The heating element inan electric drier operates on 240 V and generates heat at the rateof 2.0 kW. The heating element shorts out and, in repairing it, theowner shortens the Nichrome wire by 10%. (Assume the temperature isunchanged. In reality, the resistivity of the wire will depend onits temperature.) What effect will the repair have on the powerdissipated in the heating element? Please work out, Thanks! The heating element inan electric drier operates on 240 V and generates heat at the rateof 2.0 kW. The heating element shorts out and, in repairing it, theowner shortens the Nichrome wire by 10%. (Assume the temperature isunchanged. In reality, the resistivity of the wire will depend onits temperature.) What effect will the repair have on the powerdissipated in the heating element? Please work out, Thanks!
Explanation / Answer
R1 = L1 / A R2 = L2 / A = .9 L1 / A R1 / R2 = 1.11 P = I2 R = V2 / R P2 / P1 = R1 / R2= 1.11 so the power used will increase by11 %