TOPIC 18 : ELECTRIC CURRENT AND
DIRECT-CURRENT CIRCUITS
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
18.9
Electrical Conduction
Ohm’s law and Resistivity
Variation of resistance with temperature
Electromotive force (emf), internal
resistance and potential difference
Electrical energy and power
Resistors in series and parallel
Kirchhoff’s Laws
Potential divider
Potentiometer and Wheatstone Bridge
1
SUBTOPIC :
18.1 Electrical Conduction
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Describe microscopic model of current.
b) Define and use electric current formulae, I 
dQ
dt
2
18.1 Electrical Conduction
• Conductors contain many free electrons and move randomly.
• If a continuous wire is connected to the terminal of a battery, the
potential difference between the terminals of the battery sets up an
electric field inside the wire and parallel to it, directed from the positive
toward the negative terminal.
• Thus free electrons are attracted into the positive terminal (are
forced to drift in one direction).
• This direction is in the direction opposite to the field, E.
• The velocity of these free electrons is called drift velocity.
3
battery
18.1 Electrical Conduction
• The drift velocity vd of the free electrons is the mean velocity
of the electrons parallel to the direction of the electric field
when a potential difference is applied.
• Consider the circuit such as that in figure below, the battery
creates an electric field within and parallel to the wire,
directed from the positive toward the negative terminal.
wire
battery
4
18.1 Electrical Conduction
• Thus free electrons at one end of the wire are attracted into
the positive terminal, and at the same time, electrons leave
the negative terminal of the battery and enter the wire at the
other end.
• There is a continuous flow of electrons through the wire that
begins as soon as the wire is connected to both terminals.
• However, when the conventions of positive and negative
charge were advised two centuries ago, it was assumed that
positive charge flowed in a wire.
• For nearly all purposes, positive charge flowing in one
direction is exactly equivalent to negative charge flowing in
the opposite direction.
• Today we still use the historical convention of positive current
when discussing the direction of a current. So when we speak
of the current in a circuit, we mean the direction positive
5
charge would flow.
5.1 Electrical Conduction
Electric Current, I
• Electric current is defined as the amount of charge that
passes through the wire’s full cross section at any point per
unit time ( the rate of charge flow through a conductor).
Q
•The average current , I is defined as I 
.
t
•The instanstaneous current is defined as
dQ
I
dt
•The steady current is defined as
Q
I
t
• Unit of I is A (ampere).
6
Example 18.1
5.1 Electrical Conduction
A wire carries a current of 1.5 A.
a) How much charge flows through a point in the wire
in 5.0 s
b) How many electrons cross a given area of the wire
in 1.0 s ?
7
SUBTOPIC :
18.2 Ohm’s Law and Resistivity
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) State and use Ohm’s Law.
RA
b) Define and use resistivity formulae,  
l
8
18.2 Resistivity and Ohm’s Law
18.2 Ohm’s Law and Resistivity
Ohm’s Law
• Ohm’s law states that the potential difference across a
conductor, V is directly proportional to the current, I
through it, if its physical conditions and the temperature
are constant.
V I
V
 constant
I
V
 R  V  IR Ohm's Law
I
9
18.2 Resistivity and Ohm’s Law
V
V
 constant
I
V
R
I
Figure A
V
I
Figure B
I
• Ohmic conductors are conductors which obey Ohm’s
law. Examples: pure metals. (Figure A)
• Non-ohmic conductors do not obey Ohm’s law.
Example: junction diode. (Figure B)
10
18.2 Resistivity and Ohm’s Law
Resistivity
• Resistivity is a measure of a material’s ability to oppose
the flow of electric current through the material.
• Resistivity is defined as the resistance of a sample of the
material of cross-sectional area 1 m2 and of length 1m.
• It is a constant value.
RA


