A wire of uniform cross section and length l has a resistance of 16 ohm. It is cut into 4 equal parts each part is stretched uniformly to length l
Answers
A wire of uniform cross section area and length has a resistance of 16 ohm. It is cut into 4 equal parts and each part is stretched uniformly to length 'l' and all 4 parts are connected in parallel. Calculate the total resistance of combination so formed.
A wire of length l and cross sectional area A has a resistance of 16Ω.
it is cut into 4 equal parts and each part is stretched uniformly to length l.
as area of each piece becomes 1/4th of A.
use formula , R = ρl/A
resistance of initial wire , R = ρl/A = 16Ω
resistance of each piece , R' = ρl'/A'
here, l' = l and A' = A/4
so, R' = ρl/(A/4) = 4(ρl/A) = 4R = 64Ω
hence, resistance of each piece of wire is 64Ω.
now all pieces are joined in parallel combination.
so, 1/Req = 1/R' + 1/R' + 1/R' + 1/R'
= 4/R'
= 4/64Ω = 1/16Ω
hence, Req = 16Ω
Answer:16 ohms
Explanation:
The resistance of each part,before stretching ,is 4 ohms and after stretching to 4 times its length,becomes 4×(4)^2=64 ohms.