Question

The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.

Answer 1

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L$\frac{di(t)}{dt}$             -----------(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) = $10e^{-t/2}$A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = $(10*10^{-3})\frac{d(10e^{-t/2})}{dt}$

Solve the differential

v(t) = $(10*10^{-3})\frac{-1*10}{2} (e^{-t/2})$

v(t) = -0.05 $e^{-t/2}$

At t = 8s

v(t) = v(8) = -0.05 $e^{-8/2}$

v(t) = v(8) = -0.05 $e^{-4}$

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) = $10e^{-8/2}$

i(8) = $10e^{-4}$

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

Answer 2

Answer:

The voltage is - 0.9158 mV

The power is - 0.1677 mW

Explanation:

Given;

current through the inductor, i(t)  = $10e^{-t/2}$ -----equation (1)

inductance, L = 10 mH

given time, t  = 8 s

The voltage across the inductor is given by;

$V_L = L\frac{di}{dt} \\\\V_L = (10 *10^{-3})\frac{d}{dt} (10e^{-t/2})\\\\V_L = (10 *10^{-3})\frac{10}{-2}(e^{-t/2})\\\\ V_L = -0.05e^{-t/2} \ ----equation (2)$

when t = 8 s, the voltage will be ;

$V_L = -0.05 e^{-t/2}\\\\V_L = -0.05 e^{-8/2}\\\\V_L = -0.05 e^{-4}\\\\V_L = -9.158 *10^{-4} \ V\\\\V_L = -0.9158 \ mV$

Power is given by;

P = I V

When t = 8, the current "I" is given by;

$i(t) = 10e^{-t/2}\\\\i(8) = 10e^{-8/2}\\\\I = 10e^{-4}\\\\I = 0.1832 \ A$

P = 0.1832 x (-9.158 x 10⁻⁴)

P = -1.677 x 10⁻⁴ W

P = -0.1677 mW