(a) For a conductor in the form of a wire of uniform cross-sectional area, give an equation which relates its resistance to the resistivity of the material of the conductor. Define the symbols used in the equation.
(b) An electrical heating element, made from uniform nichrome wire, is required to dissipate 500 W when connected to the 230 V mains supply.
(i) The cross-sectional area of the wire is 8.0 × 10–8 m2. Calculate the length of nichrome wire required. resistivity of nichrome = 1.1 × 10–6 Ω m
P = IV
∴ P = V2/R 
and R = V2/P = 2302/500 = 106
R = 110 Ω
R = ρ L/A so
L = RA/ρ
L = 106 x 8.0 × 10–8/1.1 × 10–6
L = 7.7 m
(ii) Two heating elements, each rated at 230 V, 500 W are connected to the 230 mains supply (A) in series, (B) in parallel.
Explain why only one of the circuits will provide an output of 1 kW.
When they are connected in series, the voltage across each will be 115 V because the potential difference from the supply will be shared out between them.
Power = V2/R so the power output from each wil be less than (a quarter of ) 500W
In parallel, the voltage across each will be the full operating voltage of 230 V . Therefore each heater will give a power output of 500W, making a total of 1kW.
OR
When they are connected in series the high resistance from the combined resistance (1) will lower the current. Power output being I2R this will reduce the power output from each (to a quarter of the value).
When they are connected in parallel, resistance is lower (half), therefore a higher current (double) passes through the circuit .This splits to give the operating current through each heater, resulting in the full operating power output for each heater - 2 x 500W = 1kW
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