Answer:
C = 0.3 F
Explanation:
The expression for a voltage in a capacitor with an initial voltage V₀, as a function of time, is given by the following equation:
When V = =.37*V₀, we have the following expression:
Applying ln to both sides, we have:
⇒ t = R*C
if t= 300 s, and R = 10³ Ω, we can solve for C as follows:
So, the required value for C is 0.3 F.
Answer:
voltage = -0.01116V
power = -0.0249W
Explanation:
The voltage v(t) across an inductor is given by;
v(t) = L -----------(i)
Where;
L = inductance of the inductor
i(t) = current through the inductor at a given time
t = time for the flow of current
From the question:
i(t) = A
L = 10mH = 10 x 10⁻³H
Substitute these values into equation (i) as follows;
v(t) =
Solve the differential
v(t) =
v(t) = -0.05
At t = 8s
v(t) = v(8) = -0.05
v(t) = v(8) = -0.05
v(t) = -0.05 x 0.223
v(t) = -0.01116V
(b) To get the power, we use the following relation:
p(t) = i(t) x v(t)
Power at t = 8
p(8) = i(8) x v(8)
i(8) = i(t = 8) =
i(8) =
i(8) = 10 x 0.223
i(8) = 2.23
Therefore,
p(8) = 2.23 x -0.01116
p(8) = -0.0249W