Capacitors/Resistors in parallel

[megr_ftw]

Homework Statement

Consider the circuit in the figure below, in which V = 110 V, R = 30 , and the switch has been closed for a very long time.

What is the charge on the capacitor?
The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?

Homework Equations

Capacitance= Q/V
V=IR
Q(t)=Q₀*e^(-t/$$\tau$$)

The Attempt at a Solution

I found the resitance equivalence for the whole circuit to be 67.5 ohms and the total current to be 1.63A.

Is the voltage drop on the 10 ohm resistor the same as the 2$$\mu$$F capacitor?

 

[willem2]

The Attempt at a Solution

I found the resitance equivalence for the whole circuit to be 67.5 ohms and the total current to be 1.63A.

when the switch has been closed for a long time, there's no more current through the 10ohm resistance

Is the voltage drop on the 10 ohm resistor the same as the 2$$\mu$$F capacitor?

No. The voltage drops across two circuit elements that are parallel are the same. Not if they are
in series.

 

[megr_ftw]

so i should find the voltage drop for the 10ohm resistor and then the capacitor. I am starting to get confused now, how should I find the charge for the capacitor? or should I find the voltage first?

 

[Jasso]

so i should find the voltage drop for the 10ohm resistor and then the capacitor.

Yes. Two things to remember: V=IR, and the voltage drop across a series circuit is the sum of the voltage drops of each element.

Im starting to get confused now, how should I find the charge for the capacitor? or should I find the voltage first?

Refer to the equations you gave, its right there.