anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
the requested heat energy E is given by the subsequent formula: \[\Large E = R{I^2}\Delta t = 75 \times {2^2} \times 120 = ...Joules\]
Michele_Laino
  • Michele_Laino
since 2 mins= 120 seconds

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anonymous
  • anonymous
75 x 4 = 300?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
36k?
Michele_Laino
  • Michele_Laino
yes! 36,000 Joules
anonymous
  • anonymous
And the next part?
Michele_Laino
  • Michele_Laino
it is simple, since the requestes amount Q of charge is equal to: \[\Large Q = I\Delta t = 2 \times 120 = ...Coulombs\]
Michele_Laino
  • Michele_Laino
requested*
anonymous
  • anonymous
240 C.
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Help me more?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • 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?
anonymous
  • anonymous
Current = 0.24 A?
Michele_Laino
  • Michele_Laino
here, we have to apply the subsequent formula: \[\Large P = VI\] where P is the electrical power. So the requested current is: \[\Large I = \frac{P}{V} = \frac{{60}}{{250}} = ...amperes\]
anonymous
  • anonymous
0.24 Ampheres...
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
Next part?
Michele_Laino
  • Michele_Laino
when the voltage is 250 volts, the resistance of our lamp is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{250}^2}}}{{60}} = ...ohms\]
Michele_Laino
  • Michele_Laino
since: \[\Large P = \frac{{{V^2}}}{R}\] is another formula for electrical power
anonymous
  • anonymous
ohms???? They what the answer in Watts.....
Michele_Laino
  • Michele_Laino
R is the electrical resistance of our lamp
anonymous
  • anonymous
Wait, "how is the power consumed by the lamp affected?"
anonymous
  • anonymous
Power is in Watts...
Michele_Laino
  • Michele_Laino
yes! power is measured with watts, nevertheless our lamp has an electrical resistance
anonymous
  • anonymous
But we want the power not the resistance...
Michele_Laino
  • Michele_Laino
I know it, please follow my steps
anonymous
  • anonymous
Ok so 1041.6 ohm?
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
now if we apply a voltage of 200 volts, the new current is: \[\Large {I_1} = \frac{{{V_1}}}{R} = \frac{{200}}{{1041.6}} = ...amperes\]
anonymous
  • anonymous
0.192 A
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
and the new absorbed power is: \[\Large {P_1} = {I_1}{V_1} = 0.192 \times 200 = ...watts\]
anonymous
  • anonymous
38.4 W
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
THnkas. Help me more??
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
An electric bulb is rated at 220V,100V Calculare:Resistance ans safe current?
Michele_Laino
  • Michele_Laino
I think: "An electric bulb is rated at 220V,100W" am I right?
anonymous
  • anonymous
Yes, My typos t_t
Michele_Laino
  • Michele_Laino
ok! the requested resistance is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{200}^2}}}{{100}} = ...ohms\]
Michele_Laino
  • Michele_Laino
oops.. \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{220}^2}}}{{100}} = ...ohms\]
anonymous
  • anonymous
484 ohm
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
the safe current values are given by the subsequent condition: \[\Large I \leqslant \frac{P}{V} = \frac{{100}}{{220}} = ...amperes\]
anonymous
  • anonymous
That symbol after I ???
Michele_Laino
  • Michele_Laino
it means "less or equal to"
anonymous
  • anonymous
0.45
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
Help me more and thanks?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
A bulb rated 12V,24W rans for 20 mins Calculate: energy consumed
Michele_Laino
  • Michele_Laino
the requested energy E, is given by the subsequent computation: \[\Large E = P\Delta t = 24 \times 20 \times 60 = ...Joules\] since: \[\Large \Delta t = 20 \times 60{\text{ seconds}}\]
anonymous
  • anonymous
28,800J
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Thanks and I still have about 5 more questions can u help?
Michele_Laino
  • Michele_Laino
ok! Please post in Physics section
anonymous
  • anonymous
Ok!

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