#### Method 1 - Solving the DE

The formula is: Ri+L(di)/(dt)=V

After substituting: 50i+(di)/(dt)=5

We re-arrange to obtain:

(di)/(dt)+50i=5

This is a first order linear differential equation.

We'll need to apply the formula for solving a first-order DE (see Linear DEs of Order 1), which for these variables will be:

ie^(intPdt)=int(Qe^(intPdt))dt

We have P=50 and Q=5.

We find the integrating factor:

"I.F."=e^(int50dt)=e^(50t)

So after substituting into the formula, we have:

(i)(e^(50t))=int(5)e^(50t)dt =5/50e^(50t)+K =1/10e^(50t)+K

When t=0, i=0, so K=-1/10=-0.1.

This gives us: i=0.1(1-e^(-50t))

The transient current is: i=0.1(1-e^(-50t))\ "A".

The steady state current is: i=0.1\ "A".

#### Method 2: Using the Formula

NOTE: We can use this formula here only because the voltage is constant. This formula will not work with a variable voltage source.

We have the following general formula:

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

So in this case:

i=5/50(1-e^(-50t)) =0.1(1-e^(-50t))

Open image in a new page

Graph of the current at time t, given by i=0.1(1-e^(-50t)).

In this example, the time constant, TC, is

tau=L/R=1/50=0.02

So we see that the current has reached steady state by t = 0.02 \times 5 = 0.1\ "s".

#### Method 3: Using Scientific Notebook's Solve ODE

If you have Scientific Notebook, proceed as follows:

This DE has an initial condition i(0) = 0. We set up a matrix with 1 column, 2 rows.