• Its formulae is given by
.
l
where l = length of the conductor (m)
A = area of cross-section of the conductor (m-2)
•
•
•
•
Its unit is Ωm.
Its value depends on the material.
All conductors have smaller resistivity.
Insulators have larger resistivity.
11
Example 18.2
5.2 Resistivity and Ohm’s Law
A wire (length=2.0 m, diameter=1.0 mm) has a resistance of
0.45 Ω. What is the resistivity of the material used to make the
wire ?
12
Example 18.3
5.2 Resistivity and Ohm’s Law
What voltage will be measured across a 1000-Ω resistor in a
circuit if we determine that there is a current of 2.50 mA flowing
through it ?
13
SUBTOPIC :
18.3 Variation of Resistance with
Temperature
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Explain the effect of temperature on electrical
resistance in metals and superconductors.
b) Define and use temperature coefficient of resistivity, α.
c) Apply resistance R= Ro [1 +α (T-To)].
14
18.3 Variation of Resistance with
Temperature
Metal
• The resistance of a metal (conductor) depends on
a) the nature of the material,
(, resistivity)
b) the size of the conductor,
(l, the length and A, cross-sectional area)
c) the temperature of the conductor.
• The resistance of metals increases with increasing
temperature. (T↑, R↑)
• As temperature increases, the ions of the conductor
vibrate with greater amplitude.
15
18.3 Variation of Resistance with Temperature
• More collisons occur between free electrons and ions.
• These electrons are slowed down thus increases the
resistance.
• The resistance of a metal can be represented by the
equation below
R=Ro[1+α(∆T)] , R-Ro = ∆R
∆R=Ro+α∆T
where R = the resistance at temperature T,
Ro= the resistance at temperature To = 20o C or 0oC,
 = the temperature coefficient of resistance ( oC-1)
16
18.3 Variation of Resistance with Temperature
• Temperature coefficient of resistance , α is defined as the
fractional change in resistance per Celsius degree.
R
Ro

T
• α is a constant value and it is depends on the material.
17
18.3 Variation of Resistance with Temperature
Example 18.4
A platinum wire has a resistance of 0.50 Ω at 0oC. It is
placed in a water bath, where its resistance rises to a
final value of 0.60 Ω . What is the temperature of the
bath ? (α = 3.93 x 10-3 oC -1)
18
18.3 Variation of resistance with temperature
Example 18.5
A narrow rod of pure iron has a resistance of 0.10 Ω at 20oC.
What is its resistance at 50 oC ? (α = 5.0 x 10-3 oC -1)
19
18.3 Variation of Resistance with Temperature
Superconductor
• As the temperature decreases, the resistance at first
decreases smoothly.
(T↓, R↓)
• At a certain critical temperature Tc (4.2 K for mercury)
the resistance suddenly drops to zero.
R
R
superconductor
metal
T
T
Graph of resistance R against temperature T
20
SUBTOPIC :
18.4 Electromotive Force (emf), Internal
Resistance and Potential Difference
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Define emf , ε.
b) Explain the difference between emf of a battery and
potential difference across the battery terminals.
c) Apply formulae V= ε –Ir.
21
18.4 Electromotive Force (emf), Internal
Resistance And Potential Difference
What is electromotive force,emf (ε or ξ)?
• The e.m.f of a source is the work done per unit charge.
• The e.m.f of a source is the p.d across the terminals of
the source in open circuit (no current is flowing, I = 0).
• The e.m.f of a battery is the potential difference across
its terminal when it is not connected to a circuit.
• The e.m.f of a source is defined as the electrical energy
that generated by a source so that the charges can flow
from one terminal to another terminal of the source
through any resistor.
• SI unit : Volt (V)
22
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
What is internal resistance?
• In a cell or battery, the negative ions are attracted by
anode and the positive ions are attracted by the cathode.
• The flow of these ions produces current.
• However the collisions between the ions and the
recombination of opposite ions reduce the flow of
current. This resistance in the cell is called internal
resistance, r.
23
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
• Suppose a battery of emf, ε and internal resistance r
is connected to an external resistor, R.
• Total resistance in the circuit is (R + r).
24
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
• The e.m.f of this battery is given as
  I (R  r)
  Ir  IR
  Ir  Vab
Vab  Vdc
Vab = Vb – Va = terminal voltage (potential difference across
the battery terminals)
R = external resistance
r = internal resistance
25
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
• In a circuit diagram, this symbol
represents a resistor in a circuit that dissipates electrical
energy.
• A straight line
represents a conducting wire
with negligible resistance.
26
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
  Ir  Vab
Vab    Ir
terminal voltage (potential
difference across terminals)
emf
potential difference
across internal resistance
Notes:
a) Vab < ε when the battery of emf ε is connected to the
external circuit with resistance R.
b) Vab > ε when the battery of emf ε is being charged by
other battery.
c) Vab = ε when the battery of emf ε has no internal
resistance (r =0) and connected to the external circuit
with resistance R..
27
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
Example 18.6
A battery with a terminal voltage of 11.5 V when delivering
0.50 A has an internal resistance of 0.10 Ω. What is its emf?
28
18.4 Electromotive Force (Emf), Internal Resistance and Potential Difference
Example 18.7
The battery in a circuit has an emf of 9.0 V. It is attached to a
resistor and an ammeter that shows a current of 0.10 A. If a
voltmeter across the battery’s terminals reads 8.9 V, what is
its internal resistance ?
29
SUBTOPIC :
18.5 Electrical Energy and Power
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Apply formula P=IV and electrical energy, W=VIt.
30
18.5 Electrical Energy and Power
• Electric energy is useful to us because it can be transformed
into other forms of energy (thermal energy, light).
• According to the conversation of energy , all the energy
delivered to the charge carriers by the battery must be lost
in the circuit .
• That is, a charge carrier traversing the circuit must lose
all the electrical potential energy it gained from the battery
when that carrier returns to the negative terminal of the
battery.
31
18.5 Electrical Energy and Power
•The electrical (potential) energy, W is the energy gained by
the charge Q from a voltage source (battery) having a
terminal voltage V.
•W= QV (the work done by the source on the charge)
• But Q=It, then W= VIt
• Unit : Joule (J)
• The rate of energy delivered to the external circuit by the
battery is called the electric power given by,
W QV

