An engineer is designing a heating element for a laboratory kettle, to be made from nichrome wire of resistivity = 1.10×10^-6 m and uniform cross-sectional area A = 4.00×10^-8 m^2. The element must have resistance R = 55.0 when connected to the 230 V mains supply. (a) Calculate…

Answer: ((a)) I = 4.18 A ((b)) L = 2.00 m ((c)) P = 962 W ((d)) Resistance doubles to 110, so at the same 230 V the power dissipated falls to about 481 W (roughly half of the value in (c)).. Part (a): Current in the element Using Ohm's law, I = V/R, with V = 230 V and R = 55.0: I = 230/55.0 = 4.18 A (3 s.f.) Part (b): Length of wire needed Rearranging R = L/A for L: L = RA/ Substituting R = 55.0, A = 4.00×10^-8 m^2 and = 1.10×…

AS Level Physics 9702: Designing a Heating Element with Resistivity and Power

An engineer is designing a heating element for a laboratory kettle, to be made from nichrome wire of resistivity $\rho = 1.10\times10^{-6}\,\Omega\,\text{m}$ and uniform cross-sectional area $A = 4.00\times10^{-8}\text{ m}^2$ . The element must have resistance $R = 55.0\,\Omega$ when connected to the $230\text{ V}$ mains supply.

(a) Calculate the current $I$ that flows in the element when it is connected to the $230\text{ V}$ supply. [2]

(b) Calculate the length $L$ of nichrome wire needed to give the required resistance $R = 55.0\,\Omega$ , using $R = \rho L / A$ . [2]

(c) Calculate the power $P$ dissipated by the heating element when connected to the $230\text{ V}$ supply, using $P = V^2/R$ . [2]

(d) The wire is now replaced by a nichrome wire of the same length but with half the cross-sectional area. State and explain the effect this has on the resistance and, hence, on the power dissipated at the same $230\text{ V}$ supply. [2]

Part (a): Current in the element

Using Ohm’s law, $I = V/R$ , with $V = 230\text{ V}$ and $R = 55.0\,\Omega$ : $I = \frac{230}{55.0} = 4.18\text{ A} \ (3\text{ s.f.})$

Part (b): Length of wire needed

Rearranging $R = \dfrac{\rho L}{A}$ for $L$ : $L = \frac{RA}{\rho}$

Substituting $R = 55.0\,\Omega$ , $A = 4.00\times10^{-8}\text{ m}^2$ and $\rho = 1.10\times10^{-6}\,\Omega\,\text{m}$ : $L = \frac{55.0 \times 4.00\times10^{-8}}{1.10\times10^{-6}} = \frac{2.20\times10^{-6}}{1.10\times10^{-6}}$ $L = 2.00\text{ m}$

Check: $\rho L/A = (1.10\times10^{-6} \times 2.00)/(4.00\times10^{-8}) = 2.20\times10^{-6}/4.00\times10^{-8} = 55.0\,\Omega$ , which matches the required $R$ . $\checkmark$

Part (c): Power dissipated

Using $P = \dfrac{V^2}{R}$ with $V = 230\text{ V}$ and $R = 55.0\,\Omega$ : $P = \frac{230^2}{55.0} = \frac{52900}{55.0} = 962\text{ W} \ (3\text{ s.f.})$

Check using $P = I^2R$ with $I = 4.1818\text{ A}$ (unrounded) and $R = 55.0\,\Omega$ : $P = (4.1818)^2 \times 55.0 = 17.49 \times 55.0 = 962\text{ W}$ , which agrees. $\checkmark$

Part (d): Effect of halving the cross-sectional area

Since $R = \rho L/A$ with $\rho$ and $L$ unchanged, resistance is inversely proportional to area. Halving the cross-sectional area therefore doubles the resistance: $R_{\text{new}} = 2 \times 55.0 = 110\,\Omega$

At the same $230\text{ V}$ supply, using $P = V^2/R$ : $P_{\text{new}} = \frac{230^2}{110} = \frac{52900}{110} = 481\text{ W} \ (3\text{ s.f.})$

This is roughly half the power found in (c), the thinner wire offers a narrower path for the charge carriers, giving it a higher resistance, so it draws less current from the fixed $230\text{ V}$ supply and hence dissipates less power.

Final answers

(a) Current $=$ $4.18\text{ A}$
(b) Length of wire needed $=$ $2.00\text{ m}$
(c) Power dissipated $=$ $962\text{ W}$
(d) Resistance doubles to $110\,\Omega$ , so the power dissipated at the same $230\text{ V}$ falls to about $481\text{ W}$ (roughly half of the value in (c))