In an alarm clock, there needs to be a delay of 1 minute between closing switch and circuit being active. A 2200microF capacitor is charged through a resistor from a 5V supply. The pd across capictor must rise to 4.3V for alarm to become active. Calculate size of resistor. The answer is 14000ohms. I need to use the equation: V=Vo * e^(-t/RC) with t = 60s but im not sure what to use for the value of the 2 V's.