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R and 2R are in parallel. 4R and 3R are in parallel.

I dont know where to go after calculating the the parallel Rs. Do you become in series? If so, how?

It is right after the switch is closed. So capacitor acts as a wire.

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closed as off-topic by John Rennie, Jon Custer, Kyle Kanos, Aaron Stevens, M. Enns Jun 14 at 17:44

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  • $\begingroup$ $R$ and $2R$ are not in parallel, because the bifurcation does not converge back in one point. IT would be parallel if a main wirte was split in two and then rejoined, but it does not rejoin again directly. $\endgroup$ – FGSUZ Jun 13 at 11:08
  • $\begingroup$ @FGSUZ If before the switch is closed, if there is no voltage across the capacitor, then when the switch is first closed ($t=0$) the capacitor looks like a short-circuit. At that instant R is in parallel with 2R and 4R is in parallel with 3R. For $t>0$ they are not. $\endgroup$ – Bob D Jun 13 at 15:59
  • $\begingroup$ Please post the actual question - you've only given us the setup and some of your ideas. What are you actually trying to find out? $\endgroup$ – Oscar Bravo Jun 14 at 8:17
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I'm assuming you've been asked to find the the voltage across the capacitor. Is that the question?

If so, re-arrange the circuit so that R and 4R are actually a voltage divider. Same for 2R and 3R. Now you've got a capacitor across the mid-points of two voltage dividers. You can then calculate the relative voltage at each side of the cap and so get the $V_{diff}$ across it.

If that's what you require...

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  • $\begingroup$ Oh sorry voltage is provided at 100v $\endgroup$ – jee Jun 13 at 16:16
  • $\begingroup$ Im trying to find equvalent resistance to find out current $\endgroup$ – jee Jun 13 at 16:17
  • $\begingroup$ Equivalent resistance? Combine the series resistances first, then combine the resulting parallel resistances. If that's all they wanted you to do, why did they bother with the capacitor? Was it a trick? $\endgroup$ – Oscar Bravo Jun 14 at 8:21
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I don't know where to go after calculating the the parallel Rs. Do you become in series? If so, how?

It depends on what is it you are trying to determine?

If you want to find the current delivered by the battery to the circuit the instant the switch is closed you can treat the capacitor as a short circuit if it has no initial charge on it (no initial voltage across it). Then determine the equivalent resistance of the parallel R and 4R resistors and the equivalent resistance of the parallel 2R and 3R resistors. I assume you know how to do that. The two equivalent resistances are then in series so just add them. Now you have the equivalent resistance of the entire circuit as seen by the battery. At that point apply Ohms law $I=\frac{e}{R_{eq}}$.

All of this assumes you want the current delivered by the battery the instant the switch is closed.

Hope this helps

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  • $\begingroup$ He has only given us the setup of the question but has not stated what the actual question is... My guess is that he's being asked to find the steady-state voltage across the capacitor (so once the capacitor has charged and it's all settled down). Then it's a simple voltage divider problem (R & 4R form a voltage divider, so does 2R & 3R). $\endgroup$ – Oscar Bravo Jun 14 at 8:16
  • $\begingroup$ @Oscar Bravo perhaps, but he did say “It is right after the switch is closed” $\endgroup$ – Bob D Jun 14 at 9:09
  • $\begingroup$ Aha... I see your point. Then your solution is spot on... Who knows what it's about - perhaps the OP will be good enough to enlighten us one day. $\endgroup$ – Oscar Bravo Jun 14 at 12:26
  • $\begingroup$ @OscarBravo I agree. Let's just say you have it covered if it's when the switch is closed a long time and I have it covered when the switch is first closed. So I'm up voting yours. $\endgroup$ – Bob D Jun 14 at 12:43
  • $\begingroup$ To whomever. Why the down vote? $\endgroup$ – Bob D Jun 14 at 12:44

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