CBSE Class 11-science Answered
f r of a wire is increased by 50 and potential difference decreased by 50% then find the percentage change in current
Asked by imayushanand223 | 29 Dec, 2021, 08:49: AM
Expert Answer
Resistance of wire , R = ρ × ( L / A) = ρ × [ L / ( π r2 ) ]
where ρ is resistivity of material , L is length of wire , A is area of crs section and r is radius of wire
hence, resistance of wire is iversly proportional to quare of radus.
R = C / r2
where C is constant if material of wire and length of wire are fixed .
If radius is increased by 50% , then new radius r' = 1.5r
New resistance = R ' = C / (1.5 r )2 = ( 4 / 9 ) R
If voltage is decreased by 50% , then current with new resistance R ' is determined as
I ' = [ (1/2) V ] / [ ( 4 / 9) R ] = ( 9 / 8 ) ( V / R ) = 1.125 I
where I is current with initial voltage V and initial radius of wire r .
Hence we see that current increases by 12.5 %
Answered by Thiyagarajan K | 29 Dec, 2021, 10:25: AM
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