Ans 7

Substituting R = 10, L = 3 and V = 50 gives:

First, we separate the variables.

Integrate.

Since i(0) = 0

$\qquad$

So, substituting for K:

$\qquad$

$-\dfrac{1}{10}\ln$

Put log parts together.

Multiply both sides by 10

$\dfrac{5-i}{5}$

$5e^{-10t/3}=5-i$

The graph shows that the current builds up and levels out at a maximum value of 5 A.

NOTE: Another way of solving this for i could have been:

$-\dfrac{1}{10}\ln$

$\ln$

Raising both sides as a power of e: