1
\$\begingroup\$

I have the following question:

enter image description here What I've done so far:

  1. drew a vector diagram with the known voltages: enter image description here
  2. used the vector diagram to come up with 2 equations using pythagoras and used them to solve for BC and CD
  3. calculated coil impedance from 180 V and 2.4 A and used that as the "resistance" part
  4. calculated XL using the value I calculated for CD
  5. calculated the phase difference using sin (VL/180), where VL is CD

My final answers are BC = 38.3 V; CD = 175.88 V; "R" = 75 ohm; XL = 73.28 ohm; phase difference = 77.72 deg. My 2 equations for BC and CD: enter image description here

I'm pretty sure I went wrong somewhere but I can't figure out exactly where I went wrong. I THINK its in the calculation of CD, I can't remember where I got the 125 from. Any help would be appreciated!

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

There is a fault in your problem. If the total supply voltage is 240 volts then this is made up from the resistor voltage and the inductor voltage in series and in quadrature i.e.: -

Vtotal = \$\sqrt{V_R^2+V_L^2}\$ = \$\sqrt{90^2+180^2}\$ = 201.2 volts

This is not 240 volts so the problem in your question is flawed.

\$\endgroup\$
3
  • \$\begingroup\$ Hi, yes I also noticed that the voltages do not add up, but did the solution with the given voltage in case the marker is finicky about it. Would you recommend I do it with the correct total voltage? \$\endgroup\$
    – Naynay
    Commented Mar 15, 2018 at 12:42
  • \$\begingroup\$ I am not an assessor so my opinion is invalid but, to progress, I would make the point that you are calculating your answer based on the input supply voltage being 201.2 volts. \$\endgroup\$
    – Andy aka
    Commented Mar 15, 2018 at 12:45
  • \$\begingroup\$ Your other question (posted at about the same time) is totally ambiguous and although I tried to assume this or that, nothing really made much sense. Can we assume that your faculty has an idiot setting questions LOL. \$\endgroup\$
    – Andy aka
    Commented Mar 15, 2018 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.