Answer:
a. Voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)
b. Power = -59.3μW
c. Inductor is delivering power.
d. Energy = 5934.3nJ
e. Time = 100ms; Energy = 1095941.025nJ
Explanation:
Given
Current; iL=18te^(−10t)A or t≥0.
L.= inductor = 50μH
a. The voltage, V across the inductor for t>0 is calculated as follows;
V = L(di/dt)
Where L = 50μH
di/dt = 18(e^-10t + (-10)te^-10t)
di/dt = 18e^-10t(1 - 10t)
Substitute 50μH for L and 18e^-10t(1 - 10t) for di/dt in V = L(di/dt)
V = 50μH * 18e^-10t(1 - 10t)
V = 50 * 10^-6(18e^-10t(1 - 10t))
V = 0.9e^-10t(1-10t)
Hence, the voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)
b. Find the power (in microwatts) at the terminals of the inductor when t=200 ms.
Given that t = 200ms = 200 * 10^-3s = 0.2s
Power, p is calculated using the following formula;
p = Li(di/dt)
p = 50 * 10^-6(18te^-10t)18e^-10t(1-10t)
p = 50 * 10^-6 * (18 * 0.2 * e^-(10*0.2)) * (18 * e^(-10 * 0.2) * (1-10*0.2)
p = -5.93E5W
p = -59.3μW
c. Is the inductor absorbing or delivering power at 200 ms?
Because of the negative sign, the inductor is delivering power.
d. Find the energy (in microjoules) stored in the inductor at 200 ms.
Energy is calculated as ½Li²
= ½ * 50 * 10^-6 * (18te^-10t)²
= ½ * 50 * 10^-6 * (18 * 0.2 * e ^ (-10 * 0.2))²
= 0.0000059342669999498J
= 5934.3nJ
e. Find the maximum energy (in microjoules) stored in the inductor and the time (in milliseconds) when it occurs.
Calculating the derivation in (a)
di/dt = 0
18e^-10t(1-10t) = 0
1 - 10t = 0
-10t = -1
t = 1/10
t = 100ms
To calculate the energy, first we need to calculate the current
I(t=100) = 18 * 0.1 * e^(-10(0.1)
I = 0.662182994108596
I = 6621.82mA
The energy is calculated as follows;
w = ½ * 50 * 10^-6 * (6.621)²
w = 0.001095941025
w = 1095941.025nJ
