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anonymous

  • one year ago

A current of 2 A is passed through a coil of resistance 75 ohm for 2 minutes,(a)how much heat energy is produced/ (b)how much charge is passed through the resistance? @butterflydreamer @perl

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    the requested heat energy E is given by the subsequent formula: \[\Large E = R{I^2}\Delta t = 75 \times {2^2} \times 120 = ...Joules\]

  3. Michele_Laino
    • one year ago
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    since 2 mins= 120 seconds

  4. anonymous
    • one year ago
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    75 x 4 = 300?

  5. Michele_Laino
    • one year ago
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    yes!

  6. anonymous
    • one year ago
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    36k?

  7. Michele_Laino
    • one year ago
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    yes! 36,000 Joules

  8. anonymous
    • one year ago
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    And the next part?

  9. Michele_Laino
    • one year ago
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    it is simple, since the requestes amount Q of charge is equal to: \[\Large Q = I\Delta t = 2 \times 120 = ...Coulombs\]

  10. Michele_Laino
    • one year ago
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    requested*

  11. anonymous
    • one year ago
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    240 C.

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

  13. anonymous
    • one year ago
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    Help me more?

  14. Michele_Laino
    • one year ago
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    ok!

  15. anonymous
    • one year ago
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    Calculate the current through a 60 W lamp rated for 250 Volts.If the voltage line falls to 200 Volts,how is the power consumed by the lamp affected?

  16. anonymous
    • one year ago
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    Current = 0.24 A?

  17. Michele_Laino
    • one year ago
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    here, we have to apply the subsequent formula: \[\Large P = VI\] where P is the electrical power. So the requested current is: \[\Large I = \frac{P}{V} = \frac{{60}}{{250}} = ...amperes\]

  18. anonymous
    • one year ago
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    0.24 Ampheres...

  19. Michele_Laino
    • one year ago
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    correct!

  20. anonymous
    • one year ago
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    Next part?

  21. Michele_Laino
    • one year ago
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    when the voltage is 250 volts, the resistance of our lamp is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{250}^2}}}{{60}} = ...ohms\]

  22. Michele_Laino
    • one year ago
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    since: \[\Large P = \frac{{{V^2}}}{R}\] is another formula for electrical power

  23. anonymous
    • one year ago
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    ohms???? They what the answer in Watts.....

  24. Michele_Laino
    • one year ago
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    R is the electrical resistance of our lamp

  25. anonymous
    • one year ago
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    Wait, "how is the power consumed by the lamp affected?"

  26. anonymous
    • one year ago
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    Power is in Watts...

  27. Michele_Laino
    • one year ago
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    yes! power is measured with watts, nevertheless our lamp has an electrical resistance

  28. anonymous
    • one year ago
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    But we want the power not the resistance...

  29. Michele_Laino
    • one year ago
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    I know it, please follow my steps

  30. anonymous
    • one year ago
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    Ok so 1041.6 ohm?

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

  32. Michele_Laino
    • one year ago
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    now if we apply a voltage of 200 volts, the new current is: \[\Large {I_1} = \frac{{{V_1}}}{R} = \frac{{200}}{{1041.6}} = ...amperes\]

  33. anonymous
    • one year ago
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    0.192 A

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

  35. Michele_Laino
    • one year ago
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    and the new absorbed power is: \[\Large {P_1} = {I_1}{V_1} = 0.192 \times 200 = ...watts\]

  36. anonymous
    • one year ago
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    38.4 W

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

  38. anonymous
    • one year ago
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    THnkas. Help me more??

  39. Michele_Laino
    • one year ago
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    ok!

  40. anonymous
    • one year ago
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    An electric bulb is rated at 220V,100V Calculare:Resistance ans safe current?

  41. Michele_Laino
    • one year ago
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    I think: "An electric bulb is rated at 220V,100W" am I right?

  42. anonymous
    • one year ago
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    Yes, My typos t_t

  43. Michele_Laino
    • one year ago
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    ok! the requested resistance is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{200}^2}}}{{100}} = ...ohms\]

  44. Michele_Laino
    • one year ago
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    oops.. \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{220}^2}}}{{100}} = ...ohms\]

  45. anonymous
    • one year ago
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    484 ohm

  46. Michele_Laino
    • one year ago
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    correct!

  47. Michele_Laino
    • one year ago
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    the safe current values are given by the subsequent condition: \[\Large I \leqslant \frac{P}{V} = \frac{{100}}{{220}} = ...amperes\]

  48. anonymous
    • one year ago
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    That symbol after I ???

  49. Michele_Laino
    • one year ago
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    it means "less or equal to"

  50. anonymous
    • one year ago
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    0.45

  51. Michele_Laino
    • one year ago
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    correct!

  52. anonymous
    • one year ago
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    Help me more and thanks?

  53. Michele_Laino
    • one year ago
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    ok!

  54. anonymous
    • one year ago
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    A bulb rated 12V,24W rans for 20 mins Calculate: energy consumed

  55. Michele_Laino
    • one year ago
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    the requested energy E, is given by the subsequent computation: \[\Large E = P\Delta t = 24 \times 20 \times 60 = ...Joules\] since: \[\Large \Delta t = 20 \times 60{\text{ seconds}}\]

  56. anonymous
    • one year ago
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    28,800J

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

  58. anonymous
    • one year ago
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    Thanks and I still have about 5 more questions can u help?

  59. Michele_Laino
    • one year ago
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    ok! Please post in Physics section

  60. anonymous
    • one year ago
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    Ok!

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