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Question

We have a copper wire resistance R this wire is pulled so it's length get double find the new resistance in term of original resistance.


Solution

R1=pL1/a1

volume of wire renains same so

A1*L1=A2*2L1

A1=2A2

A1/2=a2

Rnew=p2L1/A1/2

   =p4L1/A2

    =4pL1/A2

    =4*R1

so new resistance is 4 times that of old resistance

Another method 

To do this, we the following formula

R=ρlA .....(1)
where R is resistance wire of length l, A its cross-sectional area and ρspecific resistance of material of wire, copper in the given question.

When length is doubled, volume remains same. If Anew is new area of cross section then

l×A=(2l)×Anew
⇒Anew=lA2l=A2

Remembering that ρ remains same, inserting values in (1) we get

Rnew=ρ2lA2
⇒Rnew=4ρlA

Using (1) we get

Rnew=4R

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