If a conductor is folded 8 times then the resistance will be
Answers
Answer:
Let L be the length and A be the cross sectional area
R=ρ
A
l
=8Ω
l
′
=l/2
A
′
=A/2
R
′
=ρ
A
′
l
′
=ρ
2A
l/2
=
4
1
ρ
A
l
=
4
1
×8=2Ω
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Answer:
When a conductor is folded 8 times it's resistance will become 1/32 of it's initial resistance.
Explanation:
Let's take the resistance of the wire (conductor) before folding to be 256Ω.
Let the length of the wire be L and the cross-section area of the wire be A.
Since,
R (Resistance) = ρ L/A = 256Ω
Now,
L' ( length after folding the wire 8 times) = L/8
A' ( area after folding the wire 8 times) = 8A
Therefore,
R' ( resistance after folding the wire 8 times)
= ρ L'/ A'
= ρ L/8 ÷ 8A
= 1/32 × ρ L/A
= 1/32 × 256Ω
= 8Ω
Conclusion:
8Ω is 1/32 of 256Ω (resistance before folding the wire).
Therefore, When a conductor is folded 8 times then the resistance will become 1/32 of it's initial resistance.