(t) = V0e^-t/rc where V(t) is the voltage across the capacitor after it's been discharging for an amount of time t, V0 is the voltage it started at, C is the capacitance of the capacitor, and R is the resistance of the resistor it's discharging through. One thing we can take away from this is that, after it's been discharging for an amount of time t = RC, the voltage across the capacitor will be V0e^−1, which is about 37% of the voltage it started at. In other words, RC, the resistance times the capacitance, is the amount of time it takes the capacitor to discharge from 100% voltage to 37% voltage. (The units check out, by the way: 1Ω⋅1F=1sec) If you have an RC circuit with a 10 mF capaci
V(t) = V0e^-t/rc
where V(t) is the voltage across the capacitor after it's been discharging for an amount of time t, V0 is the voltage it started at, C is the capacitance of the capacitor, and R is the resistance of the resistor it's discharging through.
One thing we can take away from this is that, after it's been discharging for an amount of time t = RC, the voltage across the capacitor will be V0e^−1, which is about 37% of the voltage it started at.
In other words, RC, the resistance times the capacitance, is the amount of time it takes the capacitor to discharge from 100% voltage to 37% voltage. (The units check out, by the way: 1Ω⋅1F=1sec)
If you have an RC circuit with a 10 mF capacitor and a 2 kΩ resistor, about how long will it take the capacitor to discharge to 37% of its original voltage?
Given that capacitor is C= 10 mF
Resistor R = 2 k ohm
Final value of voltage across the capacitor is 37 % of input voltage.
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
