electric heater problem

[searley]

an electric heater dissipates a power of 3kw when connected to a 230v a.c. supply.the resistance of the heater is.

can somebody please show me how to work this out.cheers

 

[kme]

an electric heater dissipates a power of 3kw when connected to a 230v a.c. supply.the resistance of the heater is.can somebody please show me how to work this out.cheers

ohms law:

P=IV & R=V/I.......

therefore P/V=I......

then V/I=R

Try that, and post your answer, I`ll tell you if it`s right then ]

HTH

KME

 

[SPECIAL LOCATION]

an electric heater dissipates a power of 3kw when connected to a 230v a.c. supply.the resistance of the heater is.can somebody please show me how to work this out.cheers

Well a question like this.. where you only have TWO variables..

is just a matter of thinking what formula can I use to get the required value...

We know

POWER = VOLTS x CURRENT or

POWER = CURRENTsquared x RESISTANCE.

we also known that

RESISTANCE = VOLT / CURRENT

so it would help if can get the current..

P=VxI so I=P/V 3000/230 = 13.04A

R = 230/13 = 17.63ohm.

 

[SPECIAL LOCATION]

KME..

pushed and shoved so much...

you changed your AVATAR...

SNEEKY disguise!!!

:^O

 

[searley]

ohms law=IV & R=V/I.......

therefore P/V=I......

then V/I=R

Try that, and post your answer, I`ll tell you if it`s right then ]

HTH

KME

cheers that was tops easy when somebody goes throw it.top man.

 

[a1]

KME..pushed and shoved so much...

you changed your AVATAR...

SNEEKY disguise!!!

:^O

:^O :^O:^O

 

[Theorysparky]

P= IsqrdR

just another one

or

http://www.angelfire.com/pa/baconbacon/page2.html