Question

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

Answer 1

@Michele_Laino

Answer 2

the requested heat energy E is given by the subsequent formula:

\[\Large E = R{I^2}\Delta t = 75 \times {2^2} \times 120 = ...Joules\]

Answer 3

since 2 mins= 120 seconds

Answer 4

75 x 4 = 300?

Answer 5

yes!

Answer 6

36k?

Answer 7

yes! 36,000 Joules

Answer 8

And the next part?

Answer 9

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

Answer 10

requested*

Answer 11

240 C.

Answer 12

that's right!

Answer 13

Help me more?

Answer 14

ok!

Answer 15

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?

Answer 16

Current = 0.24 A?

Answer 17

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

Answer 18

0.24 Ampheres...

Answer 19

correct!

Answer 20

Next part?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

R is the electrical resistance of our lamp

Answer 25

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

Answer 26

Power is in Watts...

Answer 27

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

Answer 28

But we want the power not the resistance...

Answer 29

I know it, please follow my steps

Answer 30

Ok
so
1041.6 ohm?

Answer 31

that's right!

Answer 32

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

Answer 33

0.192 A

Answer 34

that's right!

Answer 35

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

Answer 36

38.4 W

Answer 37

that's right!

Answer 38

THnkas.
Help me more??

Answer 39

ok!

Answer 40

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

Answer 41

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

Answer 42

Yes,
My typos t_t

Answer 43

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

Answer 44

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

Answer 45

484 ohm

Answer 46

correct!

Answer 47

the safe current values are given by the subsequent condition:

\[\Large I \leqslant \frac{P}{V} = \frac{{100}}{{220}} = ...amperes\]

Answer 48

That symbol after I
???

Answer 49

it means "less or equal to"

Answer 50

0.45

Answer 51

correct!

Answer 52

Help me more
and thanks?

Answer 53

ok!

Answer 54

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

Answer 55

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

Answer 56

28,800J

Answer 57

that's right!

Answer 58

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

Answer 59

ok! Please post in Physics section

Answer 60

Ok!