## 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

1. anonymous

@Michele_Laino

2. Michele_Laino

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

since 2 mins= 120 seconds

4. anonymous

75 x 4 = 300?

5. Michele_Laino

yes!

6. anonymous

36k?

7. Michele_Laino

yes! 36,000 Joules

8. anonymous

And the next part?

9. Michele_Laino

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

requested*

11. anonymous

240 C.

12. Michele_Laino

that's right!

13. anonymous

Help me more?

14. Michele_Laino

ok!

15. anonymous

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

Current = 0.24 A?

17. Michele_Laino

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

0.24 Ampheres...

19. Michele_Laino

correct!

20. anonymous

Next part?

21. Michele_Laino

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

since: $\Large P = \frac{{{V^2}}}{R}$ is another formula for electrical power

23. anonymous

ohms???? They what the answer in Watts.....

24. Michele_Laino

R is the electrical resistance of our lamp

25. anonymous

Wait, "how is the power consumed by the lamp affected?"

26. anonymous

Power is in Watts...

27. Michele_Laino

yes! power is measured with watts, nevertheless our lamp has an electrical resistance

28. anonymous

But we want the power not the resistance...

29. Michele_Laino

30. anonymous

Ok so 1041.6 ohm?

31. Michele_Laino

that's right!

32. Michele_Laino

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

0.192 A

34. Michele_Laino

that's right!

35. Michele_Laino

and the new absorbed power is: $\Large {P_1} = {I_1}{V_1} = 0.192 \times 200 = ...watts$

36. anonymous

38.4 W

37. Michele_Laino

that's right!

38. anonymous

THnkas. Help me more??

39. Michele_Laino

ok!

40. anonymous

An electric bulb is rated at 220V,100V Calculare:Resistance ans safe current?

41. Michele_Laino

I think: "An electric bulb is rated at 220V,100W" am I right?

42. anonymous

Yes, My typos t_t

43. Michele_Laino

ok! the requested resistance is: $\Large R = \frac{{{V^2}}}{P} = \frac{{{{200}^2}}}{{100}} = ...ohms$

44. Michele_Laino

oops.. $\Large R = \frac{{{V^2}}}{P} = \frac{{{{220}^2}}}{{100}} = ...ohms$

45. anonymous

484 ohm

46. Michele_Laino

correct!

47. Michele_Laino

the safe current values are given by the subsequent condition: $\Large I \leqslant \frac{P}{V} = \frac{{100}}{{220}} = ...amperes$

48. anonymous

That symbol after I ???

49. Michele_Laino

it means "less or equal to"

50. anonymous

0.45

51. Michele_Laino

correct!

52. anonymous

Help me more and thanks?

53. Michele_Laino

ok!

54. anonymous

A bulb rated 12V,24W rans for 20 mins Calculate: energy consumed

55. Michele_Laino

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

28,800J

57. Michele_Laino

that's right!

58. anonymous

Thanks and I still have about 5 more questions can u help?

59. Michele_Laino

ok! Please post in Physics section

60. anonymous

Ok!