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Question

a) Write the formula for the resistance of a wire of length L and cross-section A.
b) A piece of wire is redrawn by pulling it until its length is doubled. Compare the new
resistance with the original value.
c) How many 176Ω resistors (in parallel) are required to carry 5 A in 220 V line?
[5 MARKS]

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Solution

part a: 1 Mark
part b: 2 Marks
part c: 2 Marks

a) R = ρLA
b)The volume of the material of wire remains same. So, when length is doubled, its area of cross - section will get halved. So, if l and a are the original length and area of cross-section of wire,

Original value of the resistance, R=ρ×la

and,

New value of the resistance,

R=ρ×2la2=ρ1a×4=4R

c) Here, I = 5A, V = 220 V.

Resistance required in the circuit, R = VI = 220V5A = 44 Ω, Resistance of each resistor, r = 176 Ω

If n resistors, each of resistance r, are connected in parallel to get the required resistance R, then R = rn or n = rR = 17644 = 4


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