[SOLVED] Current through each capacitor

[preethi19]

Hi i have attached an image of a very simple ckt.

I able able to find the Charge Q of the capacitor and the Voltage across the two capacitors. Can anyone pls let me know what is the current through each capacitor 12uF and 6uF. I know current through series capacitor will be the same but how to calculate individually for both and see that both the current calculations are turning out to be the same.
Adding both the capacitors the equivalent capacitance is 4uF. And voltage across 12uF is 1.33 and that of the 6uF is 2.67.
The formula for current is I= C dV/dt. But in this case voltage is not varying so i guess it simply becomes I=CV.

So current across the 4uF capacitor (the equivalent to 6 and 12uF) is I= 4uF x 4V. (since Votlage at point X is 4V) so I= 16uA

So now for individual capacitors the current through 12uF is
I=12uF x 1.33 = 16.02uA

And current through 6uF is I= 6uF x 2.67 =15.96uA..
So current is similar in both through 12uF and 6uF.
Is this the right way??? can anyone pls correct me if wrong.

 

[Aussie Susan]

If the voltage is invariant then dV/dt = 0.
Once the circuit has reached a stable state (after being connected for the first time) then the current through the capacitors will be zero - if you think about it, a capacitor passes AC but blocks DC.
Also this is really just a simple network analysis problem but you need to remember that the impedance for a capacitor is
Z= 1/(jwC)
where j is the imaginary number and w = 2(pi)f. Therefore if w is zero (i.e. a purely DC situation) then Z will be
'-j(infinity)'
and infinite impedance means zero current.
Susan