(a) We solve it using the formula:

`i=V/R(1-e^(-(R"/"L)t))`

We have:

`i=100/50(1-e^(-5t))`

`=2(1-e^(-5t))`

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Graph of the current at time `t`, given by `i=2(1-e^(-5t))`.

(b) At `t=0.5,`

`i=[2(1-e^(-5t))]_(t=0.5)=1.8358`

(c) VR and VL are given by:

`V_R=iR `

`=2(1-e^(-5t))xx50 `

`=100(1-e^(-5t)) `

`V_L=L(di)/(dt) `

`=10d/(dt)2(1-e^(-5t)) `

`=100e^(-5t)`

(d) To find the required time, we need to solve when `V_R=V_L`.

`V_R=V_L` when

`100(1-e^(-5t))=100e^(-5t)`

`1-e^(-5t)=e^(-5t)`

`2e^(-5t)=1`

` e^(-5t)=0.5`

`-5t=ln\ 0.5=-0.69315`

So

`t=(-0.69315)/(-5)=0.13863s`

Substituting this value into VR gives:

`V_R=V_L` `=[100e^(-5t)]_(t=0.13863)` `=50.000\ "V"`

The graph of VR and VL is as follows:

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Graph of the voltages `V_R=100(1-e^(-5t))` (in green), and `V_L=100e^(-5t)` (in gray).

The time constant, TC, for this example is:

`tau=L/R=10/50=0.2`

NOTE (just for interest and comparison): If we could not use the formula in (a), and we did not use separation of variables, we could recognise that the DE is 1st order linear and so we could solve it using an integrating factor.

We use the formula:

`Ri+L(di)/(dt)=V`

The required DE is:

`10(di)/(dt)+50i=100`

`(di)/(dt)+5i=10`

`"I.F."=e^(int5dt)=e^(5t)`

`ie^(5t)=10inte^(5t)dt=` `10/5e^(5t)+K=` `2e^(5t)+K`

Since `i(0)=0`, we have `K=-2`.

So `i=2(1-e^(-5t))`

It works :-)

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