please solve these 2 questions
i.e question 8,and 9
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Resistance ‘R = 6ohm’ is related to length ‘L’, area of cross-section ‘A’ and resistivity ‘ρ’ as,
R = ρL/A
When L is doubled ,
As we know that the volume of the wire remains same.
old volume = new volume
AL = A'L' ....(i) ( as volume = area x length )
L' = 2L
So, from (i)
A' = A/2
So the new resistance is ,
R' = ρL'/A'
or
R' = ρ(2L)/(A/2) = 4 (ρL/A)
=> R' = 4R = 4 x 6ohm = 24 ohm .
Hope it helps u
Thanks....
R = ρL/A
When L is doubled ,
As we know that the volume of the wire remains same.
old volume = new volume
AL = A'L' ....(i) ( as volume = area x length )
L' = 2L
So, from (i)
A' = A/2
So the new resistance is ,
R' = ρL'/A'
or
R' = ρ(2L)/(A/2) = 4 (ρL/A)
=> R' = 4R = 4 x 6ohm = 24 ohm .
Hope it helps u
Thanks....
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