Question

What is the current I(3τ), that is, the current after three time constants have passed? The current in the circuit will approach a constant value Ic after a long time (as t tends to infinity). What is Ic?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I(\tau)=0.051 A$

b

$I(3 \tau)=0.076 A$

c

$I_c= 0.08 A$

Explanation:

From the question we are told that

                $I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

From the question we are told to find $I(\tau)$ when t=0  equals the time constant ($\tau$)

That is to obtain $I(\tau)$.This  is mathematically represented as

                   $I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{\tau}{\tau} })$

             Substituting 12 V for $\epsilon$ and 150Ω for R

                     $I(\tau) = \frac{12}{150} (1- e^{-1})$

                            $=0.051 A$

From the question we are told to find $I(3 \tau)$ when t=0  equals the 3 times the  time constant ($\tau$)

That is to obtain $I(3\tau)$.This  is mathematically represented as

                 $I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })$

                  $I(\tau) = \frac{12}{150} (1- e^{-3})$

                        $=0.076 A$

As tends to infinity $\frac{\infty}{\tau} = \infty$

So $I_c$ would be mathematically evaluated as

               $I_c=I(\infty) = \frac{12}{150} (1- e^{- \infty})$

                   $= \frac{12}{150}$

                   $= 0.08 A$