I need help understanding how to answer these types of problems: Which of the following wires would have the LEAST RESISTANCE? See below. .
a)an aluminum wire 5 cm in length and 5 cm in diameter
b)an aluminum wire 10 cm in length and 3 cm in diameter
c)an aluminum wire 10 cm in length and 5 cm in diameter
d)an aluminum wire 5 cm in length and 3 cm in diameter
Question #2: Which of the following wires would have the GREATEST resistance?
a)a copper wire 10 cm in length at 32 degrees celcius
b)a copper wire 10 cm in length at 10 degrees celcius
c)a copper wire 5 cm in length at 32 degrees celcius
d)a copper wire 5 cm in length at 10 degrees celcius
I need help understanding how to answer/do these questions. Thanks ;)
2 Answers | Add Yours
for the second part,
The resistivity of a material increases with the increasing temperature and it can be linearly approximated as follows.
rho = rho_0(1+alpha(T-T_0))
or simple we can say Rho is directly proportionate to T
So rho = kT
T in kelvins,
R = kT * L /A
assuming all the copper wires have the same cross sectioanl area, A then R = C * T * L
Therefore the wire with the greatest ressitance would be one with the greatest T*L value,
The wire with greates T*L value is wire (1) (Which has 305*10)
Therefore the wire with greatest resistance is wire (1)
This question can be approached as below.
The resistance (R) is related to resistivity (rho), length (L) and cross sectional area (A) by the following equation,
R = rho * L /A
first part -- least resistance, since these are same material (Al) we can ignore it and conside about L/A ratio. Therefore the least resistance wire would be the one with the lowest L/A ratio.
We can calculate the L/A ratio as follows for each wire, We can take the diameter*diameter as the area since (A = pi * d^2 /4)
(i) L/A = 5/25 = 0.2
(ii) L/A = 10/9 = 1.11
(iii) L/A = 10/25 = 0.4
(iv)L/A = 5/9 = 0.555
Therefore the least resistance wire will be (i) with 5 cm length and 5 cm diameter.
You are only supposed to ask 1 question at a time. So I will answer only question 1 here and 2nd one I will add my answer.
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