A nichrome wire has a resistance of 10 ohm .find the resistance of another nichrome wire,whose length is three times and area of cross-section four times the first wire.
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we know that , if R1 be the resistance of nichrome wire ,
then R1= pL1/A1; where p= (raw i.e material of the wire) , L1 = length of the wire , A1= area of cross section of wire .
then,
let R2 be the resistance of new nichrome, L2 be thelength , A2 be the area of cross section
given that, L2=2L1 and A2=4A1 ,
then R2 = pL2/A2
note : material will be same as before .so p(raw) remains same.
R1=10 (given)
R1/R2= [pL1/A1 ] / [ pL2/A2]
here , replace L2=3L1
A2=4A1
10/R2= [pL1/A1] / [p3L1/4A1]
10/R2=4/3
R2=30/4=7.5 ohm
then R1= pL1/A1; where p= (raw i.e material of the wire) , L1 = length of the wire , A1= area of cross section of wire .
then,
let R2 be the resistance of new nichrome, L2 be thelength , A2 be the area of cross section
given that, L2=2L1 and A2=4A1 ,
then R2 = pL2/A2
note : material will be same as before .so p(raw) remains same.
R1=10 (given)
R1/R2= [pL1/A1 ] / [ pL2/A2]
here , replace L2=3L1
A2=4A1
10/R2= [pL1/A1] / [p3L1/4A1]
10/R2=4/3
R2=30/4=7.5 ohm
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