, Q  It
t
t
P  IV @ P  I 
P
•Unit : watt ( 1 W = 1J/s)
32
18.5 Electrical Energy and Power
• The energy dissipated per second in an electric device
(rate of energy dissipated) is given as
W VIt
P

 VI
t
t
for any device
• A passive resistor is a resistor which converts all the
electrical energy into heat. For example, a metal wire.
P  VI but V  IR
P  I 2R
or
V2
P
R
only for resistor
33
Example 18.8
18.5 Electrical Energy and Power
Calculate the resistance of a 40 W automobile headlight
designed for 12 V?
34
18.5 Electrical Energy and Power
Example 18.9
The current through a refrigerator of resistance 12 Ω is 13
A. What is the power consumed by the refrigerator?
35
18.5 Electrical Energy and Power
Example 18.10
An electric iron with a 15-ohm heating element operates at
120 V. How many joules of energy does the iron convert to
heat in 1.0 h ?
36
SUBTOPIC :
18.6 Resistors in series and parallel
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Deduce and calculate effective resistance of resistors
in series and parallel.
37
18.6 Resistors in series and parallel
Resistors in Series
I
R1
R2
R3
V1
V2
V3
I
V
battery , r=0
•
The properties of resistors in series are given below.
o The same current I flows through each resistor
where
I  I1  I 2  I 3
38
18.6 Resistors in series and parallel
o The sum of the voltages around a circuit loop (that
is, the gains and losses with + and - ,respectively) is
zero.
V   i Vi   0
V   i Vi 
V  V1  V2  V3
total potential difference
(Assuming that the connecting wires have no resistance)
bur V1  IR1; V2  IR2 ; V3  IR3 ; V  IReq
IReq  IR1  IR2  IR3
Req  R1  R2  R3
39
where
Req : equivalent (effective ) resistance
18.6 Resistors in series and parallel
Resistors in Parallel
I
I3
R3
I2
V3
R2
I1
V2
R1
V1
V
•
The properties of resistors in
parallel are given below.
o There is the same potential
difference, V across each
resistor where
V  V1  V2  V3
I
o Charge is conserved, therefore
the total current I in the circuit
is given by
I  I1  I 2  I 3
40
18.6 Resistors in series and parallel
V
V
V
V
but I1  ; I 2 
; I3  ; I 
R1
R3
Req
R2
V
V V V
 

