Question

An aluminum wire has a resistance of 48 Ω.  What is the new resistance if:          

a.) Length of wire is doubled.

b.) Cross-sectional area is halved.

c.) Length of wire is halved and area is quartered.

d.) Length is doubled and area is doubled.

Answer 1

a. R=pL/A

R1 = p2L/A

=2pL/A  Resistande is doubled = 96 ohm

b. R2= pL/.5 A

=2pL/A Resistance is doubled = 96 ohm

c. R3 =( pL/2)/A/1/4

=4pL/2A

=2pL/A Resistance is doubled = 96 ohm

d.R4 = p2L/2A

=pL/A  Resistance stays the same = 48 ohm

Answer 2

We know that the resistance of the wire is 48 ohms.

Our formula is R= PL/A

a) R is directly proportional to the length hence the R will be double: R= 2R which is  2*48 = 96 ohms

b) Resistance inversely proportional to the area, therefore: R= PL/A

= PL/A/2

= PL* 2/A,

= 2R Which is 2*48= 96 ohms

c) = 1/2PL / A/4

= PL/2* 4/A

= 2R , 2*48= 96 Ohms

d) R= 2PL/2A , R = 48 Ohms