Solveeit Logo

Question

Science Question on Electricity

(A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R).
(ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate:
(a) power rating of the oven
(b) current drawn by the oven
(c) resistance of the oven when it is red hot

OR

(B) (i) Write the relation between resistance R and electrical resistivity ρ of the material of a conductor in the shape of a cylinder of length 𝑙 and area of cross-section A. Hence derive the SI unit of electrical resistivity.
(ii) The resistance of a metal wire of length 3 m is 60 Ω. If the area of cross-section of the wire is 4×1074 \times 10^{-7} m2, calculate the electrical resistivity of the wire.
(iii) State how would electrical resistivity be affected if the wire (of part ‘ii’) is stretched so that its length is doubled. Justify your answer.

Answer

(i). Electric power is the rate at which electrical energy is consumed or converted into another form of energy.
** Formula:** P=V2RP = \frac{V^2}{R}, where PP is power, VV is potential difference, and RR is resistance.

**(ii).Power rating of the oven: **
Energy consumed in 5 hours=11units (1 unit = 1 kWh)=11×1000=11000Wh\text{Energy consumed in 5 hours} = 11 \, \text{units (1 unit = 1 kWh)} = 11 \times 1000 = 11000 \, \text{Wh} Power rating, P=Energy consumedTime=110005=2200W\text{Power rating, } P = \frac{\text{Energy consumed}}{\text{Time}} = \frac{11000}{5} = 2200 \, \text{W}
Current drawn by the oven: P=VI    I=PV=2200220=10AP = VI \implies I = \frac{P}{V} = \frac{2200}{220} = 10 \, \text{A}
Resistance of the oven when it is red hot: R=V2P=22022200=22ΩR = \frac{V^2}{P} = \frac{220^2}{2200} = 22 \, \Omega


(i). R=ρlA,where ρ=RAlR = \rho \frac{l}{A}, \, \text{where } \rho = \frac{RA}{l}
** SI Unit of Resistivity: **ρ=Resistance×AreaLength=Ω×m2m=Ωm\rho = \frac{\text{Resistance} \times \text{Area}}{\text{Length}} = \frac{\Omega \times m^2}{m} = \Omega\cdot m

(ii). ρ=R×Al=60×4×1073=8×106Ωm\rho = \frac{R \times A}{l} = \frac{60 \times 4 \times 10^{-7}}{3} = 8 \times 10^{-6} \, \Omega \cdot m

(iii). Electrical resistivity is a material property and does not depend on the dimensions of the conductor. Stretching the wire will not change its resistivity as it depends only on the material, not its shape or size.