The formula is: `Ri+L(di)/(dt)=V`
After substituting: `50i+(di)/(dt)=5`
We re-arrange to obtain:
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:
We have `P=50` and `Q=5`.
We find the integrating factor:
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"`.
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:
So in this case:
Graph of the current at time `t`, given by `i=0.1(1-e^(-50t))`.
In this example, the time constant, TC, is
So we see that the current has reached steady state by `t = 0.02 \times 5 = 0.1\ "s".`
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.
For the answer: Compute → Solve ODE... → Exact
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