Answer :
Answer:
voltage = -0.01116V
power = -0.0249W
Explanation:
The voltage v(t) across an inductor is given by;
v(t) = L[tex]\frac{di(t)}{dt}[/tex] -----------(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) = [tex]10e^{-t/2}[/tex]A
L = 10mH = 10 x 10⁻³H
Substitute these values into equation (i) as follows;
v(t) = [tex](10*10^{-3})\frac{d(10e^{-t/2})}{dt}[/tex]
Solve the differential
v(t) = [tex](10*10^{-3})\frac{-1*10}{2} (e^{-t/2})[/tex]
v(t) = -0.05 [tex]e^{-t/2}[/tex]
At t = 8s
v(t) = v(8) = -0.05 [tex]e^{-8/2}[/tex]
v(t) = v(8) = -0.05 [tex]e^{-4}[/tex]
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) = [tex]10e^{-8/2}[/tex]
i(8) = [tex]10e^{-4}[/tex]
i(8) = 10 x 0.223
i(8) = 2.23
Therefore,
p(8) = 2.23 x -0.01116
p(8) = -0.0249W
Answer:
The voltage is - 0.9158 mV
The power is - 0.1677 mW
Explanation:
Given;
current through the inductor, i(t) = [tex]10e^{-t/2}[/tex] -----equation (1)
inductance, L = 10 mH
given time, t = 8 s
The voltage across the inductor is given by;
[tex]V_L = L\frac{di}{dt} \\\\V_L = (10 *10^{-3})\frac{d}{dt} (10e^{-t/2})\\\\V_L = (10 *10^{-3})\frac{10}{-2}(e^{-t/2})\\\\ V_L = -0.05e^{-t/2} \ ----equation (2)[/tex]
when t = 8 s, the voltage will be ;
[tex]V_L = -0.05 e^{-t/2}\\\\V_L = -0.05 e^{-8/2}\\\\V_L = -0.05 e^{-4}\\\\V_L = -9.158 *10^{-4} \ V\\\\V_L = -0.9158 \ mV[/tex]
Power is given by;
P = I V
When t = 8, the current "I" is given by;
[tex]i(t) = 10e^{-t/2}\\\\i(8) = 10e^{-8/2}\\\\I = 10e^{-4}\\\\I = 0.1832 \ A[/tex]
P = 0.1832 x (-9.158 x 10⁻⁴)
P = -1.677 x 10⁻⁴ W
P = -0.1677 mW