Charge after three time constants?

AI Thread Summary
The discussion revolves around calculating the charge on a capacitor after three time constants in an RC circuit. The time constant (T) is determined by the product of resistance (R) and capacitance (C), resulting in T = 22 milliseconds for a 10kΩ resistor and a 2.2μF capacitor. After three time constants, the voltage across the capacitor can be calculated using the formula V(t) = V0(1 - e^(-t/T)), where V0 is the supply voltage. At three time constants, the capacitor will charge to approximately 95% of the supply voltage, which is about 9.5V. Understanding this concept is essential for solving similar problems in circuit analysis.
Keeeen
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Homework Statement


A 10V supple is connected to a 2.2μF capacitor in series with a 10kΩ resistor. To what value will it have charged after three time constants?


Homework Equations


I know that T=RC but I have no idea beyond this point


The Attempt at a Solution


T=RC
T=2.2μF*10kΩ
T=22

the rest is unknown to me, if someone could explain it to me that would be awesome.
 
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Either find the expression for a charging capacitor or derive it yourself. It's a simple linear, first-order, constant-coefficient differential equation. (Sum currents to zero at the capacitor-resistor junction.)
 
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