emergency electrical help needed >_<

• WoAiMeiMei

  31 Mar 08, 20:10

  Still having problem with these 2 >__<

  

   

   

  Superceeded ( ignore everything below - kept for reference)

  

   

  Superceeded2 (reference 2)

  

  

  

   

   

   

  Superceeded questions below (for reference)

   

  

• eagle

  31 Mar 08, 21:53

  Q3

  For V1 to be 0, the potential in between R1 and R2 must be the same as the potential between R3 and R4

  Thus, R1/R2 = R3/R4
  Hence, R4 = R3 * R2 / R1 = 86.89 ohms

   

  Q4

  I’m not very sure what is node to datum analysis. I use my own method.
  V1 = Vs = 9V

  By Kirchoff’s Current Law,
  (V1 – V2)/Ra = V2 / Rb + Is
  (9 – V2)/10 = V2/4 + 3
  V2/4 + V2/10 = 9/10 -3
  V2 = -6V

  Also, (V2 – V3)/Rc = Is
  -6 – V3 = 30
  V3 = -36V

• eagle

  31 Mar 08, 21:54

  I dun really understand what is meant by power supplied or absorbed by a current or voltage source...

• WoAiMeiMei

  31 Mar 08, 22:05

  here's an example similar to 1..

  http://i105.photobucket.com/albums/m225/nickykeeng/album2/example1.png

   

  dun worry about it though..

   

  i will check the rest.. every time it's wrong i have a new question >__<

• watershed85

  31 Mar 08, 22:08

  It's been a while since I last touched this.

  Hope it's right!

  Q1)

  current is 1A throughout

  clockwise:

  voltage across resistor V=RI=14V (voltage drop)

  voltage across Vs=3V (increase)

  voltage across Is=11V (increase)

  power given off by resistor = I2R=14W

  power "absorbed" by Vs=VI=3W

  power "absorbed" by Is=VI=11W

  (not sure if my interpretation of absorb is right)

• WoAiMeiMei

  31 Mar 08, 22:11

  Question 3 was correct, my failed attempt at question 1 >__< 66%..

   

  i need to re submit questions here again

• 798

  31 Mar 08, 22:11

  absorted is something like having negative power due to opp current from power supply rite?

• watershed85

  31 Mar 08, 22:15

  Q2)

  Let the 10V thingy be Vs. 10ohm resistor be r10, 9ohm resistor be r9.

  According to Kirchoff's current law, iunknown =0A

  First look at the loop on the right.

  current through r10=1A(anticlockwise)

  voltage drop across r10=Ri=10V

  Next look at the loop on the left.

  current through r9=1A (anticlockwise)

  voltage drop across r9=Ri=9V

  voltage across Is1=19V

  therefore power across Is1=Vi=19W

• eagle

  31 Mar 08, 22:16

  Originally posted by WoAiMeiMei:

  Question 3 was correct, my failed attempt at question 1 >__< 66%..

   

  i need to re submit questions here again

  Err... if u saw my answer for Q4, it was

  9V, -6V and -36V respectively

  You entered 9V, -36V and 36V

• eagle

  31 Mar 08, 22:17

  And erm... for question 1... if you can get the first 2, the final one must be such that the total power is 0. Conservation of energy still applies.

• WoAiMeiMei

  31 Mar 08, 22:19

  the trouble is finding the power across resistor for me >_<

  resistors ALWAYS absorb power

• WoAiMeiMei

  31 Mar 08, 22:19

  oh lol mistake :x

• eagle

  31 Mar 08, 22:22

  Originally posted by WoAiMeiMei:

  the trouble is finding the power across resistor for me >_<

  resistors ALWAYS absorb power

  See Watershed's method for Q1. He's correct, but he used the previous values. You could try to apply it to the new values.

• WoAiMeiMei

  31 Mar 08, 22:27

  stress... sign convention is a killer

• WoAiMeiMei

  31 Mar 08, 22:30

  Q1.. R=39, V=6, -84?
  Q2.. I (unknown) = -1 A
  Q4.. V1=8 V2=1.25 V3=-0.75
  
  ignore above >_<

   

  thanks to ur method i got Q4 correct ^^

  now got 2 more to work through.

   

   Q1 - for some reason i cant get it fully correct.

  Q2- is tricky