tuned circuits assignment

Discussion in 'Homework Help' started by ninjaman, Feb 4, 2015.

May 18, 2013
306
1
hello

im having a hard time with my tunded circuits assignment. my tutor has decided to copy an assignment from another tutor who did this assignment last year, all the same questions. that tutor has left and now my tutor is using a copy of a previous students coursework to find the answers as he is unsure about how to complete the assignment. i have been struggling with this course since starting and thought that i would like to get ahead and study when im not at college. i handed in my coursework last week thinking it was ok. now i find i have missed things out. my tutor didn't really do a good job of explaining it to me and im a little agitated by it. i would like to lodge a complaint about him but i think that may make things worse.
i have three branches.

i have to calculate the phase angle
then i have to find the j notation
im not sure how to find the phase angle
for the first branch (cap 1uf, res 5600) i got the impedance z sqrt R^2+Xc^2,
so 5600^2+(-318.3^2) = 5590.94671

then from what my tutor has told me, (using another students coursework!)
i have to use voltage over impedance, so im using 7.07vrms over 5590.94761 to get the 1.26mA in branch one. so this is correct so far.
then i have to find the phase angle, he showed me the previous students coursework(though he admitted that some of it was wrong)
the formula for phase angle was cos^-1 R/Z which i did , i cant get an answer out of my calculator. i asked for help and by this point he was talking to me as though i should know how to do this. which i didn't really find helpful.
any help or a push in the right direction would be great

thanks
simon

2. crutschow Expert

Mar 14, 2008
13,509
3,385
Here's how I would do it:
Calculate the real and reactive currents in each circuit branch (rectangular notation).
Add these currents together (with the proper reactive polarity) to get the total real and reactive current.
Calculate the phase angle from that.

3. WBahn Moderator

Mar 31, 2012
18,089
4,917
Step 1: Go to a real school. I mean that. It sounds like the school you are going to is NOT going to get you anything resembling a decent education and that there is a good chance you will end up with a either a huge student loan debt or a big hole in your savings in exchange for a degree that may actually hinder your chances of getting employed -- I've seen it happen where students racked up over \$60k in debt to get a degree only to find that having that school's name on their resume actually discouraged potential employers from even granting them an interview. So if this is the kind of instruction you are getting, take some time to think through whether it is worth it.

Step 2: Recognize that all three branches are independent since they are in parallel and being driven by an ideal voltage source. That makes things a lot simpler because you can tackle things one at a time.

Step 3: Work with "j notation" from the start. Write down the impedance of each component and then calculate the current for each branch. As crutschow suggested, keep the result in rectangular form. Then add the currents for all three branches to get the total. The convert the total to polar form. You know have the phase angle for the current and the phase angle for the voltage (which appears to be zero based on your screen capture), from which you can find the phase angle between them.

4. shteii01 AAC Fanatic!

Feb 19, 2010
3,515
515
I have a question. Why is 318.3 negative?

5. WBahn Moderator

Mar 31, 2012
18,089
4,917
capacitive reactance is (or should be) negative, while inductive reactance is positive.

The alternative is to have reactance always be positive but then to have to forever track and chase down what the flavor of the reactance is (capacitive or inductive) and use formulas that have things like (XL - XC) scattered throughout them. The downside of this becomes apparent as soon as you start working with systems where these are unknowns and you don't have specific numbers to throw at them. At that point you don't know if the combination of elements is capacitive or inductive and so you get stuck an a complicated web of "if this is bigger than that then use this formula" rules.

6. shteii01 AAC Fanatic!

Feb 19, 2010
3,515
515
I finally figured out your problem here.

You have two sides of right triangle, first side is 5600, second side is 318.3. z is the hypotenuse of the right triangle. What do we know about hypotenuse? We know that hypotenuse is longer than any one side. Now look at your numbers! You hypotenuse (z) is actually smaller than you largest side (5600). How the eF did you manage that?
That is why you inverse cosine does not work.

5600^2+(-318.3)^2=31360000+101314.89=31461314.89
sqrt(31461314.89)=5609.039
5600/5609.039=0.99839
cos^-1(0.99839)=3.253°