How to find Qmax of a capacitor in parallel RC circuit?

[woaname]

If you're looking for Ic's maximum value, that occurs the instant the switch is closed and the capacitor is completely uncharged. How does an uncharged capacitor behave (what does it "look like" to the rest of the circuit)?

the capacitor will act like a short, and the circuit would look like only R3 and R2 in parallel, with the rest of the loop intact. if we look back at part 1 of the question, the current in I4 was found at t=0. so would that current apply here? same current passes through R1 and the simplified R23 set since it's all in series now. correct? what's next then?

What are you looking to determine? Why is a time value required?

right, the maximum current would be at t=0, so the exponential equals 1. slipped my mind. solved it

 

[gneill]

the capacitor will act like a short, and the circuit would look like only R3 and R2 in parallel, with the rest of the loop intact. if we look back at part 1 of the question, the current in I4 was found at t=0. so would that current apply here? same current passes through R1 and the simplified R23 set since it's all in series now. correct? what's next then?

You need to determine the current through R2. R3 and R2 form a current divider.

 

[woaname]

understood. got the correct answer, but from a technical error i made in calculating. so we know that the voltage through the two branches of r3 and r2 should be same, and that the current through the branches should be the sum of the current through r3 and r2, right?
so i set up the equation " I23 = (V2/REQ)+(V3/REQ) "
since v2 and v3 are equal, i set I23*Req=2*v2
HERE is my dilemma: i forgot to divide the answer by 2 to get v2, but the current i got for i2 in the answer was right! why? should the current in the capactor not be equal to i2?

 

[gneill]

understood. got the correct answer, but from a technical error i made in calculating. so we know that the voltage through the two branches of r3 and r2 should be same, and that the current through the branches should be the sum of the current through r3 and r2, right?

The voltage ACROSS the two resistors is the same because they are in parallel. The sum of their individual currents should equal the current entering at the top of the pair (via R1) which also equal s the current leaving the pair and passing on to R4.

so i set up the equation " I23 = (V2/REQ)+(V3/REQ) "
since v2 and v3 are equal, i set I23*Req=2*v2

What is this REQ? You have two different resistances; they both share the same potential drop; you know the total current that gets divided between them.

HERE is my dilemma: i forgot to divide the answer by 2 to get v2, but the current i got for i2 in the answer was right! why? should the current in the capactor not be equal to i2?

Yes, the initial current through the capacitor will be the current through R2. I can't explain the coincident of your finding a correct result even though your method is flawed, not without seeing your calculations in detail.