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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
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4the 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
 one year ago
Best ResponseYou've already chosen the best response.4since 2 mins= 120 seconds

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4yes! 36,000 Joules

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4it is simple, since the requestes amount Q of charge is equal to: \[\Large Q = I\Delta t = 2 \times 120 = ...Coulombs\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Calculate 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4here, we have to apply the subsequent formula: \[\Large P = VI\] where P is the electrical power. So the requested current is: \[\Large I = \frac{P}{V} = \frac{{60}}{{250}} = ...amperes\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4when the voltage is 250 volts, the resistance of our lamp is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{250}^2}}}{{60}} = ...ohms\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4since: \[\Large P = \frac{{{V^2}}}{R}\] is another formula for electrical power

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohms???? They what the answer in Watts.....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4R is the electrical resistance of our lamp

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, "how is the power consumed by the lamp affected?"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Power is in Watts...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4yes! power is measured with watts, nevertheless our lamp has an electrical resistance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But we want the power not the resistance...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I know it, please follow my steps

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now if we apply a voltage of 200 volts, the new current is: \[\Large {I_1} = \frac{{{V_1}}}{R} = \frac{{200}}{{1041.6}} = ...amperes\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4and the new absorbed power is: \[\Large {P_1} = {I_1}{V_1} = 0.192 \times 200 = ...watts\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THnkas. Help me more??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0An electric bulb is rated at 220V,100V Calculare:Resistance ans safe current?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I think: "An electric bulb is rated at 220V,100W" am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4ok! the requested resistance is: \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{200}^2}}}{{100}} = ...ohms\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4oops.. \[\Large R = \frac{{{V^2}}}{P} = \frac{{{{220}^2}}}{{100}} = ...ohms\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4the safe current values are given by the subsequent condition: \[\Large I \leqslant \frac{P}{V} = \frac{{100}}{{220}} = ...amperes\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That symbol after I ???

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4it means "less or equal to"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Help me more and thanks?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A bulb rated 12V,24W rans for 20 mins Calculate: energy consumed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4the 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks and I still have about 5 more questions can u help?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4ok! Please post in Physics section
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