NOTA TUTORIAL SEM 2 CHAPTER 3(Electric Current And Direct-current Circuit)

TOPIC 3 : ELECTRIC CURRENT AND Thus free electrons are attracted into the
DIRECT CURRENT positive terminal (are forced to drift in one
3.1 Electrical Conduction direction).
3.2 Ohm’s law and Resistivity This direction is in the direction opposite
3.3 Variation of resistance with temperature to the field, E.
3.4 Electromotive force (emf) internal The velocity of these free electrons is
called drift velocity.
resistance and potential difference
The drift velocity vd of the free
3.5 Resistors in series and parallel electrons is the mean velocity of the
3.6 Kirchhoff’s Rules electrons parallel to the direction of
3.7 Electrical energy and power the electric field when a potential
3.8 Potential divider difference (battery) is applied.
3.9 Potentiometer

3.1 Electrical Conduction
a) Describe microscopic model of
current.
I  dQ
b) Define electric current, dt
I  dQ
c) c) Use electric current, dt , Q=ne

**Emphasize on the flow of free electrons in a
metal. Include concept of drift velocity.

3.1 a) Describe microscopic model of
current

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.

3.1 b) Define 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).

I ins  dQ
dt

The steady current is defined as

Current I is the rate at which charge moves IQ
through an area, A such as the cross-section t Unit of I is A (ampere).
of a wire. Conventional current is defined to
move in the direction of the electric field. 3.1 c) Use electric Current
a) Positive charges move in the direction of
the electric field and the same direction as I  dQ , Q= ne
the conventional current. dt
b) Negative charges moves in the direction
opposite to the electric filed. Example 3.1

The direction of current flow – from A wire carries a current of 1.5 A.
the positive terminal to the negative
one – was decided before it was a) How much charge flows through a
realized that electrons are negatively point in the wire in 5.0 s
charged.
Therefore, current flows around a b) How many electrons cross a given
circuit in the direction a positive area of the wire in 1.0 s?
charge would move; electrons move
the other way. Solution:
However, this does not matter in most a) I  Q
circuits.
t
Q  It  (1.5)(5.0)  7.5C

b) Q  ne, dQ  Idt

n dQ  Idt 1.5  1.0   9.38 1018 electrons
e e
 1.6 1019

3.2 Ohm’s Law and Resistivity

a) State and use Ohm’s Law.

  RA
l
b) Define and use resistivity,

c) Sketch V-I graph

3.2 a) State and use Ohm’s Law

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 Its formula is given by

V  constant Ohm's Law   RA
I l
V  R  V  IR where
I R = resistance
l = length of the conductor (m)
The resistance of R offered by a conductor A = area of cross-section of the
depends on following factors: conductor (m-2)
Its unit is Ω m.
i) length, l

ii) cross-sectional area, A of the
conductor

iii) nature of the material

iv) temperature of the conductor

3.2 b) Define and use resistivity The material from which a wire is made
is also affects its resistance: current flows
Resistivity, ρ is a measure of a easily through conductors and only with
material’s ability to oppose the flow of difficulty through insulators.
electric current through the material.
The quantitative measure of a material’s
Resistivity, ρ is defined as the opposition to the flow of current is called
resistance of a sample of the material resistivity.
of cross-sectional area 1 m2 and of
length 1m.

It is a constant value depends on the
types of materials.

It depends only on the chemical Example 3.3
composition of the material, not its shape
or size. A wire (length=2.0 m, diameter=1.0 mm) has
a resistance of 0.45 Ω. What is the resistivity
All conductors have smaller
resistivity. of the material used to make the wire?
Insulators have larger resistivity.
Solution:
3.2 c) Sketch V-I graph
RA   d 2  1x10 3 2
Ohmic conductors are conductors l  2  2
which obey Ohm’s law.   , A  r 2     7.86x107 m2
Examples: pure metals. (Figure A)
Non-ohmic conductors do not obey   0.45(7.86x107 )
Ohm’s law. 2
Example: junction diode. (Figure B)
  1.77x107 m
Example 3.2
What voltage will be measured across a 1000 3.3 Variation of Resistance with
Ω resistor in a circuit if we determine that Temperature
there is a current of 2.50 mA flowing through
it? a) Explain the effect of temperature on
Solution: electrical resistance in metals.

