# Math Help - Electrical Systems Paper (Mainly Trig)

1. ## Electrical Systems Paper (Mainly Trig)

Hi guys, thanks for the help given with the last paper I was looking for help with, I think I just may have past that one, however next comes the hard paper, this one is Electrical Systems and there is nothing "Electrical" about it in my view, the entire paper is math theory and the majority of it is trigonometry (with which i am increadibly uncomfortable).

I will start from the beginning with Q2 and Q3 of the paper. (Q1 was simple) any help on what i need to look for and how to manipulate the questions for the answer would be much appreciated, thanks.

Q2:

A circuit breaker is designed to stop fault current flowing in a circuit at a point in time where the current in the circuit is 0.

If the fault curent in the circuit is: $i(t) = 50cos(314t - 60^\circ)$ Amperes

i) At what time would the first current 0 in the circuit occur?

ii) Given the circuit breaker will take 40mS to operate, at what time would the next current 0 occur after that?

For the first part I am assuming i should make i(t) = 0 so:

$50cos(314t - 60^\circ) = 0$

Now since I am unsure how one should correctly manipule the cos trig function in this case, i would do the following ... is this correct?:

$314t-60^\circ = 50cos(0)$

or should i first have divided both sides by 50 to cancel the 50 out?

2. Originally Posted by Alias_NeO
...
Q2:

A circuit breaker is designed to stop fault current flowing in a circuit at a point in time where the current in the circuit is 0.

If the fault curent in the circuit is: $i(t) = 50cos(314t - 60^\circ)$ Amperes

i) At what time would the first current 0 in the circuit occur?

ii) Given the circuit breaker will take 40mS to operate, at what time would the next current 0 occur after that?

For the first part I am assuming i should make i(t) = 0 so:

$50cos(314t - 60^\circ) = 0$

Now since I am unsure how one should correctly manipule the cos trig function in this case, i would do the following ... is this correct?:

...
Hello,

you start 100% correctly but you should go on maybe like this. (First a personal remark: I assume that the 314 is a degree value too - and that is quite a unusual notation)

$50cos(314t - 60^\circ) = 0$ Divide by 50

$cos(314t - 60^\circ) = 0$ . Now you should know that the first (positive) zero of the cosine function is at α = 90°. Use the arccos function (or maybe you say the cos^(-1) function or the inverse function of the cosine):

$314^\circ \cdot t -60^\circ = 90^\circ~\Longrightarrow~ 314^\circ \cdot t = 150^\circ~\Longrightarrow~t\approx 0.4777\ s$

to ii)

The next positive zero occurs at 270°. Your equation becomes now:

$314^\circ \cdot t -60^\circ = 270^\circ~\Longrightarrow~ 314^\circ \cdot t = 330^\circ~\Longrightarrow~t\approx 1.0560\ s$

The first operation took place at 0.4777 s. The operation time is 40 ms = 0.04 s. The next operation takes place at 1.0560 s. The elapsed time is:

1.0560 s - (0.4777 s + 0.04 s) = 0.5383 s

3. ## Thanks

Hey, thanks for the reply, unfortunately I sat the paper yesterday before i got the reply and failed it miserably

I think it was unfair because he (the lecturer - who is hated by students and staff alike) gave us no study material, no past papers or anything, then to make it worse, he totally changed the structure of this exam which hasn't been done by any of the lecturers in a long time.

Not to worry, what's done is done.