×
Follow Us
Results 1 to 10 of 10

Determine total reactive resistance of this circuit

 Jump to latest post
    #1
  1. Join Date
    Mar 2018
    Posts
    18
    Reputation

    Determine total reactive resistance of this circuit

    This question appeared on the NETA III exam.

    Find the total reactance in this circuit. I distinctly remember it because it was the very first question to hit me and I said "Gee thanks alot".

    I wasn't quite ready for it.

    Click image for larger version. 

Name:	NETA Schematic.jpg 
Views:	16 
Size:	23.9 KB 
ID:	384

  2. #2
  3. rofo42 is offline Member Pro Subscriber
    Join Date
    Apr 2018
    Posts
    26
    Reputation
    Quote Originally Posted by whiskers View Post
    This question appeared on the NETA III exam.

    Find the total reactance in this circuit. I distinctly remember it because it was the very first question to hit me and I said "Gee thanks alot".

    I wasn't quite ready for it.

    Click image for larger version. 

Name:	NETA Schematic.jpg 
Views:	16 
Size:	23.9 KB 
ID:	384
    I'll take a stab at this...

    Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
    Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

    Capacitive Reactance (Xc) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
    Xc = 76

    Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
    Inductors in series are added together. 500 mH + 125 mH = 625 mH

    Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
    Xl = 236

    Xc + Xl (76 + 236) = 312 ohms

    I would be surprised if I did that correctly. Someone correct me if needed!

  4. #3
  5. Join Date
    Sep 2015
    Posts
    1
    Reputation
    Quote Originally Posted by rofo42 View Post
    Xc + Xl (76 + 236) = 312 ohms

    I would be surprised if I did that correctly. Someone correct me if needed!
    I think you're correct up to this point.

    Looking at the equation sheet, it should be XL - XC

    So doing that I get 160 ohms.

    I'd be willing to bet that both 160 and 312 are possible answers on the test.

  6. #4
  7. Warrengarber's Avatar
    Warrengarber is offline
    If you ignore it, it will fail.
    Seasoned Member Pro Subscriber
    Join Date
    Jul 2017
    Location
    Jacksonville, FL
    Posts
    57
    Reputation
    Your math all works out to me. A time saver when performing this and from my experience with similar questions on the test is that they will use two identical, in this case capacitors in series and reactors in parallel, values. When this is the case the calculated value is half of one of the identical capacitors.

    Capacitor Example: Look at Rofo42's capacitors 20ufx20uf=400uf over the sum of 20uf+20uf= 40uf, 400uf/40uf=10uf or half of one of the 20uf capacitors. (Capacitors in series)

    Inductor Example: 1/250mH+1/250mH=0.004mH+0.004mH=0.008mH then 1/0.008 with a result of 125mH or half of one of the two parallel 250mH reactors (Reactors in Parallel)

    It is all about the time it takes to answer the questions on the test. Remembering these shortcuts during the stress of taking the test and knowing that that one wrong answer may be the make or break is not easy. A question like this could take 5 minutes to answer when doing all of the math but when you only have an average of 72 seconds to answer a question shortcuts count. I hope that helps and great job Rofo42



    Quote Originally Posted by rofo42 View Post
    I'll take a stab at this...

    Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
    Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

    Capacitive Reactance (Xc) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
    Xc = 76

    Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
    Inductors in series are added together. 500 mH + 125 mH = 625 mH

    Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
    Xl = 236

    Xc + Xl (76 + 236) = 312 ohms

    I would be surprised if I did that correctly. Someone correct me if needed!
    Warren Garber
    Have a great day!

  8. #5
  9. Join Date
    Mar 2018
    Posts
    18
    Reputation
    Quote Originally Posted by Warrengarber View Post
    Your math all works out to me. A time saver when performing this and from my experience with similar questions on the test is that they will use two identical, in this case capacitors in series and reactors in parallel, values. When this is the case the calculated value is half of one of the identical capacitors.

    Capacitor Example: Look at Rofo42's capacitors 20ufx20uf=400uf over the sum of 20uf+20uf= 40uf, 400uf/40uf=10uf or half of one of the 20uf capacitors. (Capacitors in series)

    Inductor Example: 1/250mH+1/250mH=0.004mH+0.004mH=0.008mH then 1/0.008 with a result of 125mH or half of one of the two parallel 250mH reactors (Reactors in Parallel)

    It is all about the time it takes to answer the questions on the test. Remembering these shortcuts during the stress of taking the test and knowing that that one wrong answer may be the make or break is not easy. A question like this could take 5 minutes to answer when doing all of the math but when you only have an average of 72 seconds to answer a question shortcuts count. I hope that helps and great job Rofo42
    Yes the math does work out.

