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?


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

  • 1
    $\begingroup$ 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. $\endgroup$ – Manishearth Mar 12 '12 at 3:14
  • $\begingroup$ 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!) $\endgroup$ – David Z Mar 12 '12 at 3:19
  • $\begingroup$ Yeah, that's exactly what I answered. Didn't feel like looking through all those equations. I hate solving simultaneous equations. $\endgroup$ – Manishearth Mar 12 '12 at 3:22

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.

  • $\begingroup$ 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)] $\endgroup$ – ParoX Mar 12 '12 at 11:49
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    $\begingroup$ @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 $\endgroup$ – Manishearth Mar 12 '12 at 11:55
  • $\begingroup$ Really helpful answer thanks... but can you please explain the case where I have 2 capacitors and I need to find the time constant for each of them in a similar complex circuit? In that case do we ignore the other capacitor when we are calculating time constant for one of them.....or the other capacitor is also involved in getting the answer? $\endgroup$ – Shubham Oct 27 '15 at 17:15

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