Suppose a capacitor of C=2.0 uF with no excess charge on it is suddenly connected in series with a resistor of R=447 ohms , and an ideal battery of voltage V=25 V . How long will the capacitor take to gain 80% of the charge it will eventually have when current stops flowing through the battery?
C = Q/Vc
so Q = CVc = integral i dt
so Vc = (1/C) int i dt
Vc = 0 at start
25 = i R + Vc
25 = i R + (1/C) int i dt
0 = R di/dt + i/C
di/dt = -i/(RC)
di/i =-dt/(RC)
ln i = -t/(RC)
i = e^[-t/(RC)]
when t = 0, Vc = 0 and i = V/R
i = (V/R) e^[-t/(RC)]
Remember from above
Vc =(1/C) int i dt
so
Vc = (V/RC) int e^[-t/(RC)] dt
Vc = (V/RC) (-RC) e^[-t/(RC)] + c
Vc = -V e^[-t/(RC)] + c
at t = 0 Vcapacitor = 0 so C = V
Vc = V (1 - e^[-t/(RC)] )
.8 = (1 -e^[-t/(RC)])
.2 = e^[-t/(RC)]
-1.61 = -t/(RC)
put in R and C to get t
To determine how long it will take for the capacitor to gain 80% of the charge it will eventually have, we can use the concept of RC time constant.
The RC time constant is given by the product of the resistance (R) and the capacitance (C) in the circuit. It represents the time it takes for a capacitor to charge or discharge to approximately 63.2% of its final value.
In this case, the resistor and capacitor are connected in series, so the time constant (τ) is given by:
τ = RC
Substituting the given values:
R = 447 ohms
C = 2.0 uF = 2.0 x 10^(-6) F
τ = (447 ohms) * (2.0 x 10^(-6) F) = 8.94 x 10^(-4) seconds
To find the time it takes for the capacitor to gain 80% of the charge it will eventually have, we need to multiply the time constant by a factor of ln(1/0.8) or approximately 0.2231.
time = τ * ln(1/0.8) = 8.94 x 10^(-4) seconds * ln(1/0.8) ≈ 6.11 x 10^(-4) seconds
So, the capacitor will take approximately 6.11 x 10^(-4) seconds to gain 80% of the charge it will eventually have when current stops flowing through the battery.