(i) Derivation of the expression for drift velocity
Deduction of Ohm's law
(ii) Name of quantity and justification
Let an electric field E be applied the conductor.
Acceleration of each electron is
a=−‌

eE

m

Velocity gained by the electron
v=−‌

eE

m

t
Let the conductor contain n electrons per unit volume. The average value of time ' t ', between their successive collisions, is the relaxation time, ' τ '.
Hence average drift velocity
vd=‌

−eE

m

τ
The amount of charge, crossing area A, in time ∆t is neAvd∆t=I∆t
Substituting the value of vd, we get
I∆t=neA(‌

e2Eτ

m

)∆t
∴‌‌I=(‌

e2Aτn

m

)
J=σE,(σ=‌

e2τn

m

‌ is the conductivity ‌)
But I=JA, where J is the current density
⇒J=(‌

e2τn

m

)E
⇒J=σE
This is Ohm's law
[Note: Credit should be given if the student derives the alternative form of Ohm's law by substituting E=‌

V

ℓ

]
(ii) Electric current well remain constant in the wire.
All other quantities, depend on the cross sectional area of the wire.
OR
(i) Statement of Kirchhoff's laws Justification
(ii) Calculation of (A) current drawn and (B) Power consumed
(i) Junction Rule: At any Junction, the sum of currents, entering the junction, is equal to the sum of currents leaving the junction.
Loop Rule: The algebraic sum, of changes in potential, around any closed loop involving resistors and cells, in the loop is zero.
Σ(∆V)=0
Justification : The first law is in accord with the law of conservation of charge.
The Second law is in accord with the law of conservation of energy.
(ii) Equivalent resistance of the loop
R=‌

r

3

Hence current drawn from the cell
I=‌

E

r 3 +r

=‌

3E

4r

Power consumed
P‌=I2(‌

r

3

)
‌=‌

9E2

16r2

×‌

4r

3

=‌

3E2

4r

[Note: Award the last 1

1

2

marks for this part, if the calculations, for these parts, are done by using (any other) value of equivalent resistance obtained by the student.)