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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
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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