Discussion in 'Homework Help' started by zaidqais, Mar 6, 2009.

1. ### zaidqais Thread Starter Member

Feb 14, 2009
16
0
Hi every body

I am stuck on question from the book Introductry circuits Analysis

I tried to do the following

To find T1 I need to count the time needed v to be 0

So my solution is :

v = 200 Sin(2π1000t+60)

0= 200 Sin(360000+60)

Sin(360000t+60) = 0

Sin־1 (0) = 360000t+60

t = 60/360000 = 0.16666 ms

But the book I get the question from said that this answer is 1/3 ms ,, can you please tell me where is the mistake in my solution

Thank you

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Haven't used the TI but I suspect the problem is the mixing of radians and degrees in the equation. If the calculator can be set in radian mode try converting the 60° to radians first and then solve as you have already done. Hope that is the solution!

3. ### Ratch New Member

Mar 20, 2007
1,068
3
zaidqais,

w*t = 2*pi*f*t= 2*pi*1000*t ===> f = 1000 Hz ===> P = 1 ms

They want the time of (120/360) period = 1/3 period which is 1/3 ms = 0.333 ms.

Ratch

Last edited: Mar 7, 2009
4. ### zaidqais Thread Starter Member

Feb 14, 2009
16
0

I have a question
in the solution i put it in my first post what is the mistake ? I suppose it must calculate T1 or not !!

and in the equation
Code ( (Unknown Language)):
1.
2. w*t = 2*pi*f*t= 2*pi*1000*t ===> f = 1000 kHz ===> P = 1 ms
3.
Pi is 180 degree or 3.14 ?
note f = 1000 Hz and not 1000 kHz

Thank you

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782

keeping in mind that sin(x) = 0 at x=0 , pi [180°], 2 x pi [360°], etc

solve
v = 200 Sin(2π1000t+60) = 0 as you did!
Consider the cases where
Sin(2π1000t+60)= 0 , 180°, etc

case 1:
200 Sin(2π1000t+60) = 0
2π1000t+60 = 0 (strictly 2π1000t + π/3 = 0)
360000t = -60 (using degrees)
t = - 0.1667 ms - wrong solution as t is less than zero

case 2:
360000t=180-60
t = (180-60)/360000
t = 0.333 ms (1/3 ms exactly) - first positive time at which function is zero - the required solution. t1

Feb 14, 2009
16
0
You rock

7. ### Ratch New Member

Mar 20, 2007
1,068
3
t_n_k,

Why solve for 200 Sin(2π1000t+60) = 0 ? That was 1/6 ms previously. That is not what the problem asked.

2π1000t + π/3 = 0 , absolutely, no other way. When you introduce time into trig functions, then all frequencies and phase degrees have to be converted to radians. 2n1000t = -n/3 =====> t = -1/6 ms. Isn't that when the function was zero previously?

You have to use radians with time. 2n1000t + n/3 = n ====> 2n1000t = 2n/3 ====> t = 1/3 ms, which is when the next zero value will occur.

Arithmetic wrong, 2n1000t = 180-60 = 120 =====> t = 0.019099, which is off base because radians were not used.

Ratch

Last edited: Mar 7, 2009
8. ### Ratch New Member

Mar 20, 2007
1,068
3
t_n_k,

Upon further reflection, I realize you were right and I was wrong. I did not realize that you were converting radians/sec into Hz/sec. Doing it that way you can work exclusively with Hz instead of radians. Sorry for the mixup. At least we both got the same answers.

Ratch

9. ### zaidqais Thread Starter Member

Feb 14, 2009
16
0
Really , you helped me so much

Thank you very much