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wire is folded and doubled upon itself from centre.what is percentage change in its resistance

Posted by Raghvi Sharma 6 years ago

CBSE > Class 10 > Science

2 answers

Divyanshu Tripathi 6 years ago

Let initial length and area of wire be 'l' and 'A'.Then,initial resistance,R=p (l÷A).--------(1) Now,when the wire is folded and doubled,its length becomes half and area gets doubled.Thus,l'=l÷2 and A'=2A.Then, New resistance=p l'÷A' =p l/2÷2A =p l÷4A =p l÷A ×1/4 =R/4 (from 1) Thus,change in resistance=R-R/4 =3R/4 Percentage change=change in resistance/initial resistance×100 =3R/4÷R×100 =3/4×100 =75%(decrease)

2Thank You

Venkada Laxmi 6 years ago

Remains same

3Thank You

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