Calculate the time constant (RC circuit)

1. Feb 3, 2006

hellraiser

Hi.
How am I supposed to calculate the time constant for the enclosed circuit diagram.
I know time constant t = 1/RC
In this case do i take only one resistance or the combined value (series value)
Assume any values. I only want to understand how it is to be done.

PS: Sorry for the stupid drawing. I used MSPAINT to draw it. I think I should have practised drawing more as a kid. :)

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2. Feb 4, 2006

Hootenanny

Staff Emeritus
I can't see the picture but as far as I know you use the combined resistance of the timing circuit using the rules for the sumation of parallel and series resistances.

3. Feb 4, 2006

hellraiser

Ok i will try to draw it in ASCII.
__res______
| | |
| | |
bat cap res
| | |
|____|____|
res is resistor
cap is capacitor
bat is battery (+ve terminal up)

4. Feb 4, 2006

hellraiser

Obviously I failed :(

5. Feb 4, 2006

Hootenanny

Staff Emeritus
If I guessed right, the resistor is on the top line between the battery and the capacitor and the resistor is on the leg to the right of the capacitor??

6. Feb 4, 2006

hellraiser

Yes. You got it right.

Why is the attachment not being cleared. It's been nearly 24 hrs since I posted it. Anyone listening?

7. Feb 5, 2006

Hootenanny

Staff Emeritus
So now you calculate the total resistance of the circuit. Remember one is in parallel while the other is in series.

8. Feb 5, 2006

hellraiser

How? Could you show me?
I know R(series) = R1 + R2
R(parallel) = 1/R1 + 1/R2

9. Feb 5, 2006

Hootenanny

Staff Emeritus
Correct. You want to know the resistance acting through the capacitor. Think about it logically, if one resistor is in series with the capacitor and one is in parallel with it then could you write an equation of the total resistance acting through the circuit base on what you just said?

10. Feb 6, 2006

hellraiser

I don't know how? Should I apply Ohm's Law and solve for an expression with time variation of current? But that would be quite lengthy.

11. Feb 6, 2006

Hootenanny

Staff Emeritus
No. Simply use $$R_{total} = R_{series} + \frac{1}{R_{parallel}}$$

12. Feb 6, 2006

hellraiser

So the ans is R1 + R2 + (R1*R2)/(R1+R2)

13. Feb 6, 2006

Hootenanny

Staff Emeritus
Ive given you the answer to calculate the total resistance above.

14. May 5, 2009

andrewjohnsc

I'm doing the same problem--how did you get that expression for Rtotal?

15. May 6, 2009

davieddy

Witchcraft or maybe prayer?

Let the battery have emf Vo, and label the resistors R and r.

Answer these simple questions:

a) When Vc = 0, what is Ic?

b) When Ic = 0, what is Vc?

How long would it take the current in a) (if it stayed constant)
to charge (hint!) the capacitor C to the voltage in b)?
This is the time constant.

David

Last edited: May 6, 2009
16. May 6, 2009

davieddy

Here's a better presented problem.
(Ignore the random stabs at answers)

Having said that, the first question should read:

1) Determine the current in each resistor and charge on the capacitor
immediately after the switch is closed

https://www.physicsforums.com/showthread.php?t=312098

In addition to what you are asked for, answer this:

What is the time constant during the charging process (switch closed)?

Last edited: May 6, 2009
17. May 6, 2009

rl.bhat

Let R1 is R series and R2 is the resistance in parallel with capacitor. You can interchange R2 and C with affecting the circuit diagram. At the instant the switch is on, the voltage across R2 is E*R2/(R1 + R2). This voltage charges the capacitor. There is no resistance in series with the capacitor. Since the time constant t = RC ( not 1/RC as you have mentioned.) and R = 0 , the time constant is zero.

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