Question

Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3te−2t −2 A. Be certain to simplify your expression.

Answer 1

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208