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Homework Help: Calculating time from two voltages

  1. Oct 12, 2013 #1
    I've gotten this for homework from my Electrotechnics class

    1. The problem statement, all variables and given/known data

    You are given two voltages: u1=50 sin(314t + ∏/3) and u2=100 sin(314t - ∏/6). You need to calculate the phase difference (pretty easy, already did that it's 90 degrees) and time when the two voltages have the same value. The solution in the book says 3,16ms.

    2. Relevant equations

    General formula for calculating the voltage is u = Um sin(ωt +- θ).

    3. The attempt at a solution

    So, because it says that I need to find the time both voltages are the same value, u1 must be equal to u2. So, u1=u2 50 sin(314t + ∏/3)=100 sin(314t - ∏/6). Now the problem is how do I get the 't' out of the sine and aren't they going to cancel each other out because they're same?

    Please ask me anything else you need to solve this problem. THank you.
  2. jcsd
  3. Oct 12, 2013 #2
    Hi Spreco! Welcome to PF!

    What cancels what? You have 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) or sin(314t + ∏/3)=2sin(314t - ∏/6). You cannot solve this or is it something else that troubles you?
  4. Oct 12, 2013 #3
    Hi Spreco

    Welcome to Physicsforums!!!

    Rewrite u1=50 sin(314t + ∏/3) such that instead of sine ,you get a cosine .I mean convert sine into cosine.

    What do you get ?
    Last edited: Oct 12, 2013
  5. Oct 12, 2013 #4
    Hi Spreco, do this: u1=u2, 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) => sin(Pi t + Pi/3)=2 sin(Pi t - Pi/6),
    sin(a+b) = sin a cos b + cos a sin b, then:
    sin(Pi t) cos(Pi/3) + cos(Pi t) sin(Pi/3) = 2 sin(Pi t) cos(Pi/6) - 2 cos(Pi t) sin(Pi/6)
    sin(Pi t)= a, cos(Pi t)= b,
    sqrt(3) / 2 a + 1/2 b = sqrt(3) a - b
    -1/2 a = 3/2 b
    -a = 3 b
    sin x = -3 cos x, now specify any value of cos x s.t. 3 cos x<1, which means cos x< 1/3, -70.5< x < 70.5°, or 109< x <251 and then the equation has a solution. In particular you find the values of t from x= 180 * t. The
  6. Oct 17, 2013 #5
    The task was to find the time, and after some thinking and ideas I solved it and got the solution. Thanks for all the help and ideas you gave me.
    http://i.imgur.com/EwR4GOp.jpg (The picture is high resolution, you can zoom in really close to see the details)
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