The voltage across the terminals of a 0.2 F capacitor is V= 150V when t <= 0 V=
ID: 1797251 • Letter: T
Question
The voltage across the terminals of a 0.2 F capacitor is
V= 150V when t <= 0
V= (A1)(t)(e-5000t)+(A2)(e-5000t) when t >= 0
The initial current in the capacitor is 250 mA. Assume the passive sign convention.
What is the initial energy stored in the capacitor?
This I correctly determined to be 150V * 250mA = 2.25mJ
Evaluate the coefficient A1
I am stuck on this question! Unsure how to solve. Answer will use units of V/s
Evaulate the coefficient A2
I detemined this by setting 150=(A1t+A2)e-5000t when t=0 and got a correct answer of 150V
What is the expression for the capacitor current? Express your answer in terms of t, where t is in seconds.
I am also stuck on this part simply because I cannot determine A1. Once I can I know that ic= C*(dV/dt) and I can simply plug in the values for A1 and A2
Any and all help is appreciated!! Please be clear in your answer, I will be awarding lifesaver to whomever can help me.
Explanation / Answer
Please ask if you have any doubt.I will help you.
The expression for current : Ic = CdV/dt
Now differentiate the expression V and get Ic
Ic = (0.2F)(e-5000t)(A1 - 5000A1t - 5000A2 )
substitute t = 0 and Ic = 0.25 A.
0.25 = (0.2)(A1) - (0.2)(5000)(150) .
A1 = (0.25 + 0.15)/.2 = 2(106) V/s.
Now the current expression will be :
Ic = (e-5000t)(0.4 - 1000t - 0.15)
Ic(t) = (e-5000t)(0.25 - 1000t) A
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