# Anyone can check if these homework questions are correct

Discussion in 'Recreational Math' started by BIGEYE, Dec 6, 2004.

1. ### BIGEYEGuest

I have a couple of homework questions, appreciate if someone can take the
time to check my answers.
Q1 Given that voltages v1 = 12 sin 100 pi t and v2 = 20 sin ( 100 pi t +
(pi / 3) ). State the minimum value and the phase angle (relative to v1) of
the resultant voltage v1 + v2 by writing the sum as a single sinusoid.

My answer 28 sin ( 100 pi t + 0.667 )

Q2 In an ac circuit i = 100 sin 20 pi t amperes and v = 50 sin (20 pi t -
( pi / 6 )) volts

Instantaneous power p, is given by p = vi watts

Use "products-to-sums" formulae to express p in a form involving only one
sinusoid and hence state the maximum power.

My answer 2165 - 2500 cos ( 20 pi t - (pi / 6) )

TIA

BIGEYE, Dec 6, 2004

2. ### Guess whoGuest

It is always possible to get a right answer by wrong methods. Can you
show your work on how you came by those answers?

Guess who, Dec 6, 2004

3. ### BIGEYEGuest

OK Q1
sin(A+B) = sinAcosB+cosAsinB
to make it easier to write out I made 100pi = w
20sin(wt + (pi/3))
20 sin wt cos (pi/3) + 20 cos wt sin (pi/3)
= 20 sin wt * 0.5 + 20 cos wt * 0.866
= 10 sin wt + 18.32 cos wt
=> 12 sin wt + 10 sin wt + 18.32 cos wt
= 22 sin wt + 17.32 cos wt
I then put into R = wt + a (where a =alpha)
R^2 - SQRT 22^2 + 17.32^2
R = 28
tan a = 17.32/22
a = 0.667 radians
therefore v1 + v2 = 28 sin (wt + 0.667)
substituting 100pi back
v1 + v2 = 28 sin (100 pi t + 0.667)

Q2
w = 20 pi
50 sin (wt - (pi/6)) * 100 sin wt
= 5000 sin (wt - (pi/6)) sin wt
sinA sinB = 0.5[cos (A - B) - cos (A + B)]
=> 5000 * 0.5[ cos (wt - (pi/6) - wt) - cos (wt - (pi/6) + wt)]
= 2500[cos ( -pi/6) - cos (2wt - (pi/6))]
= 2500[0.866 - cos (2wt - (pi/6))]
= 2165 - 2500 cos (2wt - (pi/6))
Substituting 20 pi back for w
2165 - 2500 cos (20 pi t - (pi/6))

BIGEYE, Dec 6, 2004
4. ### BIGEYEGuest

In Q1
R^2 - SQRT 22^2 + 17.32^2