Download QUESTIONS ON OHM`S LAW AND KIRCHHOFF`S LAW A

QUESTIONS ON OHM’S LAW AND
KIRCHHOFF’S LAW
A - EASY TYPE
01What is drift velocity? Define mobility.
02.Define the term ‘resistivity’. Give its SI
unit.
03.Define temperature coefficient of
resistance. What is super conductivity?
04.Define ‘critical temperature’ and ‘critical
field’ for a super conductor.
05.What are thermistors? Mention its uses.
• 06.Define e.m.f. and internal resistance of
a cell.
• 07.State Kirchhoff’s law on electrical
network.
• 08.State Ohm’s law and mention one of its
limitations. 09.What is the condition for the
balanced state of a wheat stone’s
network?
• 10.Give the SI unit of mobility
• 11.What is a node? What is an electrical
loop?
B - APPLICATION TYPE
01.How does resistance of a i) conductor
ii) thermistor vary with the increase in
temperature?
02.What is the colour of the 2nd and 3rd
band in a coded resistor ( 5.4 X 103 )Ω?
03.When current is drawn from a battery,
terminal P.D. is less than e.m.f. Why?
04.What happens to the balanced state of a
wheat stone’s network when cell and the
galvanometer are interchanged
galvanometer is replaced by another
galvanometer of different resistance.
05.A resistor of resistance R is drawn so
that its length is doubled. What is it’s
new resistance.
06.How is resistance of a conductor related
to i) length
ii) cross sectional area.
07.Mention the principle behind Kirchhoff’s
1st and 2nd law
1. Using Ohm’s law, derive the expression
for the electric current through an
external resistor when it is connected to a
source of e.m.f.
2. Derive the expression for branch current
when two resistor are in parallel.
3. Using Kirchhoff’s law, arrive at the
condition for the balanced state of a
wheat stone’s network.
C - QUESTIONS FOR LONG ANSWERS
1. Define ‘strength of the electric current’.
Derive a relation between electric current
and drift velocity.
2. Assuming the expression for electric
current in terms of drift velocity, deduce
Ohm’s law. Define SI unit of resistance.
3. What is effective resistance? Derive the
expression for effective resistance when
three resistors are in series/parallel
NOTE
1. Drift velocity is the average velocity with
which the free electrons are drifting under
the applied electric field in a conductor. As
the temperature increases the collision
becomes more frequent than earlier. i.e.
average relaxation time decreases and
resistivity increases and hence resistance
of a conductor increases.
2. Drift velocity per unit electric field applied
is called mobility.Strength of the electric
current is directly proportional to drift
velocity.
3. Resistance is the opposition offered by
the conductor for the flow of charge.
4.Resistivity of the material of the conductor
is the resistance offered by the conductor
of unit length and unit cross-sectional
area.
SI unit: Ohm meter
5. Resistivity of the material of the
conductor increases as the temperature
increases.
• 06.The ratio of change in resistance per 0C
rise in temperature and its original resistance
at 00C is called temperature coefficient of
resistance. If the value of is negative, this
means as the temperature increases,
resistance decreases.
• 07.Critical temperature is the temperature at
which the resistance of the given material
abruptly falls to zero. Critical magnetic field is
the minimum magnetic field at which a super
conductor returns to its normal state when
the temperature is kept constant (Applied
external magnetic field has an adverse effect
on super conductivity.)
• Let L be the length of the conductor, A be
the cross-sectional area of the conductor.
Let V be the P.D. across the ends of the
conductor, n be the number conduction
electrons present in the conductor per unit
volume. Let e be the charge and m be the
mass of the electron. Let Vd be the drift
velocity of the free electron and t be the
time taken by the electron to travel the
length L. We know, total charge of the free
electron =q =n A L e.
• And
,
q
nALe
I =
=
A = nAeV
t
t
L
where Vd=
t
We know,
I = n e A Vd
-----------(1)
but
Vd =
a ´T
d
• where T = average relaxation time
EeT
\ Vd =
m
F Ee
a= =
m m
Sub (2) in (1) we get
I=
but E
2
ne AET
---------- (2)
-----------(3)
m
=
V
L
numerically
\I =
2
ne AVT
mL
=
2
ne TA
mL
.V
mL
L
V = 2 I = r × I = RI
A
ne TA
V
\ I =
R
where
OR
I a
m
r= 2
ne T
is a constant called
resistivity of the material of the
conductor
V
L
R = r
A
•
•
is a constant called resistance of
the conductor
Let R1, R2, R3 be the resistances of
three resistors connected in series.
Let V1, V2, V3 be the P.D. across them
respectively. Since they are in series,
the same current I flows through each.
• For series combination of resistors,
• P.D. between the ends of the
combination is equal to the sum of the
P.D. across the individual resistors.
\
\
V= V1 +V2 +V3 But, V1 =IR1
V2 = IR2
V3 = IR3
V = I (R1 + R2 + R3) ---------------(1)
• For the effective resistor Rs, the P.D. is V
and the current flowing is I.
