L1=2m
R1=4ohms
D=2d
L2=3×2=6m
A1=πd2\4=(πr2)
A2=π(2d)2/4=π4d2/4
a wire of lengh 2meter has a resistance of 4 ohms,obrain the resistance if specific resistance is doubled,diameter is doubled and the length is made three times of the first.
2 answers
R₁=ρ₁L₁/A₁=ρ₁L₁/(πd₁²/4)
R₂=ρ₂L₂/A₂=ρ₂L₂ /(πd₂²/4)
R₂/R₁=ρ₂L₂ πd₁²/ρ₁L₁πd₂² =
=2ρ₁3L₁(d₁)²/ρ₁L₁(2d₁)² =2•3•/4= 3/2=1.5
R₂ = 1.5R₁ = 1.5•4 = 6 Ω
R₂=ρ₂L₂/A₂=ρ₂L₂ /(πd₂²/4)
R₂/R₁=ρ₂L₂ πd₁²/ρ₁L₁πd₂² =
=2ρ₁3L₁(d₁)²/ρ₁L₁(2d₁)² =2•3•/4= 3/2=1.5
R₂ = 1.5R₁ = 1.5•4 = 6 Ω