Discharging Capacitor
For R = 1000 W, C = 2000 mF, Qo= 20 mC.
(a)How long will it take before the capacitor discharges to Qo/100.
(b)What is the value of current at t = 4 sec.
I am completely lost on how to solve this question, please can anyone give me help in the right direction. Thanks in advance!
Do you read your text? Do you go to class and take notes? I am fairly certain you are not so helpless. Look in your text under RC decay.
I will be happy to critique your thinking, but honestly, I think you are answer grazing. This relation in RC circuits is pretty basic.
To solve this question, you need to use the equation that relates the charge on a capacitor to its voltage: Q = CV.
(a) To find out how long it will take for the capacitor to discharge to Qo/100, you can set up an equation using this relationship.
First, convert Qo/100 to coulombs by dividing Qo by 100: Q = (20 mC) / 100 = 0.2 mC = 0.2 × 10^-3 C.
Now, the exponential discharge equation for a capacitor is given by: Q(t) = Qo * e^(-t/RC), where Q(t) is the charge on the capacitor at time t, Qo is the initial charge, R is the resistance, C is the capacitance, and e is the base of the natural logarithm.
In this case, since we want to find the time it takes for the charge to drop to 0.2 × 10^-3 C, we can set up the following equation:
0.2 × 10^-3 C = (20 mC) * e^(-t/(1000 Ω * 2000 × 10^-6 F))
Simplifying the equation:
0.2 × 10^-3 C / 20 mC = e^(-t/(1000 Ω * 2000 × 10^-6 F))
0.01 = e^(-t/(1000 Ω * 2000 × 10^-6 F))
Now, we can solve for t by taking the natural logarithm (ln) of both sides:
ln(0.01) = -t/(1000 Ω * 2000 × 10^-6 F)
t = -ln(0.01) * (1000 Ω * 2000 × 10^-6 F)
Using a calculator:
t ≈ 13.815 seconds
Therefore, it will take approximately 13.815 seconds for the capacitor to discharge to Qo/100.
(b) To find the value of the current at t = 4 seconds, we can differentiate the charge equation Q(t) with respect to time to get the current since I = dQ/dt.
Differentiating Q(t) = Qo * e^(-t/RC):
dQ/dt = Qo * (-1/RC) * e^(-t/RC)
At t = 4 seconds, we can substitute these values into the equation:
I(4) = (20 mC) * (-1/(1000 Ω * 2000 × 10^-6 F)) * e^(-4/(1000 Ω * 2000 × 10^-6 F))
Calculating with the given values:
I(4) ≈ -0.01 A
Therefore, the value of the current at t = 4 seconds is approximately -0.01 A. Note that the negative sign indicates that the current is in the opposite direction as the initial charging direction.