Req R1 R2 R3
1
1
1
1
 

Req R1 R2 R3
41
18.6 Resistors in series and parallel
Example 18.10
2.0 
12 
4.0 
6.0 V
Calculate :
a. the total resistance of the circuit.
b. the total current in the circuit.
c. the potential difference across 4.0  resistor.
42
18.6 Resistors in series and parallel
Solution 18.10
2.0 
12 
4.0 
6.0 V
43
18.6 Resistors in series and parallel
Solution 18.10
b. Total current,
c. The potential difference across R1=2.0  is
Therefore the potential difference across R3 =4.0 
is given by
44
18.6 Resistors in series and parallel
Example 18.11
For the circuits shown below, calculate the equivalent
resistance between points x and y.
x
1.0 
y
8.0 
2.0 
1.0 
3.0 
(0.79 Ω)
16.0 
20.0 
16.0 
2.0 
x
9.0 
18.0 
y
6.0 
(8.0 Ω)
45
SUBTOPIC :
18.7 Kirchhoff’s Laws
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) State and use Kirchhoff’s Law’s.
46
18.7 Kirchhoff’s Laws
Kirchhoff’s first law (junction/current law)
• It states that the algebraic sum of the currents at any
junction of a circuit is zero,
I 0
or
I
in
  I out
47
18.7 Kirchhoff’s Laws.
Kirchhoff’s second law (loop/voltage law)
• It states that the algebraic sum of the voltages across
all of the elements of any closed loop is zero.
or
• It states that in any closed loop, the algebraic sum of
e.m.fs is equal to the algebraic sum of the products
of current and resistance.
    IR
Sign convention
-ε
-IR
R
+ε
across battery
I
+IR
across resistor
48
18.7 Kirchhoff’s Laws
Example 18.12
Using Kirchhoff’s rules, find the current in each resistor.
1  20 V
R2 = 20 Ω
R1 = 10 Ω
2  10 V
49
18.7 Kirchhoff’s Laws
Solution 18.12
1  20 V
R2 = 20 Ω
Step
I
R1 = 10 Ω
2  10 V
1. Draw current. (arbitrary)
2. Draw loop. (arbitrary)
3. Apply Kirchhoff’s laws.
I  0
,
   IR
50
18.7 Kirchhoff’s Laws
Example 18.13
Apply Kirchhoff’s rules to the circuit in figure below and find
the current in each resistor.
1  2.0 V
R1 = 3.0 Ω
R2 = 3.0 Ω
R4 = 3.0 Ω
R3 = 3.0 Ω
2  2.0 V
51
18.7 Kirchhoff’s Laws
Solution 18.13
1  6.0 V
R1 = 3.0 Ω
R2 = 4.0 Ω
R4 = 2.0 Ω
I
R3 = 5.0 Ω
2  3.0 V
2nd KL,
    ( IR)
52
18.7 Kirchhoff’s Laws
Example 18.14
Find the current in each resistor in the circuit shown below.
R2 = 4.0 Ω
R1 = 4.0 Ω
2  5.0 V
1  10 V
3  5.0 V
R3 = 4.0 Ω
I1 =3.75 A up, I2 = 1.25 A left, I3 = 1.25 A right
53
18.7 Kirchhoff’s Laws
Example 18.15
A cell of e.m.f. 4.0 V and internal resistance 1.0 Ω is
connected in series with another cell of e.m.f. 2.5 internal
resistance 0.5 Ω in a closed loop in such a way that the
current in the loop is minimum. Draw a circuit diagram to
show how the cells are connected and calculate the current.
(I = 1.0 A)
54
18.7 Kirchhoff’s Laws
Example 18.16
Calculate the currents I1,I2 and I3. Neglect the internal
resistance in each battery.
R1  1 
I1
ε1  15 V
I2
R2  0.5 
ε2  10 V
R3  0.1 
I3
ε3  3.0 V
I1  17.69 A ; I 2  14.62 A ; I 3  3.07 A
55
18.7 Kirchhoff’s Laws
Example 18.17
Given 1=8 V, R2=2 , R3=3 ,
R1 =1  and I=3 A. Ignore the internal resistance in
ε1
each battery.
I1
Calculate
a. the currents I1 and I2.
R1
ε2 R2
I2
b. the e.m.f. 2.
Ans. : 1 A, 4 A , 17 V
I
R3
56
SUBTOPIC :
18.8 Potential divider
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Explain the principle of a potential divider.
b) Apply equation of potential divider V  

1
R1
 R1  R2

V .

