Question

A wire of length l and cross section A is doubled on itself. Find the ratio of its initial resistance to it's new resistance

Answer 1

Answer:

heya❤❤✔️✔️

Explanation:

Length of wire = L

Area of cross section = A

Resistance = ρ L/A

\underline{ \bf{when \: it \: is \: doubled \: on \: itself}}

whenitisdoubledonitself

Length will reduce to half .

So, new length = L/2

Area of cross section will be doubled

So, New area of cross section=2A

New Resistance = ρ L/2/2A

Ratio of initial resistance to new resistance

=4:1

So, initial resistance was 4 times the new resistance.

❤________________________❤

Answer 2

Answer:

2:1

Explanation:

for initial resistance

Given,

Length of the wire = l

Cross section = A

As we know that,

the formula of resistance is:

R = pl / A

{Putting the given values}

=> R1 = pl / A

.°. R1 = pl/A

Now,

for new resistance

Given,

Length of the wire = l

Cross section = 2A

As we know that,

the formula of resistance is:

R = pl / A

{Putting the given values}

=> R2 = pl / A

.°. R2 = pl/2A

On dividing R1 and R2, we get

=> R / R/2

=> 2:1