W varies jointly as square of I, R; if W=48 for I=0.4,R=300,

ptebwwong

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I need help with this word problem.

The wattage rating of an appliance W varies jointly as the square of the current I & the resistance R. If the wattage is 48 watts when the current is 0.4 ampere & the resistance is 300 ohms, find the wattage when the current is 0.1 ampere & the resistance is 250 ohms.

The equation I have so far is: W=k x square root of IR
Is that right?
 
Re: word problem help

yes you are right.
You should find that K=1 when using W=48 I=.4 R=300

Then solve the problem for W given the parameters

Arthur
 
ptebwwong said:
... The equation I have so far is: W = k x square root of IR

Is that right? No, it is not. From where did the square root come?

In addition to being wrong, the phrase "square root of IR" is ambiguous.

It could mean (square root of I) * R.

It could mean square root of (I * R).

Also, please stop using the letter x as a multiplication sign; It's confusing in algebra because the letter x is so often used as a variable symbol. Use an asterisk or grouping symbols to indicate multiplication, instead.

EG:

W = k * square root of (I * R)

W = (k)(square root of I)(R)




arthur ohlsten said:
yes you are right.

You should find that K=1 ... Not using their equation, they may.



Ptebwwong:

Did you make a typographical error in your post when you typed the word "root"?

Do you know how to type exponents?

EG:

A squared = A^2

Cheers,

~ Mark :)

 
Re: word problem help

I never saw the square root. sorry
W=kI^2 R
and k should be 1

What is worse is that one of my degrees is in Electrical Engineering.

Arthur
 
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