Solve RC Circuit Problem: 45V -> 10V in 4.21ms

[alteradoplebada]

Homework Statement

I have an RC circuit with two capacitors connected in parallel. Both capacitors are initially charged to 45.0 V.
a.) How long after closing the switch S will the potential across each capacitor be reduced to 10.0V?
Knowns:
C1= 15.0 microFarads
C2= 20.0 microFarads
R1= 30.0 ohms
R2= 50.0 ohms
Here is a diagram. http://www.chegg.com/homework-help/questions-and-answers/circuit-shown-figure-capacitors-areinitially-charged-450--long-closing-switch-s-potential--q201797

Homework Equations

v=q/c
q=Q(initial)e^(-t/RC)[/B]

The Attempt at a Solution

I know that both capacitors have the same potential difference because they are connected in parallel.
I have used v=q/c to find the charge q of the first capacitor when the potential v is 10 V.
q=vc
=10V*(15*10^-6F)
=1.5*10^-4C

Next, I found the initial charge of the first capacitor.Q(initial)
Q(initial)=45V*(15*10^-6F)
=6.75*10^-4C

Lastly, I foundthe time at which the potential of the first capacitor is 10V which corresponds to the charge q at that time.
q=Q(initial)*e^(-t/RC)
solving for time t=-RC*ln(q/Q(initial))
My answer comes out as 1.8 milliseconds but the correct answer is 4.21 milliseconds. Is there something wrong in my understanding of the problem? What is going on here?

 

[Qwertywerty]

v=q/c
q=Q(initial)e^(-t/RC)

So , according to you , formula for q_t of a single capacitor in this question would follow the same function rule as in the case of a single capacitor RC circuit ?

*Hint - Solve using equivalent capacitance .

 

[SammyS]

Homework Statement

I have an RC circuit with two capacitors connected in parallel. Both capacitors are initially charged to 45.0 V.
a.) How long after closing the switch S will the potential across each capacitor be reduced to 10.0V?
Knowns:
C1= 15.0 microFarads
C2= 20.0 microFarads
R1= 30.0 ohms
R2= 50.0 ohms
Here is a diagram. http://www.chegg.com/homework-help/questions-and-answers/circuit-shown-figure-capacitors-areinitially-charged-450--long-closing-switch-s-potential--q201797

Homework Equations

v=q/c
q=Q(initial)e^(-t/RC)[/B]

The Attempt at a Solution

I know that both capacitors have the same potential difference because they are connected in parallel.
I have used v=q/c to find the charge q of the first capacitor when the potential v is 10 V.
q=vc
=10V*(15*10^-6F)
=1.5*10^-4C

Next, I found the initial charge of the first capacitor.Q(initial)
Q(initial)=45V*(15*10^-6F)
=6.75*10^-4C

Lastly, I foundthe time at which the potential of the first capacitor is 10V which corresponds to the charge q at that time.
q=Q(initial)*e^(-t/RC)
solving for time t=-RC*ln(q/Q(initial))
My answer comes out as 1.8 milliseconds but the correct answer is 4.21 milliseconds. Is there something wrong in my understanding of the problem? What is going on here?

Hello alteradoplebada. Welcome to PF !

Here is what is shown in that link:

In the circuit shown in the figure both capacitors are initially charged to 45.0

.

How long after closing the switch S will the potential across each capacitor be reduced to 15.0

?

What will be the current at that time?​

What did you use for R and C to get the time constant?

 

[alteradoplebada]

Hello. In my case the potential would be reduced to 10 V not 15V. All the values in the diagram are correct.
I used 80 ohm for the resistance. For the capacitance I used 15 microFarads. Am I wrong in my approach? I am finding the charge of the 15 microFarad capacitor when its potential is 10 V. q=vc Then i find the time at which the charge is q which corresponds to the potential being 10 V. q=Q(initial)*e^(-t/RC). I solve for t.

 

[SammyS]

Hello. In my case the potential would be reduced to 10 V not 15V. All the values in the diagram are correct.
I used 80 ohm for the resistance. For the capacitance I used 15 microFarads. Am I wrong in my approach? I am finding the charge of the 15 microFarad capacitor when its potential is 10 V. q=vc Then i find the time at which the charge is q which corresponds to the potential being 10 V. q=Q(initial)*e^(-t/RC). I solve for t.

Yes, you are wrong in your approach.

The 20 μF capacitor also discharges. You need to take into account that both capacitors are discharging. The easiest way to do that is to use equivalent capacitance.