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

`10i+3(di)/(dt)=50`

`3(di)/(dt)=50-10i`

First, we separate the variables.

`(di)/((50-10i))=(dt)/3`

Integrate.

`1/10int(di)/(5-i)=1/3intdt`

`-1/10ln(5-i)=t/3+K`

Since i(0) = 0,

`-1/10ln(5-0)=0+K`

`K=(-ln\ 5)/10`

So, substituting for K:

`-1/10ln(5-i)=t/3-(ln\ 5)/10`

Put log parts together.

`-t/3=1/10ln(5-i)-(ln\ 5)/10`

Multiply both sides by 10

`-(10t)/3=ln(5-i)-ln\ 5`

`-(10t)/3=ln\ (5-i)/5`

`e^(-10t"/"3)=(5-i)/5`

`5e^(-10t"/"3)=5-i`

`i=5-5e^(-10t"/"3)=5(1-e^(-10t"/"3))`

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

1234123456-1tiOpen image in a new page

Solution graph `i=5(1-e^(-10t"/"3))`.

NOTE: We could have solved this for `i` another way. Here it is - you may find it easier.

`-1/10ln(5-i)=t/3-(ln\ 5)/10`

`ln(5-i)=-(10t)/3+ln\ 5`

Raising both sides as a power of e:

`5-i=e^(-10t"/"3 + ln\ 5)` `=e^(-10t"/"3)e^(ln\ 5)` `=5e^(-10t"/"3)`

`i=5-5e^(-10t"/"3)` `=5(1-e^(-10t"/"3))`

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