Answer:
A) V(t) = 0.4e^-2t
B) i(t) = (25tsin5t+10) A for t>0
Step-by-step explanation:
Formula for calculating voltage across an inductor is expressed as:
V = Ldi/dt
Given L = 100mH = 100×10^-3
If i(t) = 6 - 2e^-2t A t >= 0
di/dt = (-2)(-2)e^-2t
di/dt = 4e^-2t
If t ≥ 0
V(t) = 100×10^-3 × (4e^-2t)
V(t) = 0.1×4e^-2t
V(t) = 0.4e^-2t for t≥0
B) Applying the same formula as above
V = Ldi/dt
Vdt = Ldi
V/L dt = di
On integration
Vt/L = i + C
When t = 0, i = -10A
Substituting the values into the formula
V(0)/L = -10 + C
0 = -10+C
C = 10
To get the current i(t) through the inductor for t>0,
Since Vt/L = i + C
Given V(t) = 5sin5t Volts
L = 200mH = 200×10^-3H
C = 10
On substituting
(5sin5t)t/0.2 = i + 10
25tsin5t = i + 10
i(t) = (25tsin5t-10) A for t>0