>>12242534>it says its 10 uF so you can assume its fully chargedFor an ideal capacitor this response makes no sense.
Let me help you out. From 0ms to 5ms 5mA will be applied to the capacitor. There an extremely simple relationship between current and voltage for a capacitor. I=C*dv/dt. This means the voltage must be a linear function in this time so. Vc=a*t+b. 5mA=10uF*a. This solves for the slope of charging, but you can't solve for b. Because of this you can't know the voltage over the capacitor at any time in the future. There is no full charged here either. A voltage rating on a cap is a function of dielectric breakdown strength and dielectric thickness. In an ideal circuit it doesn't exist.
