{V_0}{R}e^{-t/RC} = I_0e^{-t/RC} [/tex]

 

[lpau001]

Cepheid: Thank you, I understand that very well now. I appologize if I came off blunt in my reply to you.
Thanks for your help in showing me the derivation steps.

 

[gneill]

The basic thing to remember is that quantities in these charging/discharging type circuits are all going to follow exponential curves, either starting at some beginning value and rising to a maximum value, or starting at some maximum value and decaying to some lower one. Quite often the minimums and maximums are set by zero and the supply voltage (or current). The trick is to determine what the minimum and maximums are or will be for the parameter you're interested in.

In a series circuit, if the voltage across one component is starting at zero and rising to a maximum, you can be sure that the other component will be having the opposite happen. So you choose the appropriate form of the exponential function that fits.