how do I calculate I(t) for capacitor when R3 is also present

[PG1995]

Hi

Please have a look on the attachment, you can find my question there with other details. Thanks a lot.

Regards
PG

 

[MrAl]

Hi PG,

Well no, you have to include R3 in the analysis. R3 plays a big part in the analysis even though the cap voltage is zero to start. This might give actually you a chance to get to look at circuits in a more broad way.

Rather than calculate the current through the capacitor, imagine it is a resistor to start with. How would you calculate the current through this new resistor?
With a capacitor however you can not use the ratio of resistances because after all it's not a resistor, but there's another trick we can use here where we just think of the cap as another branch similar to how we think of a resistor.

If we knew the current through R1 and the current through R3, we could subtract the current through R3 (iR3) from the current through R1 (iR1) and the result would be the current through the capacitor, and that's because the sum of currents into the node at the top of C has to be zero.
But do we even know the current through these two resistors?

We know the voltage across both of them, therefore we know the current through both of them.

The voltage across R1 is Vin-Vc, and the voltage across R3 is simply Vc. iR1 is therefore (Vin-Vc)/R1, and iR3 is Vc/R3. So we have the two currents we needed.

But what is Vc? Once you get your previous exercise down pat you'll have the answer to that too. You can then insert that result in to the above equations for the current, and you'll then get the current through the capacitor:
iC=iR1-iR3

and note that is because the total current is iR1 and iR3 is one branch current and iC is the other branch current.

Does this make it easier now? If not just yell...