• V = I Rs -----------------------------(2)
From (1) and (2) we have,
I Rs = I (R1 +R2 +R3)
OR
Rs = R1 +R2 +R3
Let R1, R2, R3 be the resistances of three resistors
connected in parallel. Let I be the current drawn from the
cell.Let I1, I2, I3 be the currents flowing through
R1, R2, R3 respectively. Since the resistors are in
parallel, the P.D. across each resistor is same.
• For parallel combination of resistors,
• the current drawn from the cell is equal
to the sum of the currents flowing
through individual resistor.
\I = I1 + I2 + I3
V
But, I1 =
R1
V
I2 =
R2
V
I3 =
R3
•
æ 1
1
1 ö
÷÷
I = Vçç
+
+
è R1 R 2 : R 3 ø
------(1)
For the effective resistance RP ,
the P.D. is V and the current is I
V
I =
R P
---------(2)
• From (1) and (2) we have,
•
V
æ 1
1
1 ö
+
+
= V ç
÷
R2
R3 ø
RP
è R1
1
RP
=
1
1
1
+
+
R1 R 2 R 3
•
•
•
•
•
Common mistakes in the above
derivations
Use of the word ‘potential difference at a
point instead of potential difference
between two points.
Use of the word current across the
conductor instead of current flowing
through the conductor.
No mention of the direction of the current.
Students may also notice that effective
resistance is the resistance of the single
resistor which produces the same effect as
produced by the combination of
resistances.
Let R be the resistance of the external resistor
connected to a cell of e.m.f E,internal
resistance ‘r’. Let I be the strength of the current
flowing through R. Let V be the terminal P.D.
We know, work done in shifting unit positive
charge once round the closed circuit is e.m.f.
of the cell.
• Hence e.m.f. is sum of work done in
shifting unit positive charge once
through R and r.
• But, work done in shifting unit positive
charge through R is P.D. across R
and through r is P.D. across r.
• Let V and V1 be the P.D. across R
and r respectively.
• E =V + V1 (from the principle of
conservation of energy)
• From Ohm’s law, we have
• V = I R and V1 = I r
• E = I R + I r = I (R + r)
• OR
•
I=
E
R+r
Note: ( i) V= I R=
ER
R+r
is the terminal P.D.
•
V=
E
r
1+
R
• Hence, E = V when r = 0 or R = ¥
i.e. e.m.f. of a cell is measured by the P.D.
across the cell when the circuit is open ie R = ¥
(ii) A part of the e m f is used to drive the charge
through internal resistance. Since r is not
equal to zero for any cell, e.m.f. is greater
than terminal P.D. when current is drawn from the cell
•
•
•
Generally noticed mistakes:
Without explaining why E = V + V¢,
students substitute V = I R, V¢ = I r.
No mention of the value of the current
and its direction
1st LAW: Algebric sum of the electric
currents meeting at a node is zero
I1 – I2 – I3 = 0
Students generally draw large
number of currents and write the
sign of the currents wrongly in
the above equation.
2ND LAW: In a closed
electric network, the
algebraic sum of the e.m.f.
is equal to algebraic sum of
the IR products in that
network.
In the network ABCA
I1 R1 + I2 R2 - I3 R3 = - E1 + E2
•
•
•
General mistakes done by the
students.
All IR products and e.m.f.s are
taken positive without knowing
when are they negative.
Without mentioning the network IR
products and e.m.f. s are equated.
•
•
•
PRECAUTIONS:
While solving the problems on
Kirchhoff’s Law, it is better to retain the
solved values of the currents I1 & I2 in
the fraction form. This helps to check
the correctness of the current values.
In a given problem, if the cells are of
different e.m.f.s and also in parallel, that
problem can be solved using only
Kirchhoff’s Law.
•
While solving the problems on Ohm’s
law a) if internal resistance is there for a
given cell, use the equation
e.m.f.in
the
circuit
I=
R E +r
where RE = effective external resistance
b) If terminal P.D. is given,
use I
V
=
RE
• P, Q, S and R are the four resistances
connected in cyclic order to form a
Wheatstone’s network ACDB as shown.
Let I1, I 2, I 3, I 4 and I g be the currents
flowing through P, R, Q, S and the
galvanometer of resistances G
respectively in the directions as shown.
• From Kirchhoff’s 1st law we have,
• I1 = Ig+I3 at the node C – (1) and
• I2+Ig = I4 at the node D – (2)
•
•
•
•
•
•
From Kirchhoff’s 2nd law we have,
I1P + IgG – I2 R = 0 for the mesh ACDA,
--------- (3)
I3Q – I4 S – Ig G = 0 for the mesh CBDC
-----------(4)
For the balanced state of the
Wheatstone’s network, the current
flowing through the galvanometer must
be zero
• i.e. Ig = 0
• I1 = I3 from (1)
and I2 = I4 from (2)
• I1 P = I3 R from (3) and I2 Q = I4 S from (4)
I2R
• (3) gives I1 P
=
(4)
I Q I S
3
• OR
4
P R
=
Q S
•
•
•
Note:
To verify the balanced state of the
given Wheatstone’s network, match
the product of 1st and 3rd
resistances with the product of 2nd
and the 4th resistances when four
resistances are given in cyclic order.
Cell and the galvanometer can be
interchanged without effecting the
balanced condition of the network.