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Given this diagram:

enter image description here

With S1 switch closed and S2 switch left open, I am trying to find the time constant

Relevant equations

I know τ = RC for a basic circuit, but how would you calculate it for a complex circuit? Is R the equivalent resistance to the battery?

Q=VC V=IR

i = dq/dt dq/dt + Q/τ - emf / R = 0

The attempt at a solution

I start with junction rule and loop rule

I1 = I2 + I3 -emf + I1R1 + Q/C + I2R2 = 0 =emf + I1R1 + I3R3 + I3R4 = 0

At this point the teacher says I2 = dq/dt and we need to get rid of I1 so we can put something next to the C in Q/C.

I use I1 = I2+I3 in junction rule and put it into loop rule #2, getting:

-emf + (I2 + I3)R1 + I3R3 + I3R4 = 0 I2R1 + I3(R1+R3+R4) = emf I3 = (emf - I2R1) / (R1+R3+R3)

Then I put I3 in for the I1 eq.

I1 = I2 + (emf - I2R1) / (R1+R3+R3)

At this point it's so messy and confusing I think I am doing it all wrong.

The baseline she is giving us is that

dq/dt + Q/[foo] - emf/[bar] = 0

where [foo] would be the time constant. In a simple circuit she gives [foo] = RC and [bar] = R

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Usually we don't do "check my work" or "solve this" questions here on physics.SE. I'll give a hint that will probably help you solve all such sums. –  Manishearth Mar 12 '12 at 3:14
    
I think this one is just a little more than that - it's kind of buried in all the text, but the essential question seems to be how you find the time constant for a complex circuit, which is a reasonable thing to answer with an explanation of either circuit reduction rules or using Kirchoff's laws to write a differential equation. (By the way, welcome to the site BHare!) –  David Z Mar 12 '12 at 3:19
    
Yeah, that's exactly what I answered. Didn't feel like looking through all those equations. I hate solving simultaneous equations. –  Manishearth Mar 12 '12 at 3:22
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1 Answer 1

Use Thévenin's theorem.

Since Wikipedia is notorious for making simple things look complicated, I'll simplify it.

Simply put, you can calculate the $q_0$ and $RC$ in the expression $q=q_0(1-e^{-t/RC})$ rather easily provided the capacitor is discharged initially, with the following method:

  • Short all the batteries, find equivalent resistance across the ends of the capacitor by series-parallel. This gives you $R$
  • Remove the capacitor (unshort the batteries), and find potential difference across the terminals at steady state. Use this to find $q_0=CV$

This works for inductors as well, though you use it to find $i_0$, not $q_0$ in that case.

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I think all the work I was doing was trying to find the R. So you are saying R = R2+1/[1/R1 + 1/(R3+R4)] –  BHare Mar 12 '12 at 11:49
    
@BHare Yep. I realised you only needed R, but the entire theorem is so useful (and it's a great way to amaze your peers; save time in tests), that I decided to give the $q_0$ part as well =D –  Manishearth Mar 12 '12 at 11:55
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