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?

1 Answer
May 4, 2017

The solution is:

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

The steady state value is given by #I=5#, and it takes #0.046 # (time units) (2sf) to reach half this value.

The units are not defined in the question

Explanation:

We have:

# (dI)/dt+4I=20 \ \ \ \ ...... [1]#

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

# dy/dx + P(x)y=Q(x) #

Then the integrating factor is given by;

# IF = e^(int P(x) dx) #
# " " = exp(int \ 4 \ dt) #
# " " = exp( 4t ) #
# " " = e^(4t) #

And if we multiply the DE [1] by this Integrating Factor, #IF#, we will have a perfect product differential;

# (dI)/dt+4I=20 #

# :. e^(4t)(dI)/dt+4I e^(4t)=20e^(4t) #

# :. d/dt {e^(4t) I} = 20e^(4t) #

Which we can directly integrate to get:

# int \ d/dt {e^(4t) I} \ dt = int \ 20e^(4t) \ dt#
# :. e^(4t) I = int \ 20e^(4t) \ dt#
# :. e^(4t) I = 5e^(4t) + C#
# :. I = 5 + Ce^(-4t)#

Applying the initial condition, #I(0)=2#, we get:

# 2 = 5 + Ce^(0) => C=-3#

Thus the solution is:

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

For the steady state solution we look at:

# lim_(t rarr oo) I(t) = lim_(t rarr oo) {5-3e^(-4t)} #
# " " = lim_(t rarr oo) 5-3lim_(t rarr oo)e^(-4t) #
# " " = 5-3lim_(t rarr oo)e^(-4t) #
# " " = 5 #

So to find the time taken to reach half the steady state solution, we require the value of #t# such that:

# I(t)=5/2 => 5-3e^(-4t) = 5/2 #

# :. 3e^(-4t) = 5/2 #
# :. e^(-4t) = 5/6 #
# :. -4t = ln(5/6) #

# :. t = -1/4ln(5/6) #
# " " = -1/4ln(5/6) #
# " " = 0.045580 ... #
# " " = 0.046 # (2sf)