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TrojanPoem

  • one year ago

Find R if Pw on it was 20 watt

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  1. TrojanPoem
    • one year ago
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    |dw:1432754744584:dw|

  2. TrojanPoem
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    is 20 Watt the power dissipated on R?

  4. Michele_Laino
    • one year ago
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    do you know how compute the equivalent resistance of your circuit?

  5. TrojanPoem
    • one year ago
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    yeah

  6. Michele_Laino
    • one year ago
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    what is that resistance?

  7. TrojanPoem
    • one year ago
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    The total resistance it not given

  8. Michele_Laino
    • one year ago
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    we can compute the total resistance

  9. TrojanPoem
    • one year ago
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    Will be exhausting much.. think.

  10. TrojanPoem
    • one year ago
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    (40+R)* 30 / (70+R) + 5

  11. Michele_Laino
    • one year ago
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    here are the necessary steps: |dw:1432755438420:dw|

  12. TrojanPoem
    • one year ago
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    (40+R)* 30 / (70+R) + 5

  13. Michele_Laino
    • one year ago
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    that's right!

  14. TrojanPoem
    • one year ago
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    Okie, now ?

  15. Michele_Laino
    • one year ago
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    I got this: \[\Large {R_{TOTAL}} = \frac{{1550 + 35R}}{{70 + R}}\]

  16. TrojanPoem
    • one year ago
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    You got rid of 5 ? by making them on one fraction ?

  17. TrojanPoem
    • one year ago
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    Well, Now what to do ?

  18. Michele_Laino
    • one year ago
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    yes!

  19. Michele_Laino
    • one year ago
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    we can compute the current, like below: \[\Large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{\left( {70 + R} \right)V}}{{1550 + 35R}}\]

  20. TrojanPoem
    • one year ago
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    Ok , Now ?

  21. Michele_Laino
    • one year ago
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    |dw:1432755794918:dw|

  22. Michele_Laino
    • one year ago
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    we can compute the voltage drop V_AB such that: |dw:1432755868980:dw| \[\Large {V_{AB}} = V - 5I = V - \frac{{5\left( {70 + R} \right)V}}{{1550 + 35R}}\]

  23. Michele_Laino
    • one year ago
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    please, simplify that expression

  24. Michele_Laino
    • one year ago
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    whatdo you get?

  25. Michele_Laino
    • one year ago
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    oops..what do you get?

  26. TrojanPoem
    • one year ago
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    - 26250 - 375R / 1550 + 35 R + V

  27. TrojanPoem
    • one year ago
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    HUGE NUMBER !

  28. Michele_Laino
    • one year ago
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    I got this: \[\Large \begin{gathered} {V_{AB}} = V - 5I = V - \frac{{5\left( {70 + R} \right)V}}{{1550 + 35R}} = \hfill \\ \hfill \\ = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} \hfill \\ \end{gathered} \]

  29. TrojanPoem
    • one year ago
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    Is V the VB or V at 5 ohm resistor ?

  30. Michele_Laino
    • one year ago
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    it is V_AB

  31. TrojanPoem
    • one year ago
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    VAB = the current X the total resistance between AB

  32. TrojanPoem
    • one year ago
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    Or VB - V at 5 ohm resistor Or VAB = VA - VB

  33. Michele_Laino
    • one year ago
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    |dw:1432756298125:dw|

  34. TrojanPoem
    • one year ago
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    Well, now what will we do ?

  35. Michele_Laino
    • one year ago
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    we can compute the current along the resistors 40 and R, namely the current I_2: |dw:1432756485462:dw|

  36. TrojanPoem
    • one year ago
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    .... ?

  37. Michele_Laino
    • one year ago
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    so we can write: \[\Large \begin{gathered} {I_2} = \frac{{{V_{AB}}}}{{40 + R}} = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} \times \frac{1}{{40 + R}} = \hfill \\ \hfill \\ = \frac{{30V\left( {40 + R} \right)}}{{\left( {1550 + 35R} \right)\left( {40 + R} \right)}} = \hfill \\ \hfill \\ = \frac{{30V}}{{1550 + 35R}} \hfill \\ \end{gathered} \]

  38. Michele_Laino
    • one year ago
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    what do you think about that?

  39. TrojanPoem
    • one year ago
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    PW = I^2 R solve for R ,

  40. Michele_Laino
    • one year ago
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    yes! Nevertheless we don't know the value of R

  41. Michele_Laino
    • one year ago
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    Next we can compute the value of the voltage drop on resistor R, as below:

  42. Michele_Laino
    • one year ago
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    |dw:1432756998788:dw|

  43. TrojanPoem
    • one year ago
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    (30V / 1550 + 35 R)^2 * R = 20 30V/ 1550 + 35R * sqrt(R) = 2sqrt(5)

  44. TrojanPoem
    • one year ago
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    30 * 75 sqrt(R) = 2 sqrt(5) (1550 +35R)

  45. Michele_Laino
    • one year ago
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    \[\large {V_1} = {V_{AB}} - 40{I_2} = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} - \frac{{40 \times 30V}}{{1550 + 35R}} = ...\]

  46. Michele_Laino
    • one year ago
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    |dw:1432757229841:dw|

  47. TrojanPoem
    • one year ago
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    2250 sqrt(R) = 31500sqrt(5) + 70sqrt(5)R 70sqrt(5)R - 2250sqrt(R) - 31500sqrt(5) = 0 R = 15 sqrt(437) + 45sqrt(5) / 14

  48. TrojanPoem
    • one year ago
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    You miscalculated the current :(( noooooooooooooooooooooooooooooooooooo

  49. Michele_Laino
    • one year ago
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    no, my computations are correct!

  50. Michele_Laino
    • one year ago
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    I got: \[\Large {V_1} = \frac{{30RV}}{{1550 + 35R}}\]

  51. TrojanPoem
    • one year ago
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    Now, how will we get R ?

  52. Michele_Laino
    • one year ago
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    we have to use the power dissipated by R, so we can write this formula: \[\Large W = \frac{{V_1^2}}{R}\]

  53. Michele_Laino
    • one year ago
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    where W=20 Watt

  54. Michele_Laino
    • one year ago
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    after a simple substitution, we get: \[\Large W = \frac{{V_1^2}}{R} = \frac{{900R{V^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]

  55. Michele_Laino
    • one year ago
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    so we got this condition: \[\Large 20 = \frac{{900R{{\left( {75} \right)}^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]

  56. Michele_Laino
    • one year ago
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    which is a quadratic equation for R. Please solve it for R

  57. TrojanPoem
    • one year ago
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    20 ( 1550 + 35R)^2 = 900R(75)^2 2402500 + 24500R^2 + 1085000R = 5062500R 24500R^2 -3977500 R + 2402500 = 0 R = 161, 7 R= 0.7

  58. TrojanPoem
    • one year ago
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    hmm , Is that normal ?

  59. Michele_Laino
    • one year ago
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    please, wait I'm checking your computation

  60. TrojanPoem
    • one year ago
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    Take your time Pr0

  61. TrojanPoem
    • one year ago
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    Hey

  62. Michele_Laino
    • one year ago
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    I got these values: \[\Large \begin{gathered} R = \frac{{5365 + 2475}}{{98}} = 80\;Ohms \hfill \\ \hfill \\ R = \frac{{5365 - 2475}}{{98}} \cong 29.5\;Ohms \hfill \\ \end{gathered} \]

  63. TrojanPoem
    • one year ago
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    Thanks so much.

  64. Michele_Laino
    • one year ago
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    thanks! :)

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