Note Physics 2 SP025

Internal resistance,

 is defined as the resistance of the chemicals inside the
battery (cell) between the poles.

 The smaller the internal resistance for a given emf, the more
current and the more power the source (battery) can supply.

 Internal resistance can behave in complex ways. As noted, increases
as a battery is depleted (ages) and electrolyte dries out.

 But internal resistance may also depend on the magnitude and
direction of the current through a voltage source, its
temperature and type of chemical material in the battery.

The internal resistance of rechargeable nickel-cadmium cells, for example,
depends on how many times and how deeply they have been depleted.
 symbol of emf and internal resistance in the electrical circuit

OR

33

Example 3.4

Question 1:
When is the potential difference across the battery

(a) equal to its e.m.f ? in open circuit

(b) less than its e.m.f ? when battery is connected
Question 2: to resistor

A battery of e.m.f. 12 V and internal resistance 1.5 Ω
was connected to a 4.5 Ω resistor, what is the value
for current?

34

Question 3:

A battery of e.m.f. 12 V and internal resistance 1.5 Ω
was connected to a 4.5 Ω resistor, what is the value
for current?

Solution:

Question 3:

A battery of e.m.f. 12 V and internal resistance 1.5 Ω
was connected to a 6.0 Ω resistor.

(a) What is the value for current?

(a) What is the terminal voltage of the battery 35

Solution:
(a)

(b)

36

3.5 Resistors in series and parallel

Resistors in series o The same current
passes through each
resistor.

o Total voltage, is
shared by the resistors.

o

Cancel the common

o Generally, for number of resistors connected in series, the
effective resistance is given by

o Value of is always greater than any individual resistance.
37

Resistors in parallel o Voltage, same across each
resistors.

o Total current, splits into 3
separate path,

Divide out from each term :

o Generally, for number of resistors
connected in parallel, the effective
resistance is given by:

o Value of is always less than the smallest resistance in

group. 38

Flow map : Problem solving
strategy for Resistors in series
and parallel

Simplify circuit Find total Reverse process to
and find the current using calculate V and I across
equivalent
resistance, RE V = IR each capacitor
requested in question.

Series : Parallel : 1 Series : Parallel:
RE  R1  R2 1 1  I same I adds
RE   R1  R2 V adds V same


Example 3.5

Question 1:

Use the concept of equivalent resistance to determine the

unknown resistance of the identified resistor that would

make the circuits equivalent Series

Parallel

40

Series

Parallel

41

Question 2:

FIGURE 1 shows a circuit with e.m.f of 30 V, calculate the
effective resistance in the circuit

30 V 7Ω
15 Ω 4 Ω 1 Ω
FIGURE 1

42

Solution: Effective resistance in the circuit 7Ω
7Ω
I3
30 V 15 Ω 0.8 Ω

30 V 15 Ω 4 Ω 1 Ω

R2=0.8+7=7.8Ω

30 V 5.13 Ω 30 V 15 Ω 7.8 Ω

43

Question 3:
FIGURE 2 shows a circuit with e.m.f of 24 V, calculate the

(a) Effective resistance in the circuit
(b) Value of the current I1, I2 and I.
(c) Voltage across for both V1 and V2 .

24 V
I

5 Ω I1 2 Ω V1
4Ω

I2 4 Ω 44
V2

Solution: 24 V
(a) Effective resistance in the circuit
2Ω 4Ω
R1=2+4=6Ω 4Ω

Re=2.4+5=7.4Ω 5Ω 24 V
24 V 5Ω
6Ω
7.4 Ω 24 V 4Ω
5 Ω 2.4 Ω
45

(b) Value of the current I1, I2 and I. 24 V
1) calculate value total current, I 7.4 Ω

2) calculate current across 2.4 , I=3.24 A24 V
3) Voltage across 6 and 4 5 Ω 2.4 Ω
4) calculate current I2,
24 V

5 Ω I1 6Ω
4 Ω 46

(c) Voltage across for both V1 and V2 .
1) Voltage across,V2
2) Voltage across V1

47

Question 4:

A 9-V battery is hooked up to two resistors in series using wires of negligible
resistance. One has a resistance of 5 Ω, and the other has a resistance of
10 Ω. Several locations along the circuit are marked with letters, as shown in
the FIGURE 1. Which statements about this circuit are true?

A) The current is exactly the same at points
A, B, C, and D

B) The current at A is greater than the current
at B, which is equal to the current at C,
which is greater than the current at D

C) The potential at D is equal to the potential
at C.

D) The potential at B is equal to the potential
at C.

48

3.6 KIRCHHOFF’S RULES

Kirchhoff’s 1st rule (Junction rule)

 A statement of conservation of electric charge
 states that the sum of the currents entering any

junctions in a circuit must equal the sum of the
currents leaving that junction.