V  IR  (2.50x103)(1000)  2.50V b) Use resistance, R = Ro [1 +α (T - To)].
* α is at temperature 20oC

3.4 a) Explain the effect of temperature on
electrical resistance in metals.

Metal
The resistance of a metal (conductor)

depends on
a) the nature of the material,

(resistivity, )
b) the size of the conductor,

(the length, l & cross-sectional area, A)
c) the temperature, T of the conductor.

R = R [1 +α (T - T )]
oo

The resistance of metals increases
with increasing temperature. (T↑, R↑)
As temperature increases, the ions of
the conductor vibrate with greater
amplitude.
More collisions occur between free
electrons and ions.
These electrons are slowed down thus
increases the resistance.

3.3 b) Use resistance, R = Ro [1 +α (T - To)] Example 3.6:

Temperature coefficient of resistance, α is Two wires P and Q with circular cross section
defined as the fractional change in resistance are made of the same metal and have equal
per Celsius degree. length. If the resistance of wire P is three
times greater than that of wire Q, determine
R Ro the ratio of their diameters

 T

Solution:

The resistance of a metal can be represented Given
by the equation below

R=Ro[1+α(∆T)] R - Ro = ∆R ρP  ρQ  ρ;lP  lQ  l;

∆R=Roα∆T R ρl πd 2 and RP  3RQ
; A
A4
α is a constant value and it is depends on the
material.

Example 3.4 RP  3RQ
ρPlP  3 ρQlQ
A platinum wire has a resistance of 0.50 Ω at AP AQ
0oC. It is placed in a water bath, where its
resistance rises to a final value of 0.60 Ω . 4 ρl  3 4 ρl 
What is the temperature of the bath? πdP2 πdQ 2 

(α = 3.93 x 10-3 oC -1)

Solution: dP  1
dQ 3
T  R  R  Ro  0.60  0.50  51 oC dQ  3 OR
Ro Ro 3.93103 (0.50) dP

T = To +∆T

T = 0+51 = 51 oC 3.4 Electromotive force (emf), internal

Example 3.5 resistance and potential difference
a) Define emf , ε and internal
A narrow rod of pure iron has a resistance of
0.10 Ω at 20 oC. What is its resistance at resistance, r of a battery.
50 oC?
(α= 5.0 x 10-3 oC-1) b) State factors that influence internal
resistance.
Solution:
c) Describe the relationship between
Given; emf of a battery and potential
Ro = 0.10 Ω, To = 20 oC and T = 50 oC difference across the battery
R =Ro[1+α(T-To)] terminals.
R =0.10[1+5.0x10-3(50-20)]
R= 0.12 Ω d) Use terminal voltage, V= ε – Ir.

3.5 a) Define emf , ε and internal resistance, Any devices that increase the potential
r of a battery. energy of charges circulating in circuits
are sources of emf
What is electromotive force,emf (ε or ξ) A real battery has some internal
of a battery? resistance, r
Therefore, the terminal voltage is not
The e.m.f of a battery is defined as the necessarily equal to the emf
electrical energy that generated by a
battery so that the charges can flow
from one terminal to another
terminal of the battery through any
resistor.

