# The current in a circuit at time #t \ s# is given by the solution of the Differential Equation #(dI)/dt+4I=20#. Find the solution given that #I=2 \ A# when #t=0#, and find the time taken to reach half the steady state solution?

The solution is:

# I(t) = 5-3e^(-4t) #

The steady state value is given by

The units are not defined in the question

We have:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

Then the integrating factor is given by;

Which we can directly integrate to get:

Thus the solution is:

For the steady state solution we look at:

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The solution to the given differential equation is: ( I(t) = 5 - 3e^{-4t} ).

To find the time taken to reach half the steady-state solution, set ( I(t) = \frac{1}{2} \times 5 = \frac{5}{2} ) and solve for ( t ).

( \frac{5}{2} = 5 - 3e^{-4t} )

( e^{-4t} = \frac{5}{6} )

( -4t = \ln\left(\frac{5}{6}\right) )

( t = -\frac{1}{4} \ln\left(\frac{5}{6}\right) )

( t \approx 0.091 , \text{s} ) (rounded to three decimal places).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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