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

[woaname]

oh, charging vs discharging. right. thanks.

 

[woaname]

What is IC,max(closed), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is closed? A positive value for the current is defined to be in the direction of the arrow shown.

What is IC,max(open), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is open? A positive value for the current is defined to be in the direction of the arrow shown.
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for these questions, the first one would use the charging equation for current:
I(t)=(V(battery)/Rtotal)*EXP(-t/RC)
and the second would the discharging equation:
I(t)=(Qmax)/RC)*EXP(-t/RC)

is this correct?

 

[gneill]

for these questions, the first one would use the charging equation for current:
I(t)=(V(battery)/Rtotal)*EXP(-t/RC)
and the second would the discharging equation:
I(t)=(Qmax)/RC)*EXP(-t/RC)

is this correct?

In the first case where the capacitor is charging, the resistor network will complicate things a bit since there's a voltage, and hence current, division taking place. You'll have to find the current through R2 when the capacitor is replaced by a short circuit (i.e. what the capacitor "looks like" when it's initially uncharged) to determine the leading coefficient for the exponential.

For the second equation, again the correct choice for R makes all the difference, but essentially that's the correct form for the equation.

 

[woaname]

no luck trying on my own... the equations aren't working out. i haven't been given time values to substitute for t. and for the value of RC, this is what i have in understanding:
1) for Ic,max(closed), i can't simply add all the resistances, but i can't find how to include R3 in there.
2) only the smaller loop will be in action, so the RC value would include only the two live resistors in circuit. but then, the time variable stands unsolved.

 

[gneill]

no luck trying on my own... the equations aren't working out. i haven't been given time values to substitute for t. and for the value of RC, this is what i have in understanding:
1) for Ic,max(closed), i can't simply add all the resistances, but i can't find how to include R3 in there.

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)?

2) only the smaller loop will be in action, so the RC value would include only the two live resistors in circuit. but then, the time variable stands unsolved.

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

 

[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.