Respuesta :
Answer:
The ranking of current from the most to the least is [tex]I_{3} >I_{4}> I_{2}> I_{1}[/tex]
Explanation:
Case Number 1 :
Drift velocity=[tex]V_{drift}[/tex]= 5v
Radius=r= 1r
Square of radius=[tex]r^{2}[/tex]= 1
Current=[tex]I=C(V _{drift} *r^{2} )[/tex]=C(5v[tex]r^{2}[/tex])
Case Number 2 :
Drift velocity=[tex]V_{drift}[/tex]=0.5v
Radius=r=4r
Square of radius=[tex]r^{2}[/tex]=16
Current=[tex]I=C(V _{drift} *r^{2} )[/tex]= C(8v[tex]r^{2}[/tex])
Case number 3 :
Drift velocity=[tex]V_{drift}[/tex]=2.5
Radius=r=2r
Square of radius=[tex]r^{2}[/tex]=4
Current=[tex]I=C(V _{drift} *r^{2} )[/tex]= C(10v[tex]r^{2[/tex])
Case number 4:
Drift velocity=[tex]V_{drift}[/tex]=1v
Radius=r=3r
Square of radius=[tex]r^{2}[/tex]=9
Current=[tex]I=C(V _{drift} *r^{2} )[/tex]=C(9v[tex]r^{2[/tex])
Hence, The ranking of current from the most to the least is [tex]I_{3} >I_{4}> I_{2}> I_{1}[/tex]
Answer:
So in order of most to least electric current,
Ic > Id > Ib > Ia
Explanation:
Let the wires be represented as A, B, C, D
in the the given order.
So wire A has radius 1 and drift speed = 5V
wire B has radius 4 and drift speed = 0.5V
wire C has radius 2 and drift speed = 2.5V
wire D has radius 3 nd drift speed = 1V
The current in a given wire is directly proportional to the cross seectional area A(πr²) and also to the drift speed v
I = nqvA
Assuming n and q are constant,
I ∝v× r²
Ia = C×5v × r² = 5Cvr²
Ib = C×0.5v × (4r)² = 8Cvr²
Ic = C×2.5v× (2r)² = 10Cvr²
Id = C×1v×(3r)² = 9Cvr²
Where Cis a constant = nqπ
So in order of most to least electric current,
Ic > Id > Ib > Ia