Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]I(\tau)=0.051 A[/tex]
b
[tex]I(3 \tau)=0.076 A[/tex]
c
[tex]I_c= 0.08 A[/tex]
Explanation:
From the question we are told that
[tex]I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R[/tex]
From the question we are told to find [tex]I(\tau)[/tex] when t=0 equals the time constant ([tex]\tau[/tex])
That is to obtain [tex]I(\tau)[/tex].This is mathematically represented as
[tex]I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{\tau}{\tau} })[/tex]
Substituting 12 V for [tex]\epsilon[/tex] and 150Ω for R
[tex]I(\tau) = \frac{12}{150} (1- e^{-1})[/tex]
[tex]=0.051 A[/tex]
From the question we are told to find [tex]I(3 \tau)[/tex] when t=0 equals the 3 times the time constant ([tex]\tau[/tex])
That is to obtain [tex]I(3\tau)[/tex].This is mathematically represented as
[tex]I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })[/tex]
[tex]I(\tau) = \frac{12}{150} (1- e^{-3})[/tex]
[tex]=0.076 A[/tex]
As tends to infinity [tex]\frac{\infty}{\tau} = \infty[/tex]
So [tex]I_c[/tex] would be mathematically evaluated as
[tex]I_c=I(\infty) = \frac{12}{150} (1- e^{- \infty})[/tex]
[tex]= \frac{12}{150}[/tex]
[tex]= 0.08 A[/tex]
