The resistance of 2m of constantan wire is 10 Ω. What is the resistance of 4m of constantan wire whose cross-sectional area is twice as large?
My solution:
let x be the x-sectional area
2m ---> 10 Ω ---> x
4m ---> 20 Ω ---> x
4m ---> ? Ω ---> 2x
Ans: still 10 Ω.
There is no answer given but i dont think my answer is correct...can u please help me check? Thank you.
For the original wire , let the length(2m) of the wire be represented as l , while the cross sectional area of it be represented as A , which would form the original equation of R = (rho x l)/A - (1)
For the next wire, let its resistance be represented as r , while its length(4m) be represented by 2l and its cross sectional area be represented by 2A.
The resistivity( rho) of the wires are the same, since the same type of wire is used.
Thus, the resistance of the 2nd wire can be expressed as r = (rho x 2l)/(2A) = (rho x l)/ A
At this step , the '2' at the numerator and the denominator is cancelled.
By substituting the (1) at this point , an equation of r = R would be derived, while R has a value of 10 ohms .
Finally , since r which represents the resistance of the 2nd wire is equal to R , therefore the latter has a resistance of 10 ohms as well.
Hope this helps ;)
ok thanks^^