A wire of resistance 32 ohm is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire and the percentage change in resistance?
Answers
Answer:The wire of resistance 16 ohm. It is melted and drawn into a wire of half its length. Calculate the resistance of the new wire. What is the percentage change in its resistance?
Explanation:Samira, simple
R depends on L so L and R both halve
R depends on 1/A so A doubles and R halves
Now new R = 16 x 0.5 x 0.5 = 4 ohm
Change is 16–4 = 12
% change = 12/16 = 0.75 = 75%
or
The wire of resistance 16 ohm. It is melted and drawn into a wire of half its length. Calculate the resistance of the new wire. What is the percentage change in its resistance?
This is an important question in the UK.
In our power systems around buildings, we use a ‘ring main’ system of wiring.
So from the fuse box, 2 live wires, 2 neutral wires, and 2 earth wires. And these are connected so the lives form a ring, as do the neutrals and the earth wires.
Now if you measure the resistance of the live wire around the ring, you get say 0.5 ohms. But from the fuse to the furthest point on the ring, with everything connected, the resistance would be 0.125 ohms - assuming that the furthest point is actually mid-point on the wire. (In practice it varies)
What you can see is that from the fuse to the mid-point, the wire is half the total length, and because there are 2 routes round the ring, the cross sectional area is effectively double.
As to why we like ring mains in the UK - that’s ancient history - someone once told me it was due to copper shortages in the post war period, but I don’t know if this is true.
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