CHAPTER 3: ELECTRIC CURRENT & DIRECT-CURRENT CIRCUIT

3.5 Resistors in series and parallel

Learning Outcome:
a) Determine effective resistance of resistors in series and parallel.

50

3.5 a) Determine effective resistance of
resistors in series and parallel

Resistors in Series

The properties of resistors in series are given below: 51
= 1 = 2 = 3
= 1 + 2 + 3

but = → = 1 + 2 + 3

= 1 + 2 + 3

Resistors in Parallel

The properties of resistors in parallel are

given below:

= 1 = 2 = 3

= 1 + 2 + 3

but =



= 1 + 2 + 3

1 111
= 1 + 2 + 3

Note: Effective resistance is also called as equivalent/total resistance. 52

R1 R2
R3

6.0 V

Example 3.5.1

Given 1 = 2.0 Ω, 2 = 12.0 Ω and 3 = 4.0 Ω. Calculate:

a) the total resistance of the circuit.

b) the total current in the circuit.

c) the potential difference across the 4.0 Ω resistor.

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a) 1 = 1 + 1 = 1 + 1 = 1

23 2 3 12 4 3

23 = 3.0 Ω
= 1 + 23 = 2 + 3 = 5.0 Ω

b) = = 6 = 1.2
5

c) The potential difference across 1 is
1 = 1 = 1.2 2 = 2.4

Therefore, the potential difference across 3 is given by
3 = − 1
= 6.0 − 2.4

= 3.6

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Example 3.5.2

A 20  5.0 

10  10 
5.0 
B

For the circuit above, calculate the effective resistance between the
points A and B.

Solution : R1  5.0 ; R2  5.0 ; R3  10 ; R4  20 ;

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Example 3.5.2

Solution :

A R4 A R4 R12

R5 R3 R5 R3 56

B R2 R1

B

R1 and R2 are connected in series, thus R12 is

R12  R1  R2 R12  5.0  5.0  10 

Example 3.5.2

Solution : R4 Since R12 and R3 are connected in
parallel , thus R123 is given by
A R123
1  1 1 1 11
R5 R123 R12 R3 R123 10 10

B R123 and R4 are connected in series ,

A thus R1234 is given by

R5 R1234 R1234  R123  R4 R1234  5.0  20

B R1234  25 

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Example 3.5.2 A
Solution :
Reff

B

SinceR1234 and R5 are connected in parallel , therefore the
effective resistance Reff is given by 1  1  1
Reff R1234 R5
1 11
Reff 25 10

Reff  7.14 

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3.6 Kirchhoff’s Rules

Learning Outcomes:
a) State and apply Kirchhoff’s Rules.
b) Use Kirchhoff’s Laws.
* i) Maximum two closed circuit loops
ii) Use scientific calculator to solve the simultaneous equations

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3.6 a) State and apply Kirchhoff’s Rules

Kirchhoff’s First Law (junction/current rule)
• states that the algebraic sum of the currents at any junction of a

circuit is zero,
Σ = 0 at any junction
or
Σ = Σ at any junction

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Junction Junction

Σ = Σ
1 + 2 + 3 = 4 + 5

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Kirchhoff’s Second Law (loop/voltage rule)
• states that the algebraic sum of the voltages across all of the

elements of any closed loop is zero
or

• states that in any closed loop, the algebraic sum of emfs are equal to
the algebraic sum of the products of current and resistance,
Σ = Σ in any closed loop

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Example 3.6.1
Using Kirchhoff’s laws, find the current in each resistor.

Solution: 63
Step
Draw current. (arbitrary)
Draw loop. (arbitrary)
Apply Kirchhoff’s laws.

Σ = Σ
1 − 2 = 1 + 2
20 − 10 = 10 + 20
10 = 30
= 0.33

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Example 3.6.2
Apply Kirchhoff’s rules to the circuit in figure below and find the
current in each resistor.

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Solution:
Σ = Σ
2 − 1 = 1 + 2 + 3 + 4
3.0 − 6.0 = 3.0 + 4.0 + 5.0 + 2.0
−3.0 = 14
= −0.21
∴ = 0.21
(-ve sign on current means the direction that current ACTUALLY flows is
opposite the direction of current assigned on the circuit)

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Example 3.6.3
Find the current in each resistor in the circuit shown below.

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Solution:

Loop 1 Loop 2

68

Using Kirchhoff’s 1st Law: 69
Σ = Σ at any junction
1 + 3 = 2 ……….eqn. (1)

Using Kirchhoff’s 2nd Law:
Σ = Σ in any closed loop
Loop 1: 2 + 1 = 1 1
15 = 4 1
1 = 3.75
Loop 2: 2 + 3 = 3 3 + 3 2
10 = 8 3
3 = 1.25

From eqn. (1), 2 = 3.75 + 1.25 = 5.0

Example 3.6.4
Calculate the currents I1, I2 and I3. Neglect the internal resistance in
each battery.

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Solution:
Using Kirchhoff’s 1st Law:

1 = 3 + 2……….(1)
Using Kirchhoff’s 2nd Law:

Loop 1: 1 + 2 = 1 1 + 2 2
15 + 10 = 1(1) + 2(0.5) ……….(2)

Loop 2: 3 − 2 = 3 3 − 2 2
3−10 = 3(0.1) − 2(0.5) ……….(3)

Using calculator:
1 = 17.69 , 2 = 14.62 , 3 = 6.07

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*Note: Make sure to arrange your equations so that the terms are lined
together. For example:

Eqn. (1) → 1 1 − 1 2 − 1 3 = 0
Eqn. (2) → 1 1 + 0.5 2 + 0 3 = 25
Eqn. (3) → 0 1 − 0.5 2 + 0.1 3 = −7

Therefore, the coefficients are:

Eqn. (1) A B CD
Eqn. (2) 1 -1 -1 0
Eqn. (3) 1 0.5 0 25
0 -0.5 0.1 -7

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3.7 Electrical Energy and Power

Learning Outcomes:

a) Use power, P = IV , = 2 and = 2 . (Known as power loss)


b) Use electrical energy, =

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3.7 a) Use power, P = IV , = 2 and = 2 .