@ The e.m.f of a battery is the work
done per unit charge.
@ The e.m.f of a battery is the p.d
across the terminals of the source in
open circuit (no current is flowing,
I = 0).
**open circuit =circuit is not complete
SI unit : Volt (V)

emf,ε and Internal Resistance, r The terminal voltage

The source that maintains the current ΔV = - = ε – Ir
in a closed circuit is called a source of
emf For the entire circuit,

  I(R  r)
  Ir  IR
  Ir Vab

where
ΔV = terminal voltage
Ɛ = emf of the battery
r = internal resistance
R = external resistance

In a circuit diagram, this symbol Example 3.7
represents a resistor in a circuit that A battery with a terminal voltage of 11.5
dissipates electrical energy. V when delivering 0.50 A has an internal
A straight line resistance of 0.10 Ω. What is its emf?
represents a conducting wire with Solution:
negligible resistance.   Ir Vt  (0.50)(0.10) 11.5 11.55V
Example 3.8
3.4 b) State factors that influence The battery in a circuit has an emf of 9.0
internal resistance. V. It is attached to a resistor and an
ammeter that shows a current of 0.10 A.
a) The surface area of electrodes-larger If a voltmeter across the battery’s
the surface area of electrodes, less terminals reads 8.9 V, what is its internal
internal resistance. resistance?
Solution:
b) The distance between the electrodes-
more the distance between the   Ir Vt
electrodes, the greater is the internal r   Vt  9.0  8.9  1 
resistance.
I 0.10
c) The nature and concentration of the
electrolyte-less ionic the electrolyte or Example 3.9
higher the concentration of electrolyte,
greater is the internal resistance. Refer to the circuit shown in figure
above, R is an ohmic resistor of
d) The temperature of the electrolyte- resistance 10.0 Ω. Initially the switch
higher the temperature of the is open. At time T the switch is
electrolyte, less is the internal closed. The graph shows the
resistance. voltmeter reading V before and after
time T. Determine the internal
3.4 c) Describe the relationship between resistance of the battery and the
emf of a battery and potential current in the circuit when the switch
difference across the battery terminals is closed.

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.

Solution: where
Before time T, the voltmeter records Req : equivalent (effective ) resistance
the emf, Ɛ of the battery Ɛ = 6.0 V
After time T, the voltmeter records Resistors in Parallel
the terminal p.d. across R
Vt  5.0 V  IR The properties of resistors in parallel are
5.0  I (10.0) given below.
I  0.50 A
VT  V1  V2  V3
  Ir Vt
I  I1  I2  I3
6.0 = 0.50 r + 5.0
r = 2.0 Ω but I  V
R
3.6 Resistors in series and parallel
V V V V
a) Derive and determine effective Req R1 R2 R3
resistance of resistors in series and 1 111
parallel. Req R1 R2 R3
where
3.5 a) Derive and determine effective
resistance of resistors in series and Req : equivalent (effective ) resistance
parallel.

Resistors in Series

The properties of resistors in series are
given below.

I  I1  I2  I3

VT  V1 V2 V3

but V  IR
IReq  IR1  IR2  IR3

Req  R1  R2  R3

Example 3.10 R2 c. The potential difference across R1 = 2.0 
R1 R3 is V1  IR1  (1.2)(2)

6.0 V V1  2.4 V

Given 1 = 2.0Ω, 2 = 12.0Ω and Therefore, the potential difference across
3 = 4.0Ω. R3 = 4.0  is given by
Calculate:
a. the total resistance of the circuit. V3  V V1  6.0  2.4
b. the total current in the circuit.
c. the potential difference across 4.0  V3  3.6 V

resistor. Exercise 3.1
Solution:
1. x 2.0 

1.0 
1.0  2.0 

y 3.0 

(0.79 Ω)

a. 1  1  1  1  1 2. 8.0  20.0 
R23 R2 R3 12 4 16.0 
x
R23  3.0  16.0 

Rtotal  R1  R23  2  3 9.0  y
6.0 
Rtotal  5.0 
18.0 
b. Total current,
I V 6 (8.0 Ω)
Rtotal 5

I  1.2 A

3.7 Kirchhoff’s Laws Junction

a) State and describe Kirchhoff’s Laws.
b) Use Kirchhoff’s Laws.

