CHAPTER 3: ELECTRIC CURRENT AND DC-CIRCUITS

3. Electric Current &
Direct-Current Circuits

1  DP024

ANALOGY: Electric circuits

Voltage: A force that Resistance: Current:
pushes the current Friction that The actual
through the circuit (in impedes flow of “substance”
this picture it would be current through the that is
equivalent to gravity) circuit (rocks in the flowing
river) through the
2 wires of the
circuit
(electrons!)

4.1 Electric Current &
Direct-Current Circuits

(a) Define and use electric current, I = dQ

dt

(b) Define and use resistivity,

ρ = RA
l

(c) State and use Ohm’s law

(d) Sketch V-I graph for ohmic conductor

(e) Calculate the effective resistance of resistors in series and parallel by
showing the following formulae:

i. Series: = 1 + 2 + 3 + … +

ii. Parallei, 1 111 1
= 1 + 2 + 3 + ⋯ +

(Experiment 4: Ohm’s law: Resistor in Series & Parallel)

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(a) Define and use Electric
Current, I

 Electric current, I is defined as the rate of flow of charge
Mathematically,
I = dQ
Average I=Q Instantaneous dt
current t current

where Q is charge
n is the number of electrons flow
e = charge of electron =1.610−19C

 One Ampere is the flow of one Coulomb charge through an
area in one second

1ampere = 1coulomb = 1C s−1
1 second

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Electric Current, I It is a base and scalar quantity
S.I. unit: ampere (A).

 Consider a simple closed circuit consists of
wires, a battery and a light bulb as shown

 From the diagram,

 Direction of electric field or electric current:
(+)ve to (–)ve

 Direction of electron flows: (–)ve to (+)ve

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Example 4.1 (b) A silver wire carries a
current of 3.0 A.
(a) There is a current of 0.5 A Determine
in a flashlight bulb for 2
min. How much charge (i) the number of electrons
passes through the bulb per second pass through
during this time? the wire,

Solution: N = 1.88 1019 electrons s−1
t
I=Q (ii) the amount of charge
t flows through a cross-
sectional area of the wire
Q = 60C in 55 s.

Q = 165 C

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(b) Define and use resistivity

Resistivity,  (specific resistance)

 is defined as the resistance of a unit cross-
sectional area per unit length of the
material

ρ = RA where R= resistance,

l A= cross sectional area
 Scalar quantity l = length
 Unit: ohm meter ( m)

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Resistivity,  Material Resistivity,  (  m)
Silver 1.59  10−8
 It is a measure of a material’s Copper 1.68  10−8
ability to oppose the flow of 2.82  10−8
an electric current Aluminum 2.44  10−8
Gold 1010−1014
 Resistivity depends on the type Glass
of the material &
temperature Conductivity,  * EXTRA

 A good electric conductors   is defined as the reciprocal of the
have a very low resistivities, resistivity of a material

 good insulators  have very Mathematically, 1
high resistivitiesa1s4s  Scalar quantity, ρ

unit: −1 m−1 σ =

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Example 4.2 Solution:
L = 1.0 m,
A constantan wire of length A = 0.5 mm2 = 0.5 x 10–6 m2,
1.0m and cross sectional ρ = 4.9 x 10–7 Ω m
area of 0.5 mm2 has a
resistivity of 4.9 x 10–7 Ωm. Using: R =  L
Find the resistance of the
wire. A

= 4.910−7 (1.0)
0.5 10−6

R = 0.98 

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Example 4.3 RP = 3RQ and R = ρl
A
Two wires P and Q with ρPlP = 3 ρQlQ and A = πd 2
circular cross section are made AP AQ 4
of the same metal and have
equal length. If the resistance 4 ρl = 3 4 ρl 
of wire P is three times greater πd P 2 πd Q 2 
than that of wire Q, determine
the ratio of their diameters. dQ = 3 OR dP = 1
dP dQ 3
Solution:

Same metal: P = Q

Same length: lP = lQ

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Exercise 4.1

1. A wire 5.0 m long and 3.0 mm in diameter has a
resistance of 100 .A 15 V of potential difference is
applied across the wire. Determine

(a) the current in the wire,
(b) the resistivity of the wire,
(c) the rate at which heat is being produced in the wire.

(College Physics,6th edition,Wilson, Buffa & Lou, Q75,
p.589)

ANS: 0.15 A; 1.414  10−4  m; 2.25 W

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(c) State and use Ohm’s law

 Ohm’s law states that the potential difference, V
across a conductor is proportional to the
current, I flowing through it if its temperature is
constant

 V I where T is constant
where R = resistance
 Then

V = IR

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Resistance, R – A ratio of the potential difference

across an electrical component to the current passing

through it R=V
I

 Scalar quantity, unit: ohm ( ) or V A−1

Length, l Cross  In general, the
sectional resistance of
area, A a conductor
increases
Type of Temperature with
material temperature

Resistance

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(d) Sketch V-I graph for ohmic
conductor

V (V) Gradient, m  Materials that obey Ohm’s law are
0 =R materials that have constant