• Electric energy is useful to us because it can be transformed into
other forms of 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.

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• = (the work done by the source on the charge)
• But = , then =
• Unit: Joule (J)
• The rate of energy delivered to the external circuit by the battery is

called the electric power given by,

= = =



= @ =
• Unit: Watt (W) or Joule/second −1

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• The energy dissipated per second in an electric device (rate of energy
dissipated) is given as

= = = for any device



• A passive resistor is a resistor which converts all the electrical energy

into heat. For example, a metal wire.

= but =

= 2 or = 2



Only for resistor

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Example 3.7.1

Calculate the resistance of a 40 W automobile headlight designed for
12 V.

Solution:
Given, = 40 , = 12

From = 2 , = 2 = 122 = 3.6 Ω
40

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Example 3.7.2
The current through a refrigerator of resistance 12 Ω is 13 A. What is
the power consumed by the refrigerator?
Solution:
Given, R = 12 Ω, = 13
From = 2

= 13 2 12
= 2028

78

Example 3.7.3

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 hour?

Solution:
Given, R = 15 Ω, = 120 , = 60 × 60 = 3600

From = but =


= 2 = 120 2(3600) = 3.46 × 106

15

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3.8 Potential Divider

Learning Outcomes:

a) Explain the principle of a potential divider.

b) Use equation of potential divider, 1 = 1
1+ 2+..+

*Maximum three resistors

80

3.8 a) Explain the principle of a potential
divider

• A potential divider is a fraction of the voltage supplied by a source of

emf V

I R2 I

R1

V1 V2 81

Potential divider circuit

• Two resistors are connected in series. The current flows in each

resistor is the same:

= = ; = 1 + 2
1+ 2

• Potential difference across R1 is

1 = 1 → 1 = 1
1+ 2

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• Now, resistance R1 and R2 are replaced by a uniform homogeneous
wire as shown in figure below.

V

II
l1 l2
a b
c

V1 V2

• Substituting = into 1 = 1 will gives us:
1+ 2

1 = 1
1+ 2
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3.8 b) Use equation of potential divider

Example 3.8.1

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 resistor

b) 6.0 Ω resistor?

Solution:

a) 1 = 1 = 3 12 = 4
1+ 2 3+6

b) 2 = 2 = 6 12 = 8
1+ 2 3+6

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Example 3.19

Refer to the figure, calculate the
voltage drop for each resistor.

Solution:

= 1 + 2 + 3 = 12

1 = 1 2
= 12 × 24 = 4

2 = 2 4
= 12 × 24 = 8

3 = 3 6
= 12 × 24 = 12

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3.9 Potentiometer

Learning Outcomes:

a) Explain the principle of potentiometer and their applications.

b) Use related equations for potentiometer, 1 = 1
2 2

c) Determine internal resistance, r of a dry cell by using

potentiometer.

(Experiment 3: Potentiometer)

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3.9 a) Explain the principle of potentiometer
and their applications.

• A potentiometer is mainly used to measure potential difference, V.
• It consists of a uniform wire.
• Basically a potentiometer circuit consists of a uniform wire AB of

specific length (usually 100.0 cm), connected in series to a driver cell
with emf, V of negligible internal resistance.

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• 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 is
balanced, the unknown voltage
(potential difference being
measured) is equal to the
voltage across AC.

=

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Equation of potentiometer:
1 = 1 = 1

2 2 2

Potentiometer can be applied to:
i) Measure an unknown emf of a cell.
ii) Compare the emf of two cells.
iii) Measure the internal resistance of a cell. (Exp. 3: Potentiometer)

89

i) Measure an unknown emf. of a
cell

When the potentiometer is
balanced, = 0.

∴ Balanced length =

=

= ..........eqn. (1)

= ..........eqn. (2)


= ..........eqn. (3)


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. (2) ∶ = =
. (3)




= ..........eqn. (4)

= ..........eqn. (5)

Substitute eqn. (4) and (5) into eqn. (1):


=

=


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Example 3.9.1

A potentiometer consists of a
150 cm wire XY and internal
resistance of 0.5 Ω. When the
cell is connected to the
potentiometer, the balanced
length XP is 72.6 cm.
Calculate the emf of the cell,
Ɛ.

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Solution:

Total resistance of the potentiometer:

= 0.5 + 1.0 + 1.6 = 3.1 Ω
Resistance of wire XY = 0.5 Ω

For wire XY, ∝ :

= 72.6 → = 0.24 Ω
0.5 150

Using potential divider equation:

= =


= 0.24 × 2

3.1

= 0.15

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ii) Compare the emf of two cells

When the potentiometer is
balanced, = 0.

∴ Balanced length: = 1for ε1
and = 2 for ε2

From = , thus


1 = … . 1


2 = … . 2


. (2) : 2 = 1 @ 2 = 2 1
. (1) 1

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Example 3.9.2

When S is connected to X, the
balanced length is 20 cm.
Balanced length is 40 cm
when S is connected to Y.
Calculate 2 if V and 1are 5.0
V and 2.0 V respectively.

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Solution:

Find by comparing the driver cell and 1:

1 = ,

5.0 =
2.0 20

= 50

Find 2 by comparing 2 with V: ,
2
=

2 = 40
5.0 50

2 = 4.0 96

OR using 1 = 1 = 1
2 2
1 2
2
=

2 20
2 = 40
2 = 4.0

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END OF TOPIC 3