How long will it take a charged 80-ìF capacitor to lose 20 percent of its initial energy when it is allowed to discharge through a 45 ohm resistor?
i mean 80 microfarads
Oh, you want to predict the future of an 80-ìF capacitor? Well, you know, I tried getting into the fortune-telling business once, but my crystal ball kept laughing at me. As for your question, let me do some quick calculations while trying not to short-circuit my circuits... Got it! Based on my calculations, it will take approximately a little longer than a squirrel's attention span, which is about...oh, forget it. It will take about 3 and a half time constants for the capacitor to lose 20 percent of its initial energy. The time constant of this particular circuit is roughly 0.002 seconds. So, grab some popcorn and enjoy the show!
To find out how long it will take the capacitor to lose 20 percent of its initial energy when discharging through a resistor, we can use the equation for the voltage across a discharging capacitor:
V(t) = V(0) * exp(-t / RC)
where:
V(t) is the voltage across the capacitor at time t,
V(0) is the initial voltage across the capacitor,
t is the time in seconds,
R is the resistance in ohms, and
C is the capacitance in farads.
In this case, we are given:
C = 80 μF = 80 * 10^(-6) F (convert from microfarads to farads),
R = 45 Ω,
and we want to determine the time it takes for the capacitor to lose 20 percent of its initial energy, which corresponds to the time when V(t) = 0.2 * V(0).
Now, let's rearrange the equation above to solve for t:
t = -RC * ln(V(t) / V(0))
Substituting the given values:
t = - (45 Ω) * (80 * 10^(-6) F) * ln(0.2)
Using a scientific calculator or computer program, we can calculate the value of t as follows:
t ≈ - (45) * (80 * 10^(-6)) * ln(0.2)
t ≈ - (45) * (80 * 10^(-6)) * (-1.6094) (taking the natural logarithm of 0.2 results in -1.6094)
t ≈ 5.4824 seconds
Therefore, it will take approximately 5.4824 seconds for the charged 80 μF capacitor to lose 20 percent of its initial energy when discharging through a 45 Ω resistor.
The time constant for charge decay is
RC = 3.6*10^-3 s, if your ì symbol means "micro"
For 80% of the initial energy, the charge must decrease to sqrt(0.8) = 89.4% of the initial value.
exp(-t/RC) = 0.894
t/RC = .1121
t = 4*10^-4 seconds