An electrical heating element is to be designed so that the power dissipated will be 750 W when connected to the 240 V mains supply.
(a) Calculate the resistance of the wire needed.
(b) The element is to be made from nichrome ribbon 1.0 mm wide and 0.050 mm thick. The resistivity of nichrome = 1.1 * 10-6 Ω m. Calculate the length of ribbon required.
(c) Draw a circuit diagram to show how a second heating element would be connected to increase the power dissipated to 1.5 kW.
(d) State one important property of a conductor used to make heating elements.
Answers: (a) 76.8 Ω, (b) 3.49 m.
2. The attempt at a solution
(a) P = V * I and R = V / I. So we find I = P / V = 750 / 240 = 3.125 A and find R = 240 / 3.125 = 76.8 Ω.
(b) Area = (1 / 10 / 100) * (0.050 / 10 / 100) = 5 * 10-8 m2. L = (R * A) / ρ = (76.8 * 5 * 10-8) / (1.1 * 10-6) = 3.49 m.
(c) This is the part I'm not sure. This is how I see the question:
I would say that if we add another 76.8 Ω heating element the current would increase to 6.25 A (1500 W / 240 V). And then I just put the second element above or beneath the given one. And then we will have a series circuit with I = 6.25 A, two elements with R = 76.8 Ω each and V = 240 V. Is this correct?
(d) This thing I don't know. Maybe it has to do something with it's dimensions or resistivity or something else?