  At any junction : Iin  Iout

49

Kirchhoff’s 2nd rule (Loop rule)

 Follows from the laws of conservation of
energy

 states that in any loop, the sum of emfs is equal to
the sum of the products of current and resistance

 OR In any closed loop,   IR

ε1 R1

ε2 R2

50

Caution ! Connections in a circuit affect the total emf

Sign convention ewmhf enε applying Kirchoff’s 2nd law:
Direction drawn from terminal (−) to (+)

 For emf,  : Same direction of loop Opposite
direction direction with
direction of loop with loop ε loop

ε ε +‒ ε

‒+

 For product of IR: Opposite
direction with
direction of loop direction of loop loop

Same direction R  IR
I
R with loop 51
I
 IR

Flow map : Problem solving strategy
(Kirchhoff’s Laws)

–+ I for each Junction Rule
resistor(s)
Direction for ξ  Iin  Iout
Hint: Resistors at the same
branch which connected in
series will have same current, I.

Direction of Loop Loop Rule Solve equation(s)
use
   IR
mode EQN
* Direction ξ and I OPPOSITE to

loop give (–)

o From the calculation, sometimes we get negative value of current
and emf . This negative sign indicates that the direction of the

actual current and polarity of emf is opposite (reverse) to

the direction shown in circuit.

Example 3.6

FIGURE 1
Referring to the circuit in FIGURE 1, calculate the
current I1, I2 and I3.

Solution c Step 1 :
Draw direction ξ
a
e

I1 Loop 1 Loop 2

b d I3 I3

I1 f Step 4 :

Step 2: color 1 branch Choose
path with 1 color. R direction of
on the same branch loop (CW
path will have the
same value of I. @ACW)

Note : Path dbac : current I1
Path cd : current I2
Path cefd : current I3

Consider Junction c : Step 5 :
Write Kirchoff 2nd law.
 Iin   Iout NOTE : if direction of I and
I1  I2 I3 ξ opposite to direction of
loop travel, Give (–) sign
I 1  I 2  I 3  0 (1)
(2)
Refer Loop 1 ( acdba – clockwise )

   IR

30  40I 1 10I 2

Refer Loop 2 ( cefdc– clockwise )

   IR

20 10  40I 3 10I 2
10  40I 3 10I 2 (3)

Solve 3 equations simultaneously using scientific
calculator

CASIO fx 570 ES I 1  I 2  I 3  0 (1)
Select MODE 40I 1 10I 2  30 (2)

5 : EQN 10I 2  40I 3  10 (3)
2 : anX+bnY+cnZ = dn

XYZ

0
40 30

10

I1  0.6666A I 2  0.3333A I 3  0.3333A

Question 2:

FIGURE 2 shown, cell P has e.m.f 3.0 V and internal resistance battery
2.0 Ω and cell Q with e.m.f 2.0 V and internal resistance battery
3.0 Ω . If the resistance of the ammeter is 5.0 Ω, what is the ammeter
reading?

P

A

Q
FIGURE 2

57

3 V,2 Ω Consider Junction :

I1 Loop 1  Iin   Iout
I3
5Ω Refer Loop 1 (anticlockwise)
I2 A
   IR
Loop 2
Refer Loop 2 (clockwise)

   IR

2 V,3 Ω

Solve 3 equations simultaneously using scientific
calculator

58

3.7 ELECTRICAL ENERGY AND POWER

Electrical energy, W

 Electric energy, is the

amount of energy given up

Electrical device by a charge in passing

AB through an electric device

IV 

I  If the electrical device is
passive resistor (device

which convert all the

electrical energy supplied

into heat), the energy

dissipated, is given

by

 SI unit : Joule (J) 59

Power, P

 is defined as the rate of energy liberated in the
electrical device. (

 Electrical power, supplied to the electrical
device is given :

 When the electric current flows through wire or
passive resistor, the electrical power can be written
as :

OR (known as power loss)
 scalar quantity and its unit is watts (W).

60

Example 3.7

Question 1:
A 25 W lamp is connected to 240 V supply.
(a) What is the current in the lamp?
(b) What is the resistance of the filament?

Solution:
(a)

(b)

61

Question 2:
A filament lamps A and B is rated 240 V, 100 W and 240
V, 60 W respectively.
(a) Resistances of the filaments of A and B at their normal

working temperature.
(b) If each lamp are connected to a 120 V supply, what is

power dissipated from each lamp?

62

Solution:
(a) Use

(b) Use

63

Question 3:
Two filament lamps A and B is rated 12 V, 24 W and 12 V,
36 W respectively are connected in series to a 24 V
battery. Calculate the
(a) Resistances of the filaments of A and B
(b) Power dissipated from each lamp
(c) State which lamp is brighter.

