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

The purpose of a lab is to do your "expirements" the importance of the topic is so you can now what the story is about and the last one is because no one is me with my question ( what is 3 over 4 x - 9 = 27 maybe you can )