Capacitors/Resistors in parallel

1. Oct 27, 2009

megr_ftw

1. The problem statement, all variables and given/known data
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?

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

3. 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?

2. Oct 27, 2009

willem2

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

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

3. Oct 27, 2009

megr_ftw

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

4. Oct 27, 2009

Jasso

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.

Refer to the equations you gave, its right there.