I (A) resistance over a wide range of
voltage  ohmic conductor

 Materials that do not obey the
Ohm’s law  non-ohmic
conductors

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Revision

Q = It Q = ne V (V)

Ohm' s Law Ohmic conductor (metal)

V I V = IR Gradient, m
=R
ρ = RA 0
l I (A)

σ = 1
ρ

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(e) Calculate the effective
resistance of resistors in series &
parallel

Resistance of resistors in

(a) Series (b) Parallel

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Resistors in
Series

 Consider three resistors are  From the definition of
connected in series to the resistance, thus
battery
V1 = IR1;V2 = IR2;V3 = IR3;V = IReff
Characteristics:
 Same current I flows Substituting for V1, V2 , V3 and V

through each resistor IReff = IR1 + IR2 + IR3

I = I1 = I2 = I3 where Reff = R1 + R2 + R3

 Total potential difference,V Reff: effective) resistance
(Assumption: the connecting
wires have no resistance)

V = V1 + V2 + V3

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Resistors in  From the definition of
Parallel resistance, thus

 Consider three resistors are I1 = V ; I2 = V ; I3 = V ; I = V
connected in parallel to the R1 R2 R3 Reff
battery
Substituting for I1, I2 , I3 and I
Characteristics:
 Same potential V =V +V +V
Reff R1 R2 R3
difference, V across
each resistor 1 =1+1+1
Reff R1 R2 R3
V = V1 = V2 = V3

 Total current in the
circuit

I = I1 + I2 + I3

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Series VS Parallel In the series case the same current
flows through both bulbs.
 When connected to the If one of the bulbs burns out, there
same source, two light will be no current at all in the
bulbs in series draw less circuit, and neither bulb will glow
power and glow less
brightly than when they
are in parallel

In the parallel case the potential difference
across either bulb remains equal if one of the
bulbs burns out. The current through the
functional bulb remains equal and the power
delivered to that bulb remains the same .This is
another of the merits of a parallel arrangement
of light bulbs: If one fails, the other bulbs are
unaffected. This principle is used in household
wiring systems

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Example 4.4 R4 R2 A
R3
What is the equivalent R1
resistance of the resistors in
figure below? B

R2 A R34 A
R4 R3 B
R2
R1

R1

R1= R2= R3= R4= 1 Ω R234 B
A
20
R1 RE = 3 
5

B

Example 4.5 Solution:

Find the effective resistance
for the circuit shown below.

1 = 1+1 = 1 +1 =6
Rp R1 R2 10 2 10

Rp = 1.67

RE = R3 + Rp

=1.67 + 5.0

RE = 6.67

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Example 4.6

For the circuit below, calculate the effective
resistance of the circuit,

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Solution: R2 R12

R1

R3 R3
VV

a.The resistors R1 and R2 are in series, thus R12 is

R12 = R1 + R2 R12 = 4.0 + 12

R12 = 16 

Since R12 and R3 are in parallel, therefore Reff is given by

1 = 1 +1 1 = 1 +1
Reff R12 R3 Reff 16 2
Reff = 1.78 
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Example 4.7 Solution:

For the circuit above, Given R1 = 5.0, R2 = 5.0,
calculate the effective R3=10, R4 = 20 , R5=10 
resistance between the
points A and B. A R4 R1
R3
24 R5

R2

B

A R4 R12
R3
R5

B

Example 4.7 Solution:

 R1 and R2 are connected Given R1 = 5.0, R2 = 5.0,
in series, thus R12 is R3=10, R4 = 20 , R5=10 

R12 = R1 + R2 A R4 R1
R3
R12 = 5.0 + 5.0 = 10  R5

R2

B

A R4 R12
R3
R5

B

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Solution: R4 Since R12 and R3 are connected in
A R123 parallel , thus R123 is given by

R5 R1234 1 = 1 +1 1 =1+1
R123 R12 R3 R123 10 10
B R123 = 5.0 
A
R123 and R4 are connected in series ,
R5 thus R1234 is given by

B R1234 = R123 + R4 R1234 = 5.0 + 20

R1234 = 25 

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Solution:

A

Reff

B

Since R1234 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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Exercise 4.2

(a) What is the total resistance of the circuit (watch
out for the bear trap)?

(b) What is the current through each resistor?
(c) What is the total current?

ANS: 12.8; 0.40A, 0.30A, 0.24A; 0.94A

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Exercise 4.3

1. Determine the equivalent resistances of the resistors below.

2.0 

2.0  8.0  20 
2.0 
16 

2.0  16 

6.0  9.0  6.0 
18 
10 
4.0  ANS: 0.80 ; 2.7 ; 8.0 

6.0 

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Exercise 4.4

2. The circuit below includes a
battery with a finite internal
resistance, r = 0.50 .

(a) Determine the current flowing
through the 7.1  and 3.2 
resistors.

(b) How much current flows
through the battery?

(c) What is the potential difference
between the terminals of the
battery? (Physics,3th edition, James
S.Walker, Q39, p.728)

ANS: 1.1 A, 0.3 A; 1.4 A; 11.3 V

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