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/[tex]\tau[/tex]) 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[tex]\mu[/tex]F capacitor?