A given length of awire is doubled on itself and this process is repeated once again . By what factor does the resistance of the wire change ?
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Answered by
936
Resistance of a wire is directly proportional to the length of the wire. As its length increases it's resistance increases.
Let x be the resistance of a wire of length L
If length of a wires is doubled to 2L its resistance is doubled to 2x
If it is doubled again, the length of the wire becomes 4L and it's resistance becomes 4x.
Thus if a given length of a wire is doubled on itself and this process is repeated once again it's resistance increases by a factor of 4
Let x be the resistance of a wire of length L
If length of a wires is doubled to 2L its resistance is doubled to 2x
If it is doubled again, the length of the wire becomes 4L and it's resistance becomes 4x.
Thus if a given length of a wire is doubled on itself and this process is repeated once again it's resistance increases by a factor of 4
Answered by
587
Let the resistance of the wire be R.
Let L be the length of the wire.
When the wire is doubled , Then L will become L/2.
As length of the wire is reduced to L/2, area increases by 2 times.
Again the process is repeated, hence L will become L/4As length of the wire reduced by L/4, area increases by 4 times.
so new reistance will be R/4
The new resistance of the wire will be effective resistance of the resistors in parallel.
1/Reff=1/R1+1/R2+1/R3+1/R4
=1/R/4+1/R/4+1/R/4+1/R/4=16/RReff=R/16Thus the resistance of the wire reduces by 16 times
Let L be the length of the wire.
When the wire is doubled , Then L will become L/2.
As length of the wire is reduced to L/2, area increases by 2 times.
Again the process is repeated, hence L will become L/4As length of the wire reduced by L/4, area increases by 4 times.
so new reistance will be R/4
The new resistance of the wire will be effective resistance of the resistors in parallel.
1/Reff=1/R1+1/R2+1/R3+1/R4
=1/R/4+1/R/4+1/R/4+1/R/4=16/RReff=R/16Thus the resistance of the wire reduces by 16 times
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