electric heater problem

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searley

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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

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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

 

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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.

;)

 

searley

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

 
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