** i) Maximum two closed circuit loops

ii) Use scientific calculator to solve the
simultaneous equations

3.6a) State and describe Kirchhoff’s Rules ∑ = ∑

Kirchhoff’s First Law + + = +
Kirchhoff’s Second Law
(junction/current rule)

It states that the algebraic sum of the (loop/voltage rule)
currents at any junction of a circuit is
zero, It states that the algebraic sum of the voltages

across all of the elements of any closed loop

I 0 at any junction is zero or
at any junction
or It states that in any closed loop, the algebraic
sum of emfs are equal to the algebraic sum of
 Iin  Iout the products of current and resistance.

Junction

Example 3.11 Example 3.12

Using Kirchhoff’s laws, find the current in Apply Kirchhoff’s rules to the circuit in figure
each resistor. below and find the current in each resistor.

Solution: 2nd KL
Step
1. Draw current. (arbitrary)     IR
2. Draw loop. (arbitrary)
3. Apply Kirchhoff’s laws.  2  1  IR1  IR2  IR3  IR4

∑ = ∑( ) 3.0  6.0  I (3.0  4.0  5.0  2.0)
3.0  14I
1   2  I (R1  R2 ) I  0.21A
Since the current is positive,
20 10  I (10  20) then I = 0.21 A
10  30I
I  0.33A Example 3.13
Find the current in each resistor in the circuit
shown below.

Solution:

LOOP 1 LOOP 2

1st KL Solution:
∑ = ∑

(1)

2nd KL ∑ = ∑( )
LOOP 1

1st KL

∑ = ∑

= + _________( )

LOOP 1 ∑ = ∑( ) 2nd KL

LOOP 1 ∑ = ∑( )

From equation (1) LOOP 2 ∑ = ∑( )

Example 3.14 Answer:

Calculate the currents I1, I2 and I3. Neglect the Exercise 3.1
internal resistance in each battery.
1. 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 V 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.

(1.0 A) 3.7 Electrical Energy and Power
2. Given 1=8V, R2=2 , R3=3 , R1 =1 
a) Use power, P = IV , P  I 2R and
and I=3 A. Ignore the internal resistance
in each battery. PV2
R.
Calculate:
a. the currents I1 and I2. (known as power loss)
b. the e.m.f. 2.
b) Electrical energy, W = VIt
(1 A, 4 A , 17 V)
3. Determine the current through each 3.7 Electrical Energy and Power

resistor in the circuit shown below. Electric energy is useful to us because it
can be transformed into other forms of
(I1 = 2.00 A, I2 = 1.50 A, I3 = - 0.50 A) energy (thermal energy, light).

According to the conservation 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.

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,
P  W  QV , Q  It

tt
P  IV @ P  I

Unit: watt (1 W = 1J/s)

The energy dissipated per second in an Example 3.17
electric device (rate of energy
dissipated) is given as An electric iron with a 15-ohm heating
element operates at 120 V. How many
A passive resistor is a resistor which joules of energy does the iron convert to
converts all the electrical energy into heat in 1.0 h?
heat. For example, a metal wire.
Solution:

Given; R = 15 Ω , V = 120 V ,

t = 60 x 60 = 3600 s

Example 3.15 Exercise 3.2
Calculate the resistance of a 40 W
automobile headlight designed for 12 V. A resistor of 100 Ω is connected across a
Solution: battery of emf 12.0 V and internal resistance
Given; P = 40 W, V = 12 V 1.0 Ω. Determine

Example 3.16 a) The current in the circuit
The current through a refrigerator of
resistance 12 Ω is 13 A. What is the power b) The voltage terminal
consumed by the refrigerator?
Solution: c) The electrical power generated by the
Given; R = 12 Ω, I = 13 A battery

d) The power dissipated by the resistor

e) The power dissipated by the internal
resistance

f) The amount of heat dissipated by the
100 Ω resistor in one minute.

