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?

Answer 1

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

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)
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

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).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7