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 | 29th 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 | 29th Dec, 2021, 10:25: AM