(0.12 A, 11.9 V, 1.43 W, 1.42 W, 0.01 W,

85.2 J)

3.8 Potential Divider Resistance R1 and R2 are replaced by a
uniform homogeneous wire as shown in
a) Explain the principle of a potential figure below.
divider.
V
b) Use equation of potential divider,

V1   R1  . II
 R1  R2 V
 
l1 l2
a c b
3.8 Potential divider
V1 V2

A potential divider is a fraction of the R  l R1
voltage supplied by a source of e.m.f. A R1  R2

V1  V

V

Potential difference across l1 is

I R2 I
R1

V1 V2 V1   l1 l1 l2 V
Potential divider circuit 

Two resistors are connected in series. Example 3.18

The current flows in each resistor is the Resistors of 3.0 Ω and 6.0 Ω are connected in
same; series to a 12.0 V battery of negligible internal
resistance. What are the potential difference
I V I V , Req  R1  R2 across the (a) 3.0 Ω and (b) 6.0 Ω resistors?
Req R1  R2
Solution:

Potential difference across l1 or R1 is

V1  R1 V V1  ( R1 )   3 12  4V
R1  R2  9
R1  R2
V1  IR1
 ( R2 )  6 12
V2   9  8V
R1  R2

Example 3.19

Refer to the Figure above, calculate the The potentiometer is balanced when the
voltage drop for each resistor. jockey (sliding contact) is at such a position
Solution: on wire AB that there is no current through
Total resistance the galvanometer. Thus

3.9 Potentiometer Galvanometer reading = 0
a) Explain the principle of potentiometer
When the potentiometer in balanced, the
and their applications. unknown voltage (potential difference being
b) Use related equations for potentiometer, measured) is equal to the voltage across AC.

1  l1 Vx  VAC
 2 l2
**Equation for Potentiometer
3.9 Potentiometer
A potentiometer is mainly used to = =
measure potential difference, V.
It consists of a uniform wire.
Basically a potentiometer circuit consists Potentiometer can be applied to:
of a uniform wire AB of length 100.0cm, i) Measure an unknown emf of a cell.
connected in series to a driver cell with ii) Compare the emfs of two cells.
emf V of negligible internal resistance. iii) Measure the internal resistance of a cell.

i) Measure an unknown e.m.f. of a cell. Example 3.20

When the potentiometer is balanced, A potentiometer consists of a 150 cm wire XY
IG = 0 and internal resistance of 0.5 Ω. When the cell
is connected to the potentiometer, the balance-
Balance length = lAC length XP is 72.6 cm. Calculate the emf of the
V =ε cell, Ɛ.

AC Total resistance of the potentiometer,

V = IR …1 Rt = 0.5 Ω +1.6 Ω +1Ω = 3.1 Ω

AC AC Resistance of XY = 0.5 Ω

For wire XY, Rl

RXP  72.6
0.5 150
RXP  0.24

Using potential divider equation,

V XP  R XP  2.0
RT
4 and 5 into 1,
 0.24  2.0
 V   l  3.1
 RAB   l RAB 
VAC    AC  0.15V
 AB 
Emf of the cell, Ɛ = 0.15 V
  lAC V
l AB

ii) Compare the e.m.f.s of two cells. Example 3.21

When the potentiometer is balanced, When S is connected to X, the balance length
IG = 0 is 20 cm. Balance length is 40 cm when S is
connected to Y. Calculate 2 if V and 1are
Balance length, 5.0V and 2.0 V respectively.
lAC = l1 for ε1 and lCD = l2 for ε2
Solution:

S is connected to X, = 20 cm

  lAC V = 0
From lAB , thus
= 1= I

=

1  l AC V ...1 and 2  lAD V ...2 I =
l AB l AB


∴ 1= I

2 1 , 2  l AD 1 = ( ) ( )
l AC


2  l2 1 2 = 5 (20)
l1


= 50 cm

S is connected to Y, = 40 cm

= 0

= 2= I

=

I =



∴ 2= I

= ( ) ( )



2 = 5 (40)
50

= 4 V

OR

= =


=



=

=