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schematic

simulate this circuit – Schematic created using CircuitLab

Assuming i have the circuit from above and i need to find the current through every resistor and capacitor. the switch opens at t=0. I can see the symmetry that on the left flank all the resistors are 2 ohm resistors while in the right flank they are all 1 ohm, but i still can't figure how to find ALL the current's that runs through each resistor and capacitor. Any tips please?

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  • \$\begingroup\$ yes it's a dc voltage source \$\endgroup\$ – user3921 May 21 '14 at 18:44
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Go to the "end game" when all the caps can be assumed to have charged up - what will be the current through a capacitor when the voltage is not changing across it? This you should know for the formula I = C dV/dt.

Once you have concluded correctly you should be able to see what the voltages are across all the resistors. It's not a big leap of faith (given the "special" value of the resistor ratios) to make definite argument that this state of affairs also happens right at the moment when the switch closes.

Here's another tip - think wheatstone bridge.

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You know (or can assume) that initially the capacitors have no charge => the voltages across each capacitor is 0.

Now remove the capacitors for a moment. Think what the voltage across the lower (ex-) capacitor is. (If you don't see it immediately, calculate the voltage at its right side, repeat for its left side.) repeat for the upper capacitor. Conclusion?

Now calculate the current.

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  • \$\begingroup\$ Wouldn't that be a solution for thr total curren't of the circuit only for t=0? After a while curren't will differ on each branch. assuming i have found the curren't for t=0 on each resistor, how can i find the curren't for every capacitor for all t? \$\endgroup\$ – user3921 May 21 '14 at 19:25
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    \$\begingroup\$ Stop worrying and answer: what is the voltage across each capacitor at t=0? Hence: what is the current 'through' the capacitor at t=0? Then: what is different for t > 0 (trick question)? \$\endgroup\$ – Wouter van Ooijen May 21 '14 at 20:57
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Wouldn't that be a solution for thr total curren't of the circuit only for t=0?

The initial conditions are given by the DC steady state solution with the switch closed. Clearly, the voltage across each capacitor before the switch opens is zero and, thus, the capacitors are initially uncharged.

When the switch is opened, there is no transient since the capacitors are initially uncharged and remain uncharged since the circuit now has no source.

Remarkably, the same result holds if, instead, the switch is initially open and is closed at \$t=0\$.

The reason is that the DC steady state solution for the voltage across each capacitor is the same regardless of the switch setting and, thus, there is no transient; the voltage across each capacitor is constant and zero.

Of course, the voltage across each resistor does depend on the position of the switch but these voltages (and associated resistor currents) are almost trivial to solve for when the capacitors are replaced with open circuits (in order to solve for the DC steady state solution).

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