The diagram below shows two current-carrying wires joined together in the middle
ID: 1630062 • Letter: T
Question
The diagram below shows two current-carrying wires joined together in the middle. Both wires have a wider diameter at their outer edges and a narrower diameter in the middle. The wider sections have twice the diameter and twice the length of the narrower sections. The wire on the left is made of aluminum and the one on the right is made of copper. (a) Rank the current in each wire, greatest first. Explain briefly. (b) Rank the voltage across each wire, greatest first. Explain briefly. (c) Rank the electric field in each wire, greatest first. Explain briefly. (d) Rank the drift velocity in each wire, greatest first. Explain briefly.Explanation / Answer
A.
As we see in diagram, two wires are connected in series.
So, current remain same for both wires.
B.
As we know, potential drop across a wire is V=IR.
Current in each wire is same. So, potential drop across each wire depend on resistance.
R = L/A
= resistivity of wire
L = Length of wire
A = Area of cross section of wire.
As, in both wire length and area is same. So, resistance depend on resistivity
cu=1.68 *10-8 m
Al= 2.68*10-8 m
Hence, voltage drop across aluminum wire is greater than that for copper wire.
So, VAl > VCu
C.
As, J = E
= electric conductivity
J = current density
E = J/
For both the wire current density is same but electric conductivity for copper wire is greater than that for aluminum wire.
So, EAl > ECu
D.
As, I = Anevd
A = area of cross section wire
n = no. of free electron per unit volume
e = charge on electron
vd = drift velocity of electrons
As, I, A and e are constant. So, drift velocity depend on no. of free electron per unit volume
nAl > nCu
Hence vd of copper > vd of aluminum.