What Is the Time Constant and Potential Difference in This RC Circuit?

[letsgo]

Homework Statement

Consider the circuit below, where C₁=80mF, C₂=30mF, R=50ohms and E=5V.
At t=0, both capacitors are uncharged - so it's charging.

The circuit is in series, with E ---> C₁ ---> C₂ ---> R ---> (back to) E

a) Find the time constant.
b) The initial potential difference across R.
c) The current at t=3s.
d) The potential difference across C2, at t=3s.

Homework Equations

C=Q/V
V=IR
X=X_max(e^(-t/RC)) (where X is I, V or Q and is decreasing with time)
X=X_max(1 - e^(-t/RC)) (where X is I, V or Q and is increasing with time)

The Attempt at a Solution

Here are my answers, can someone check if I'm right?
a)
time constant = RC = (50 ohms)(21.8 mF) = 1.09 s
b)
V=5.0 ohms (same as E)
c)
V=RI -> 5V=(50 ohms)I -> I=0.1 A

I=I_max(e^-t/RC)
I=(0.1 A)[/SUB](e^(-3s/1.09s))
I=6.37x10^-3 A

d)
V_Ceq= V_max(1-e^-t/RC)
V_Ceq= (5 V)(1-e^-3s/1.09s)
V_Ceq= 4.68 V

V=C/Q -> 4.68V=30mF/Q -> Q=6.4x10^-3 C

 

[lewando]

τ(seconds) = resistance(ohms) * capacitance(Farads). Your result is a bit too big.

 

[lewando]

Nevermind,τ looks good.

 

[letsgo]

So are the other answers good to?

 

[lewando]

b) I know you meant V not ohms
c) good
d) looking for a voltage across C2, V_Ceq is right though.

 

[letsgo]

b) I know you meant V not ohms
c) good
d) looking for a voltage across C2, V_Ceq is right though.

Thanks. And OK, you're right, I fixed it.

V=Q/Ceq
4.68=Q/21.8mF
Q=1.02 C, which is the same across each capacitor since they're in series.

And can you please, since you seem to know what you're talking about when it comes to RC circuits, answer this question of mine as well?

https://www.physicsforums.com/showthread.php?t=648367

 

[lewando]

for d) you need to determine the voltage across C2. It is done the same way as if you were going a resistor divider.

On your other question, gneill is on it (and will advise you well).