1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus problem (electric circuit)

  1. Sep 6, 2004 #1
    A electric circuit is made of a resistance, a solenoid and a power source.
    Resistance, V= Ri
    Solenoid, V = L (di/dt)
    Power source, tension E(t)

    Hence, L(di/dt)+ Ri = E(t). If E(t)= 20sin(5t), if L= 4 henrys and R= 20 ohms. If the intensity equals 0 at the beginning, find the amperage of i(t). Show that after a certain amount of time, the electric current becomes stationary.

    I would really need help to start this problem, I'm completly lost :confused: . i was stuck at 20sin(5t)= 4(di/dt) + 20i... The answer is supposed to be
    i(t)= 1/2( sin(5t)-cos(5t) +e^-5t)
     
  2. jcsd
  3. Sep 6, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If 4 di/dt+ 20i= 20 sin(5t), then di/dt+ 5i= 5sin(5t). Since this is a "linear differential equation", the first thing I would do is look at the related homogeneous equation, di/dt+ 5i=0 which is the same as di/dt= -5i or
    (1/i) di= -5dt. Integrating both sides, ln(i)= -5t+ C or i= C' e-5t.

    We look for a solution to the entire equation of the form
    i(t)= Asin(5t)+ Bcos(5t). Then i'(t)= 5Acos(5t)- Bsin(5t). Putting those into the equation i'+ 5i= 5 sin(5t), we have 5Acos(5t)- 5Bsin(5t)+ 5Asin(5t)+5Bcos(5t)= 5sin(5t) so we must have 5A+ 5B= 0 and -5B+ 5A= 5. Adding those two equations, 10A= 5 so A= (1/2) and B= -(1/2).

    Putting those together, the general solution is i(t)= C'e-5t+ (1/2)sin(5t)- (1/2)cos(5t). i(0)= C'+ 1/2= 0 so C'= -1/2.
    That is, i(t)= (1/2)[sin(5t)- cos(5t)- e-5t].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculus problem (electric circuit)
Loading...