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**RC circuit - charge on capacitor after a "long time"**

**Circuit**: 10V goes to 100ohm resistor, then there's a branch point. One side leads to a a resistor R2 = 100ohms and the other to a capacitor with C = 1uF. The switch is located just before R1, and is closed for a "long time."

**Solution**: Ok, I first solved for the current through R1. After a long time, the capacitor becomes saturated with charge, so the current flows through R1 then R2 then back to the battery. Thus, R1 and R2 are in series and R_eq = 200ohms. By ohms law, the current must be the same through both resistors, I = V/R = 10V/200ohms = 0.05A.

**Question**: Now I am asked what the charge on the capacitor is in this situation (after a long time). The answer is 5uC. Can someone explain why?