Calculating Charge and Voltage in an RC Circuit with a Closed Switch

[the_storm]

Homework Statement

Switch S has been closed for a long time, and the electric circuit shown in the figure below carries a constant current. Take C1 = 3.00 µF, C2 = 6.00 µF, R1 = 4.00 k, and R2 = 7.00 k. The power delivered to R2 is 2.60 W.

Homework Equations

(a) Find the charge on C1.



(b) Now the switch is opened. After many milliseconds, by how much has the charge on C2 changed?

The Attempt at a Solution

I said power = I$$^{2}$$ * R
then 2.6 = I$$^{2}$$ * R then I got the current I and it was 0.0193
Then I got V1 = I R$$_{1}$$
I added V1 + V2 to get the V for the battery
then to get charge on C1 is the maximum charge because the switch has been opened for a long time then
Q = C1 * V
but I got error :S :S

 

[gneill]

The charge on C1 is to be found while the switch is still closed. What values did you get for both V1 and V2?

 

[the_storm]

value of V1 = 77.2 Volt
and V2 = 134.9 Volt

then V of the battery should be V1 + V2 = 212.1 Volt
Then I Q1 = C1*Vbattery
but it is a wrong answer :S

 

[the_storm]

I have the power at R2
since P = I^2 R2 >> I have P and I have R2 substitute in the equation and I got the current. since P = IV >> I have the current I can get v at R2
because the switch is on and for along time then R1 and R2 are connected in series then I that passes through R2 is the same as the current that passes through R1
and using ohm's law I can get the voltage across R1 v = IR1
what do u think ?

 

[Integral]

The voltage across the cap is the battery voltage less the drop across the resistor in series with the cap.

 

[the_storm]

but the caps are fully charged .. so the charges flows to R1 then to the switch then down to R2 and then go back to the battery .. ??

 

[gneill]

With the current flowing only through R1 and R2, these two resistors will form a voltage divider. Note that C1 is in parallel with R1 while the switch is closed. So the voltage on C1 should be the same as that across R1.

 

[the_storm]

Got it :)

so when the switch is open ... R1 and C2 have the same potential of the total battery... so .. then If I could get the voltage across R1 and then subtract it from the total voltage I will get the voltage on C2 correct ??

 

[gneill]

When the switch is open, you should be able to see that there are two separate parallel branches connected to the battery. Each branch consists of a resistor and a capacitor in series. What voltage do you expect to find (after a long time has passed) on a capacitor that is connected in series with a resistor and battery?

 

[the_storm]

The same voltage as the battery :D ? isn't it ?

 

[gneill]

The same voltage as the battery :D ? isn't it ?

It is indeed! When the switch is opened, the capacitors will continue charging from their "stalled" values set by the voltage divider, and continue charging until they reach the same potential as the battery.

 

[the_storm]

Got it :) .... Thank you Bro ... I really appreciate that :) :)