64

Solution:
(a) Use

(b) Current flowing in both lamp is
Power dissipated from each lamp is

(c) State which lamp is brighter.
lamp A is brighter because PA>PB

65

3.8 POTENTIAL DIVIDER

 used to obtain any smaller portion of voltage from a
single voltage source,

 This is done by connecting two or more resistors in

series as shown. V

I R1 R2 I

V1 V2

 Since the current flowing through each resistor is the
same, thus

66

 Therefore, the potential difference (voltage) across
R1 is given by:

 Put (1) into (2) :

 Similarly,


 In general, if the potential divider has n resistors
connected in series, the smaller portion of voltage
across

 OR apply Ratio = ratio 67

Example 3.7

Question 1:
Refer FIGURE 1, calculate the output voltage.

12 V 3

2  Vout

FIGURE 1

68

Solution:
The output voltage is given by

OR

69

Question 2:
Resistor of 3.0 Ω and 6.0 Ω are connected in series to a
12.0 V battery with negligible internal resistance. What
are potential difference across
(a) 3.0 Ω resistor.
(b) 6.0 Ω resistor.

Solution:

(a)

(b) OR since r in series so

70

3.9 POTENTIOMETER

Potentiometer is an instrument used to measure an
unknown voltage.

Circuit diagram of a potentiometer:

V

A + Vx- C Primary circuit
G
secondary circuit B

One terminal of battery whose It consists of a long
voltage is to be measured is resistive wire AB of
connected at point A and the other length L (about 100
terminal at any point on the cm long), a battery
resistive wire through a of known emf, V
galvanometer G and jockey. This and this forms the
forms the secondary circuit. primary circuit.

71

Working priciple of potentiometer
 The jockey is tapped along the wire AB until at such a

position, C on wire AB that there is no current through
the galvanometer (G = 0).

V (Driver cell -accumulator)

I I
I CIB

A

Wire +Vx - G Jockey
Galvanometer

Unknown voltage

 As galvanometer reads zero  potentiometer in
balanced, length is known as balance length.

72

 In this balance condition, unknown voltage, is equal to
the voltage across AC.

 From , for current and cross sectional

area, wire are constant, then

where is constant of proportionality

 voltage drop along the wire is directly proportional to

the length of the wire(

 Thus we can use ratio = ratio to obtain the value

of unknown from a balance length . 73

Potentiometer can be used to :

1. measure an unknown emf of a cell. Scan me

2. compare the emfs of two cells.

3. measure the internal resistance of a cell.

Scan me : Find unknown V Scan me : find r Scan me : compare emf 74

Example 3.7

Question 1:

Given that length AB is 1m. When the jockey is tapped at 70 cm
from point A, galvanometer reads zero, determine the value of

2V

100 cm B
70 cm

C
A

+ x- G Jockey

75

Solution: Use Tabulation
method to help
As G reads zero   x  VAC
u solve it.
Apply :
76
V l

Use ratio to obtain the VAC :
Vl

2 100 cm
ξx 70 cm

Ratio V  Ratio l
2 100

 X  70
 X  1.4 V

Question 2:

AB is a uniform wire of length 1.0 m and resistance 10.0 . When the
jockey is tapped at C, galvanometer reads zero. Determine emf .

4V , 0.5
3

10Ω B

AC

58.6 cm
G

+-



Solution:

As G reads zero    VAC

V R
4V
10+0.5+3 100 cm
VAB =13.5 58.6 cm
VAC = ξ =
10
?

Apply Ratio V = Ratio R

4  (10  0.5  3)  VAB  2.96 V
VAB 10

Apply Ratio V = Ratio l

2.96 V dibina antara jarak 100 cm

A B
C
ξ = 1.735 V
? V untuk 58.6 cm

Question 3:

AB is a uniform wire of length 1.0 m and resistance 10.0 . R1=5.0  and
R2=15.0 . E is an accumulator of e.m.f 2.0 V and negligible internal
resistance. When both the switches S1 and S2 are opened, the
galvanometer G is balanced when the length AJ is 61.5 cm. When both the
switches S1 and S2 are closed, the balanced AJ is 10.0 cm.Calculate the,

(a) e.m.f of cell E1 S2
E

(b) internal resistance of cell E1

R2

A JB
E1

G

R1

S1

Solution: S2

(a) e.m.f of cell E1 E = 2.0 V
lAB =1.0 m ,RAB = 10.0 
When switch S2 is opened, the current in R2 =15.0  B
loop is:
AJ
E1

G

When switch S1 and S2 are opened, lAJ = 61.5 cm : E1 = VAJ

Apply Ratio R = Ratio l

10 Ω bagi jarak 100 cm
? Ω untuk 61.5 cm



(b) internal resistance of cell E1 E = 2.0 V S2 B
lAB =1.0 m ,RAB = 10.0 
S1 and S2 are closed, lAJ = 10.0 cm A X
When switch S2 is closed, the current in E1
loop is: R2
J
When switch S1 and S2 are closed,
lAJ = 10.0 cm : G
VAJ = potential difference across R1,
Apply Ratio R = Ratio l R1 =5.0  S1
10 Ω bagi jarak 100 cm

? Ω untuk 10.0 cm



E = 2.0 V S2 B

A X
E1
R2
J

G

R1 =5.0  S1