Question

of uniform cross section and length L has a resistance of 4 ohm the wire is cut into 4 equal pieces each piece is when stretched to length L there after the four wires are joined in parallel calculate the net resistance

Answer 1

Answer:

Length of wire = L

resistance of wire = 16 Ohm = R ( Let)

now, wire cut four equal parts

e.g Length of each part = L/4

we know,

R = ρL/A

where ρ is resistivity ( constant )

when we cut the wire , cross section area won't be change .

hence, R is directly proportional to L

so, each part have resistance = R/4

again,

According to question ,

each part stretched and length increases L/4 to L

[ when , we stretched the wire , volume of wire is constant ]

R = ρL²/AL

AL = volume of each wire ( constant)

hence, R is directly proportional to L²

L is increased to four times so,

R' = 16( R/4) = 4R

hence, each of wire have 4× 16 = 64 ohm resistance .

now,

all wire connected in parallel .

e.g 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ = 1/Req

1/Req = 1/64 + 1/64 + 1/64 + 1/64

1/Req = 4/64

1/Req = 1/16

Req = 16 ohm

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Answer 2

Answer:

Length of wire = L

resistance of wire = 16 Ohm = R ( Let)

now, wire cut four equal parts

e.g Length of each part = L/4

we know,

R = ρL/A

where ρ is resistivity ( constant )

when we cut the wire , cross section area won't be change .

hence, R is directly proportional to L

so, each part have resistance = R/4

again,

According to question ,

each part stretched and length increases L/4 to L

[ when , we stretched the wire , volume of wire is constant ]

R = ρL²/AL

AL = volume of each wire ( constant)

hence, R is directly proportional to L²

L is increased to four times so,

R' = 16( R/4) = 4R

hence, each of wire have 4× 16 = 64 ohm resistance .

now,

all wire connected in parallel .

e.g 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ = 1/Req

1/Req = 1/64 + 1/64 + 1/64 + 1/64

1/Req = 4/64

1/Req = 1/16

Req = 16 ohm

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THANK YOU

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Explanation: