A fully charged 90-µF capacitor is discharged through a 60-Ω resistor. How long will it take for the capacitor to lose 80% of its initial energy?
I know energy is given by E(t)=q(t)^2, but beyond that I really don't know how to solve this. Any suggestions are appreciated.
Thank you!
Q(t)=Q₀•exp(-t/RC)
E~Q² =>
E=E₀•exp(-2•t/RC)
100-80=20%
0.2•E₀=E₀•exp(-2•t/R•C)
ln0.2 =- 2•t/R•C
t= - ln0.2• R•C/2 =- (-1.6) •60•90•10⁻⁶/2 =0.0043 s.
E = (1/2) c v^2
c = 90 * 10^-6
q = c v
so dq/dt = c dv/dt = -i
v = -i r = r c dv/dt
so
dv/dt = (-1/rc) v
v = Vi e^kt
dv/dt = k Vi e^kt = (-1/rc)Vi e^kt
k = -1/rc
v = Vi e^-t/rc
E = constant * v^2
so v^2 at t = .8 v^2 at t = 0
so
v at t = sqrt (.8) v at t = 0
v at t = .64 Vi
.64 = e^-(1/rc)t
ln .64 = -.446 = -(1/rc) t
t = .446 * 90 * 10^-6 * 60
sqrt .2 not .8
so
.04 = e^-(1/rc) t
-3.22 = -(1/rc) t
t = 3.22 * 90 * 10^-6 * 60
thank you!
To solve this problem, we need to use the formula for the energy stored in a capacitor, which is given by:
E(t) = 1/2 * C * V^2
Where E(t) is the energy stored in the capacitor at time t, C is the capacitance (in Farads), and V is the voltage across the capacitor.
Given that the capacitor is initially fully charged, we can assume that its initial voltage is V0 (let's say). Therefore, the initial energy stored in the capacitor is:
E0 = 1/2 * C * V0^2
We want to find the time it takes for the capacitor to lose 80% of its initial energy, which means the remaining energy at that time will be 20% of E0.
Let's start by finding the remaining energy:
E_remaining = 0.20 * E0
Now, let's substitute the formula for energy into this equation:
0.20 * E0 = 1/2 * C * V^2
Since we know that C = 90 µF, we can write:
0.20 * E0 = 1/2 * (90 × 10^(-6)) F * V^2
Now we need to find V. We can use the formula for the voltage across a charging or discharging capacitor in an RC circuit:
V(t) = V0 * e^(-t / RC)
Where V(t) is the voltage across the capacitor at time t, V0 is the initial voltage across the capacitor (which is equal to V0 since the capacitor is initially fully charged), e is Euler's number (approximately 2.71828), t is the time, R is the resistance (in ohms), and C is the capacitance (in Farads).
We want to find V when the remaining energy is 20% of E0, so we can substitute into our formula:
0.20 * E0 = 1/2 * (90 × 10^(-6)) F * (V0 * e^(-t / (60 Ω × 90 × 10^(-6) F)))^2
Now we have an equation that relates the initial energy (E0), the resistance (R), the capacitance (C), the initial voltage (V0), and the remaining energy (0.20 * E0).
To solve for time (t), we need to rearrange the equation and take the logarithm:
e^(-t / (60 Ω × 90 × 10^(-6) F)) = sqrt((0.20 * E0) / (1/2 * (90 × 10^(-6)) F))
Now, take the natural logarithm of both sides:
-ln(e^(-t / (60 Ω × 90 × 10^(-6) F))) = ln(sqrt((0.20 * E0) / (1/2 * (90 × 10^(-6)) F)))
Simplifying, we get:
-t / (60 Ω × 90 × 10^(-6) F) = ln(sqrt((0.20 * E0) / (1/2 * (90 × 10^(-6)) F)))
And rearranging for t:
t = -60 Ω × 90 × 10^(-6) F × ln(sqrt((0.20 * E0) / (1/2 * (90 × 10^(-6)) F)))
Now, plug in the given values for resistance (60 Ω), capacitance (90 µF), and adapt the equation to incorporate the given information about the initial energy:
t = -60 Ω × (90 × 10^(-6)) F × ln(sqrt((0.20 * (1/2 * (90 × 10^(-6)) F * V0^2)) / (1/2 * (90 × 10^(-6)) F)))
Finally, plug in the appropriate values for the variables and calculate t.