Question 10
(a)
State what is meant by an electric current.
[1]
(b)
A metal wire has length L and cross-sectional
area A, as shown in Fig.
6.1.
Fig. 6.1
I
is the current in the wire,
n
is the number of free electrons per unit volume in the wire,
v
is the average drift speed of a free electron and
e
is the charge on an electron.
(i)
State, in terms of A, e, L
and n, an expression for
the total charge of the free electrons in the wire. [1]
(ii)
Use your answer in (i) to show that the
current I is given by the equation
I = nAve.
[2]
(c)
A metal wire in a circuit is damaged. The resistivity of the metal
is unchanged but the cross-sectional area of the wire is reduced over a length
of 3.0 mm, as shown in Fig. 6.2.
Fig. 6.2
The wire has diameter d
at cross-section X and diameter 0.69 d
at cross-section Y.
The current in the wire is 0.50 A.
(i)
Determine the ratio
average drift speed of free electrons at cross-section Y
average drift speed of free electrons at cross-section X
[2]
(ii)
The main part of the wire with cross-section X has a resistance per
unit length of
1.7 × 10-2 Ω m-1.
For the damaged length
of the wire, calculate
1.
the resistance per unit length, [2]
2.
the power dissipated. [2]
(iii)
The diameter of the damaged length of the wire is further decreased.
Assume that the current in the wire remains constant.
State and explain
qualitatively the change, if any, to the power dissipated in the damaged length
of the wire. [2]
[Total: 12]
Reference: Past Exam Paper – November 2017 Paper 22 Q6
Solution:
(a)
Electric
current is the flow of charge carriers.
(b)
(i)
{n
is the number of electrons per unit volume
A×L = volume
So,
nAL is the number of electrons
Charge
of 1 electron = e
Total
charge = number of electrons ×
charge of 1 electron}
nALe
(ii)
(t is time taken for electrons to move length L)
Current I = Q / t
I = nALe / t
or
{Speed
v = distance / time v = L
/ t so, t = L / v}
I = nALe / (L / v)
or
I = nAvte / t and I = nAve
(c)
(i)
{ I = nAve so, v = I / nAe
The drift velocity v is inversely proportional
to the cross-sectional area A
Ratio =
v at cross-section Y / v at cross-section X
=
area at X / area at Y}
ratio =
area at X / area at Y
= [πd2 / 4] / [π(0.69d)2 /
4] or d2 / (0.69d)2 or 1 / 0.692
= 2.1
(A = πd2 so, A is proportional
to d2)
(ii)
1.
{R = ρL / A giving R/L = ρ / A Since ρ is the same for wires of the same material.}
R = ρL / A or R / L ∝ 1 / A
{Resistance per unit length: R / L ∝ 1 /
A
For wire X: 1.7 × 10-2 Ω m-1 ∝ 1 / area X eqn (1)
For wire Y: resistance per unit length ∝ 1 / area Y eqn (2)
Divide (2) by (1),}
resistance per unit length = 1.7 × 10-2 × (area at X / area at Y)
= 1.7 × 10-2 × 2.1
= 3.6 × 10-2 Ω m-1
2.
P = I2R or P = V2/ R
{Damaged
length = 3.0 mm
Resistance
per unit length = 3.6 × 10-2 Ω m-1
Resistance of damaged length = Resistance per
unit length × Damaged
length}
R = 3.6 × 10-2 × 3.0 × 10-3 (= 1.08 × 10-4 Ω)
{P = I2R or P = V2/ R }
P = 0.502 × 1.08
× 10-4 or P = (5.4 × 10-5)2 / 1.08 × 10-4
P = 2.7 × 10-5 W
(iii)
{R = ρL /
A}
The cross-sectional area decreases, so the resistance
increases
(P = I2R) So, the power increases
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