57
18.8 Potential divider
• A potential divider is used to tap a fraction of the voltage
supplied by a source of e.m.f.
• Two resistors are connected in series.
• The current flows in each
resistor is the same ;
I
V
I
, Req  R1  R2
R1
Req
V
I
R1  R2
V1
V
I
R2
V2
Potential divider circuit
• Potential difference across l1 or R1 is
V1  IR1
R1
V1 
V
R1  R2
58
18.8 Potential divider.
•
Resistance R1 and R2 are replaced by a uniform
homogeneous wire as shown in figure below.
V
I
a
l1
l2
c
V1
I
b
V2
R
l
A
R1
V1 
V
R1  R2
• Potential difference across l1 is
 l1 
V
V1  
 l1  l2 
59
18.8 Potential divider
Example 18.18
Resistors of 3.0 Ω and 6.0 Ω are connected in series to a
12.0 V battery of negligible internal resistance. What are the
potential difference across the (a) 3.0 Ω and (b) 6.0 Ω
resistors ?
4.0 V, 8.0 V
60
SUBTOPIC :
18.9 Potentiometer and Wheatstone
Bridge
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Explain the principle of potentiometer and Wheatstone
Bridge and their applications.
b) Use related equations.
61
18.9 Potentiometer and Wheatstone Bridge
Potentiometer
• A potentiometer is mainly used to measure potentiometer.
• It consists of a uniform wire.
• Basically a potentiometer circuit consists of a uniform wire
AB of length 100.0cm, connected in series to a driver cell
with emf V of negligible internal resistance.
V (Driver cell -accumulator)
I
A
I
I
C
G
I
B
Jockey
+ -
Vx
(Unknown Voltage)
62
18.9 Potentiometer and Wheatstone Bridge
The potentiometer is balanced when the jockey (sliding
contact) is at such a position on wire AB that there is no
current through the galvanometer. Thus
•
Galvanometer reading = 0
When the potentiometer in balanced, the unknown
voltage (potential difference being measured) is equal to
the voltage across AC.
•
Vx  VAC
•
Potentiometer can be used to :
i) Measure an unknown e.m.f. of a cell.
ii) Compare the e.m.f.s of two cells.
iii) Measure the internal resistance of a cell.
63
18.9 Potentiometer and Wheatstone Bridge
If the galvanometer shows deflection in one direction
only, it may be due to :
• The connections of the terminals of the cells are wrong.
The positive terminal of the cell must be connected to
the positive terminal of another cell.
• The emf of the unknown cell is more then the emf of the
cell connected across the wire of the potentiometer, AB.
• The connections are not tight and the current does not
flow in certain part of the circuit.
64
18.9 Potentiometer and Wheatstone Bridge
i) Measure an unknown e.m.f. of a cell.
V(Driver cell -accumulator)
I
A
I
I
C
G
I
•When the potentiometer is
balanced,
IG = 0
Balance length = lAC
B
VAC = ε
VAC = IRAC …1
Jockey
+ -

RAC 
(Unknown cell)
4 and 5 into 1,
 V
VAC    
 RAB
l
  AC V
l AB
  l AC


R
 
AB 
l
  AB

l AC …2 , R  l AB …3
AB
A
A
l
2
, RAC  AC RAB ...4
3
l AB
I
V
...5
RAB
65
18.9 Potentiometer and Wheatstone Bridge
ii) Compare the e.m.f.s of two cells.
V
I
A
l2
l1
I
I
C
D
J
ε1
ε2
(1)
S
(2)
G
I
B
• When the potentiometer is
balanced,
IG = 0
• Balance length,
lAC = l1 for ε1 and lCD = l2 for ε2
l AC
• From  
V , thus
l AB
l AC
lCD
1 
V ...1 and  2 
V ...2
l AB
l AB
lCD
2  1 , 2 
1
l AC
l2
 2  1
l1
66
18.9 Potentiometer and Wheatstone Bridge
Wheatstone Bridge
•
•
It is used to measure the unknown resistance of the resistor.
Figure below shows the Wheatstone bridge circuit consists of a cell
of e.m.f.  (accumulator), a galvanometer , known resistances (R1, R2
and R3) and unknown resistance Rx.
•
The Wheatstone bridge is said to be balanced when no current
flows through the galvanometer. Hence I  I  I and
AC
ε
I
I1
A
I2
R1
R3
D
1
I AD  I DB  I 2
C I1
G
CB
R2
0
I2
I
B
RX
Then , Potential at C = Potential at D
Therefore
VAC  VAD and VBC  VBD
Since V  IR thus
I 1 R1  I 2 R3 and I1 R2  I 2 RX
I 1 R1 I 2 R3

I 1 R2 I 2 RX
 R2 
RX    R3
 R1 
67
ε
I
I1
A
I2
C I1
R1
I
R2
G
0
D
I 2 RX
R3
Thick copper
strip
18.9 Potentiometer and Wheatstone Bridge
B
(Unknown
RX resistance)
R
Wire of uniform
resistance
I2
(resistance box)
I1
0G
A
RX R2

R3 R1
RX l1

R l2
I1
I
 R2 
RX    R3
 R1 
Jockey
J
ε
Accumulator
B
I
68
18.9 Potentiometer and Wheatstone Bridge
Example 18.19
An unknown length of platinum wire 0.920
mm in diameter is placed as the unknown
resistance in a Wheatstone bridge as
shown in figure below.
Resistors R1 and R2 have resistance of
38.0  and 46.0  respectively.
Balance is achieved when the switch
closed and R3 is 3.48 . Find the
length of the platinum wire if its
resistivity is 10.6 x 10-8  m.
69