Recent content by Showmaster

Right! Thanks for you reply :)!

The initial voltage across the capacitor is: [tex] \Delta V=\frac{Q}{C}=\frac{0.1}{10^{-3}} V=100 V [\tex] So if there is an initial potential difference of 100 V, and positive charge is distributed on the left side, then current will flow to the left on R1, because of the electric field...

Oh, sorry, my bad. I'm a spanish native speaker and i confused terms. In my work I assumed that the initial positive charge was on the left side of the capacitor thus leaving the negative 0.1 C on the right. If that's the case, then it DOES make sense that current flows in the opposite...

Homework Statement I have the circuit that's attached as a .bmp image. I'm asked to find the initial current that passes trough the battery (V=60 V) knowing that the capacitor is initially charged with charge q_{0}=0.1 C. The resistances are R_1=R_2=R_3=1 Ohm and C=1/1000 F. Homework...