The first part of the question is not complete and it is;
The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t
Answer:
A) initial energy = 0.3125 mJ
B) A1 = 50 and A2 = 1,800,000
C) Capacitor Current is given by the expression;
I = e^(-4000t)[0.95 - 1800t]
Explanation:
A) In capacitors, Energy stored is given as;
U = (1/2)Cv²
Where C is capacitance and v is voltage.
So initial kinetic energy;
U(0) = (1/2)C(vo)²
From the question, C = 250 nF and v = 50V
So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ
B) from the question, we know that;
A1e^(-4000t) + (A2)te^(-4000t)
So, v(0) = A1e^(0) + A2(0)e^(0)
v(0) = 50
Thus;
50 = A1
Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;
And so;
dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)
Simplifying this, we obtain;
dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]
Current (I) = C(dv/dt)
I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]
Thus, Initial current (Io) is;
Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]
We know that Io = 400mA from the question or 0.4 A
Thus;
0.4 = (250 x 10^(-9))[-4000A1 + A2]
0.4 = 0.001A1 - (250 x 10^(-9)A2)
Substituting the value of A1 = 50V;
0.4 = 0.001(50) - (250 x 10^(-9)A2)
0.4 = 0.05 - (250 x 10^(-9)A2)
Thus, making A2 the subject, we obtain;
(0.4 + 0.05)/(250 x 10^(-9))= A2
A2 = 1,800,000
C) We have derived that ;
I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]
So putting values of A1 = 50 and A2 = 1,800,000 we obtain;
I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]
I = e^(-4000t)[0.05 + 0.45 - 1800t]
I = e^(-4000t)[0.95 - 1800t]
