A72 ohm coil of wire is connected in series with an adjustable resistor (rheostat) whose resistance can be varied from 0 to 88 ohms. if the line potential of wire is 115 volts, what is the rheostat resistance when power taken by the coil is 90 watts?

30.86 ohms Using ohms law I=V/R, and watts being W=VI, you can create the formula W=V^2/R by substituting V/R for I in the watts formula: W = V^2/R Plug in the known values of 90 watts and 72 ohms. 90 = V^2/72 Solve for V by first multiplying both sides by 72 6480 = V^2 Then take the square root of both sides 80.5 = V Using ohms law, you can calculate the amperage passing through the coil I = V/R = 80.5/72 = 1.118 amperes Now you need to figure out the total resistance needed to allow for 1.118 amperes with a voltage of 115 volts. Ohms law again. I = V/R 1.118 = 115/R Multiply both sides by R 1.118 R = 115 Divide both sides by 1.118 R = 115 / 1.118 = 102.86 Now the resistance for the rheostat is 102.86 minus the resistance of the coil. Giving R = 102.86 - 72 = 30.86 ohms. The rheostat has to have a resistance of 30.86 ohms

What even is this. where is the question