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
The solution is:
# I(t) = 5-3e^(-4t) #
The steady state value is given by
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,
# (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,
# 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
# 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)