# Time constant in an RC circuit

1. Sep 25, 2010

### VenaCava

1. The problem statement, all variables and given/known data
Find the time constant

2. Relevant equations

T=RC

3. The attempt at a solution
I understand the calculation fine. But I was wondering if the resistors are considered to be in parallel with each other even though the second resistor appears to be in parallel only with the capacitor?

so T=RC=(0.5 ohm)(1x 10^-6 F) = .... etc

Thanks

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2. Sep 25, 2010

### hikaru1221

Only in the case of RC circuit (the circuit comprises of R, C and the source all connected in series) is the time constant equal to RC. The circuit in the picture is not RC circuit and thus there is a need for analyzing this circuit.
Try to solve for the current i through capacitor C or the charge q on C. The solution should contain a time-dependent exponential term which is in this format: exp(-t/T). Here T is the time constant needed.
Hint: You will see that T = RC/2, not RC.

3. Sep 25, 2010

### VenaCava

If I were to say that R= equivalent resistance of the entire circuit, would T=RC be correct?

Where R1= 1 ohm R2 = 1 ohm and R = 1/(1/1 + 1/1) = 0.5 ohm
so T=RC??? or would this still be incorrect?

4. Sep 25, 2010

### Quinzio

You apply Thevenin to solve the generator and two resistances.
The equivalent circuit is a V/2 generator in series with an R/2 resistance.
So the time constant becomes RC/2.

The capacitor voltage is

$$v_c(t) = \frac{V}{2}\left(1-e^{\frac{-2t}{RC}} \right)$$

5. Sep 25, 2010

### hikaru1221

That's just for this particular circuit. If you convert the part containing the source and the 2 resistors into its Thevenin equivalent (only C is left), the Thevenin resistor is the equivalent resistor of (R1 // R2).

6. Sep 25, 2010

### VenaCava

I am not familiar with the Thevenin equivalent...

When a capacitor is fully charged, does it act as a plain wire? I can't recall, sorry its been ages since I studied capacitor/circuit basics.

Would it be correct to say that while the battery is still connected (ie before the capacitor is discharged) that

Vbat= IR1 + Q/C

Is this equation at all useful to me?

7. Sep 25, 2010

### hikaru1221

An ideal capacitor won't. Please have a look on Wikipedia

Yes and yes. The equation shows what happens and also what must happen in the circuit.

8. Sep 25, 2010

### VenaCava

So from there, how can I use that equation to find the time constant? I don't have the voltage value or current, so I'm assuming these will have to cancel out somewhere and are therefore independent of the time constant. I'm think my the time constant will only be based on the capacitance and the ratio/values of the resistors but I'm not sure how I get to this stage.

9. Sep 25, 2010

### VenaCava

Okay, let me rephrase this. You guys say that the time constant is T=RC/2. However, if this were for two different values of R (ie. R1 and R2, which don't have the same value) how would this change the time constant and what would it equal (what is T, in terms of C, R1, and R2)?
Thanks!

10. Sep 25, 2010

### hikaru1221

Actually you have an equation for current: i = current through R parallel to C + current through C.
_ The current through C is dq/dt.
_ The current through R parallel to C = voltage across it / R, and that voltage is the same as the voltage across the capacitor.
You will then have to solve a differential equation.

In this particular problem, T = ReqC, where Req is the equivalent resistance of (R1 // R2). Solve the differential equation and you will see this

11. Sep 25, 2010

### VenaCava

My question I posted is actually based on a laboratory experiment. While the value of the resistors were the same in the lab, they had different error values associated with them, so ideally I'm trying to find a value for T that is a function of R1 and R2 (so I can calculate the error value that goes along with my calculated time constant). Is it possible to express T in terms of these two values?

By the way, thank you so much for your help so far!! I really appreciate it.

12. Sep 25, 2010

### Quinzio

If you're studying electronic in depth you have to become familiar with Thevenin. It is really udeful.

$$v_c(t) = \frac{VR_1}{R_1+R_2}\left(1-e^{\frac{-t}{\tau}} \right)$$

$$\tau = C\frac{R_1R_2}{R_1+R_2}$$

13. Sep 25, 2010

### Quinzio

Nope. As an open circuit.

14. Sep 26, 2010

### VenaCava

Thank you! This is perfect. I tried to derive it myself but was not successful. Would this be possible for me to do without Thevenin (as I have not learned it yet)?

15. Sep 26, 2010

### hikaru1221

Yes, by solving the differential equation
The 2 equations needed are:

_Voltage equation: $$V_{bat}=iR_2 + q/C$$ (*)

_ Current equation: $$i = \frac{dq}{dt} + i_{R1} = \frac{dq}{dt} + \frac{V_{R1}}{R_1}$$

But $$V_{R1} = q/C$$, so: $$i = \frac{dq}{dt} + \frac{q}{CR_1}$$ (**)

_ Substituting (**) into (*):
$$V_{bat} = (\frac{dq}{dt} + \frac{q}{CR_1})R_2 + \frac{q}{C}=R_2\frac{dq}{dt} + \frac{R_1+R_2}{CR_1}q$$

The next step is to solve this differential equation The general solution will definitely contain the term $$e^{-t/T}$$ where T is the time constant as Quinzio pointed out. But the exact solution depends on the initial conditions, which are not mentioned here, so I'll leave it aside. We only need to show that there is the exponential term in the solution

16. Sep 26, 2010

### VenaCava

So if for this particular circuit, I am discharging the capacitor, would this be correct for solving the differential eq?
(PS sorry I don't know how to do the pretty equations)

Vbat = R2 dq/dt + (R2 + R1)q/CR1 (since the circuit is being disconnected from the voltage source at t=0 vbat=0 ???)

dq/dt = -(R2 + R1)q/CR1R2

Q= Q(t=0) e(-t/ CR1R2/(R2 + R1))

therefore T= CR1R2/(R2 + R1)

17. Sep 26, 2010

### hikaru1221

You are right at deducing that V(battery)=0 when there is no battery (discharge) The same trick is used in many circuit analysis methods.

18. Sep 26, 2010

### VenaCava

Thank you so much!