    However as stated, two inductors in parallel of equal value = 1/2 the inductance of both and 2 capacitors in series of equal value, likewise = 1/2 the capacitance of both. No need for the math.

    Whiskers

  10. #6
  11. Kalbi_Rob's Avatar
    Kalbi_Rob is offline Experienced Member Pro Subscriber
    Join Date
    Feb 2018
    Location
    Richmond, VA
    Posts
    121
    Reputation
    Quote Originally Posted by rofo42 View Post
    I'll take a stab at this...

    Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
    Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

    Capacitive Reactance (Xc) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
    Xc = 76

    Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
    Inductors in series are added together. 500 mH + 125 mH = 625 mH

    Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
    Xl = 236

    Xc + Xl (76 + 236) = 312 ohms

    I would be surprised if I did that correctly. Someone correct me if needed!

    I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

    ZT2=XR2+(|XL-XC|)2

    So, since there is no XR

    ZT=(236-76)=160 Ohms

  12. #7
  13. Warrengarber's Avatar
    Warrengarber is offline
    If you ignore it, it will fail.
    Seasoned Member Pro Subscriber
    Join Date
    Jul 2017
    Location
    Jacksonville, FL
    Posts
    57
    Reputation
    Thank you for the catch. I see on the formula sheet that they have the formula for Z and you are very correct.

    Z= the square root of R squared plus (XL-Xc) squared


    Quote Originally Posted by Kalbi_Rob View Post
    I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

    ZT2=XR2+(|XL-XC|)2

    So, since there is no XR

    ZT=(236-76)=160 Ohms
    Warren Garber
    Have a great day!

  14. #8
  15. rofo42 is offline Member Pro Subscriber
    Join Date
    Apr 2018
    Posts
    26
    Reputation
    Quote Originally Posted by Kalbi_Rob View Post
    I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

    ZT2=XR2+(|XL-XC|)2

    So, since there is no XR

    ZT=(236-76)=160 Ohms
    I did not know this. Thank you for the information!

  16. #9
  17. Join Date
    Mar 2018
    Posts
    18
    Reputation
    Quote Originally Posted by rofo42 View Post
    I did not know this. Thank you for the information!
    Well, that throws a wrench in the works but, the question was not looking for the impedance of the circuit but simply the reactive component of the circuit at 60 hz.

    While the math is correct about the phase angles, I'm not sure what the exact wording of the question was or what 4 choices were available, wish I knew.

  18. #10
  19. Kalbi_Rob's Avatar
    Kalbi_Rob is offline Experienced Member Pro Subscriber
    Join Date
    Feb 2018
    Location
    Richmond, VA
    Posts
    121
    Reputation
    Quote Originally Posted by whiskers View Post
    Well, that throws a wrench in the works but, the question was not looking for the impedance of the circuit but simply the reactive component of the circuit at 60 hz.

    While the math is correct about the phase angles, I'm not sure what the exact wording of the question was or what 4 choices were available, wish I knew.
    Impedance must be calculated for each component (i.e. the capacitive impedance, inductive impedance and resistive impedance) to calculate the total reactance of the circuit. Without the frequency, you cannot determine the reactance of the circuit as inductors and capacitors react differently at different frequencies.

    This page explains it alot better than I do:
    https://www.wikihow.com/Calculate-Impedance

Subscribe

Login or register to leave a reply!


Share this thread

Related Topics

  1. Determine the approximate short circuit current available
    By rasilva in forum NETA Level 3 Exam
    Replies: 5
    Last Post: December 21, 2018, 06:25 PM
  2. Reactive Power (KVAR)
    By shamimkhan in forum Electrical Testing Talk
    Replies: 0
    Last Post: December 26, 2017, 03:09 AM
  3. What is the reactive power?
    By trhea@saberpower.com in forum NETA Level 3 Exam
    Replies: 10
    Last Post: October 5, 2017, 03:33 PM
  4. Total XC for capacitors in parallel
    By adamr8223 in forum NETA Level 3 Exam
    Replies: 6
    Last Post: December 4, 2016, 02:08 AM
  5. Total burden on current transformer
    By Sanjayo in forum NETA Level 3 Exam
    Replies: 1
    Last Post: February 19, 2016, 01:50 PM

Tags for this Thread



Related Content


Follow us


Explore TestGuy


NETA Certification Training


NICET Electrical Power Testing


Help and Support