SOLUTION: The electric power P varies jointly as the resistance R and the square of the current I. If the power is 40 watts when the resistance is 10 ohms and the current is 1.33 amps, what

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: The electric power P varies jointly as the resistance R and the square of the current I. If the power is 40 watts when the resistance is 10 ohms and the current is 1.33 amps, what      Log On


   

Question 1010408: The electric power P varies jointly as the resistance R and the square of the current I. If
the power is 40 watts when the resistance is 10 ohms and the current is 1.33 amps, what is
the power when the current is 5 amps and the resistance is 25 ohms?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
direct variation formula is y = kx

inverse variation formula is y = k/x

joint variation formula is y = kxz

k = constant of variation.

it remains the same in each equation regardless of the values of the other variables in that equation.

in your problem:

y = p
x = r
z = i^2
k = constant of variation

p = power
r = resistance
i = current

the formula of y = kxz becomes p = kri^2

when p = 40 and r = 10 and i = 1.33, the formula of p = kri^2 becomes
40 = k*10*1.33^2

solve for k to get k = 40/(10*1.33^2) which becomes k = 2.261292329.

when r = 25 and i = 5, the formula of p = kri^2 becomes:

p = 2.261292329 * 25 * 5^2.

the value of k remains the same as it was when initially calculated.

solve for p to get p = 1413.307705.