DC circuits - Automobile Society India

D
C circuits









Electrical
and

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C circuits



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D
C

CIRCUITS


INTRODUCTION


Electrical Engineering and Electronic Engineering deal with electricity and electromagnetic, the
two specialties have a slightly
different focus.

Electrical Engineering deals primarily with power generation and distribution, motors, electrical
control systems, and generally large scale, high power systems.


Electronic Engineering deals primarily with the processing and transmission
of analog and
digital information, which are usually low power systems. An exception would be high power
radio or television transmitters, which are within the domain of electronics.


Electrical Energy



Our daily life depends on electrical energy.



We use m
any electrical devices that transform electrical energy into other forms of energy.



For example, a light bulb transforms electrical energy into light and heat.



Electrical devices have various power requirements.


DC

If the charges move around a circuit in t
he same direction at all times, the current is said to be
direct current (dc),

which

is the kind produced by batteries.




Electrons flow constantly



The current from a battery is always in the same direction.



One end of the battery is positive and the othe
r end is negative.



The direction of current flows from positive to negative.



This is called direct current, or DC.


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Q.1

A dc are has voltage/ current relation expressed by V=44+ (30/1). It is connected in
series with a resistor across a 100 volts supply
. If the voltage across the arc and the resistor are
equal. Find the ohmic value of the resistor.


Ans.:

Let the unknown resistance be R. The voltage across the resistor and that across the arc are
equal.






V
R

= V
arc

=50 V




But V = 44+



… For the arc






50 = 44+





I =5 Amp.






R =









OHM’S LAW



Georg Simon Ohm (1787
-
1854), a German physicist, discovered Ohm’s law in 1826.

This is an experimental law, valid for both alternating current (ac) and direct current (dc)
circuits.

When you pass an electric current (I) through a resistance (R) there will be a potential difference
or voltage (V) created across the resistance.

Ohm’s
law gives a relationship between the voltage (V), current (I), and resistance (R) as
follows:




V = I R





Electric Potential
Difference, V

Electric Current, I

Electrical resistance

R=V/I

Ohm's Law
(Constant
Resistance)

Electrical
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UNITS

Quantity

Symbol

Unit Name

Unit Abbreviation

Current

I

ampere

A

Voltage

V

volt

V

Resistance

R

ohm

Ω


Q. 1

For the circuit shown in
Fig...Calculate the value of
current in each branch and the value of unknown
resistance

, when the total current taken by the circuit
is 2. 25 A.

Solution
. In the given circuit shown in Fig. resistances



and



are in series, similarly resistanc
es



and



are in
series. Now the two resistances of 10 Ω and (5 +

) Ω are the
parallel, as such their resultant value is,
























Now applying Ohm’s low to the circuit,









Or









*










+



















Or














Thus unknown resistance,




Current in the branch ABC with (2 + 8) Ω resistance =



Similarly current in the branch ADC with (5 + 3) Ω resistance =












= 1.25 A


Q. 2

For the circuit shown in Fig. 1. 6,

calculate the value o resistance



when the total
current taken by the network is 1. 5 A.



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Solution.

In the given circuit, the three resistances






and



are in parallel. Thus their
resultant resistance is given by,



















































































Now the resistance



and



are in series, hence the total resistance of the circuit.



















Appl
ying Ohm’s Law, i.e.





we get










(








)

Or








(








)







































Thus unknown resistance,







RESISTIVITY

Resistivity is a property of the material it is how well the
material conducts electricity.







This formula shows that the resistance will increase if


increases or the length increases.

The resistance will decrease if the area increases.








Q. 1

Calculate the resistance of a uniform wire of diameter 0.32

mm and length 5.0 m.

The resistivity of the wire = 5.0 X 10
-
7

Ω m

Solution































































Resistance = 31Ω



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RESISTIVITY OF MATERIALS

Resistivity is an inherent property of a material, inher
ent in the same sense that density is an
inherent property.


SERIES AND PARALLEL CONNECTION

Resistors connected in a circuit in series or parallel can be simplified using the following:



Series connection







Parallel connection








STAR
-
DELTA
TRANSFORMATION


This type of connection is very common for power supply transformers. It has a provision of star
point for mix loading, and a delta winding to carry third harmonic currents, which stabilize the
star point potential. Normally HV winding is d
elta connected and LV star connected, so that four
wire system can be used on LV side.























































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EFFECT OF TEMPERATURE ON RESISTANCE

Effect of temperature on resistance of pure me
tals:

The resistance R is a temperature dependent quantity. For different types of conductors the
amount of change in resistance due to change in temperature will have different values.

For pure metals, the resistance increases linearly with increase in te
mperature as shown in Fig
and for a certain range of temperature (typically 0 to 100˚C) the rate of increase of resistance
(slope of the line ) remains constant.




Fig
-

Change in resistance of pure metals due to change in temperature

The resistance of
metals reduces with reduction in temperature and it reduces to 0 Ω at a
temperature of
-
234.5˚C as shown in Fig. 1 The increase in the resistance due to increase in
temperature as follows:

At low temperature, the ions (Changed particles) inside the condu
ctor are almost stationary.




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KIRCHOFF’S LAW FOR DC CIRCUITS

Kirchoff’s First Law
–

The sum of the currents entering a junction is equal to the sum of the
currents leaving that junction.

















Kirchoff’s Second Law
–

The sum of the emf’s around any closed loop must equal the sum of
the IR drops around that same loop.


















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NODAL ANALYSIS

The nodal analysis is based on Kirchhoff’s current l
aw (KCL) unlike Maxwell’s mesh analysis
which is based on Kirchhoff’s voltage law. Nodal analysis also has the same advantage i.e. a
minimum number of equations to be written to solve the network.

For nodal analysis, node is defined as the point where mor
e than two elements are joined
together. If there are N nodes is the circuit. Then one of these node is Chosen as the reference or
datum node and equations based on KCL are written for the remaining (N
-
1) nodes.

At each of these (N
-
1) nodes, a voltage is a
ssigned with respect to the reference node. These
voltages are unknowns and are to be determined for determined for solving the network.

Referring the circuit shown is Fig.1.25 which has three nodes A, B and C. Node C is taken as the
reference node and alg
ebraic equations based on KCL are written for Node A and B Assume,
voltage at node A as V
A

and voltage at node B as V
B.


Applying KCL at node A,




























Similarly, applying KCL at node B,




























By solving Eqs (1.13) and (1.14), the value of unknown voltage


and


are determined based
on which the branch current could be calculated.


SOURCE TRANSFORMATIONS

Two sources are said to be equivalent if for any given load resistance connected to

the sources,
the same load voltage and current are produced.

SUPERPOSITION THEOREM



Used to find the solution to networks with two or more sources that are not in series or
parallel.



The current through, or voltage across, an element in a network is equal
to the algebraic sum
of the currents or voltages produced independently by each source.



Since the effect of each source will be determined independently, the number of networks to
be analyzed will equal the number of sources.



The total power delivered to a

resistive element must be determined using the total current
through or the total voltage across the element and cannot be determined by a simple sum of
the power levels established by each source.

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THEVENIN’S THEOREM

This theorem state that the current th
rough any load resistance, connected across any two points
of an active network, can be obtained by diving the potential difference between these two points
with the load resistance disconnected (equivalent Thevenin’s voltage, V
th
), by the sum of load
resi
stance and the resistance of the network measured between these points with load resistance
disconnected and sources of emf, replaced by their internal resistances (equivalent Thevenin’s
resistance).

Thevenin’s theorem will become quite clear with the foll
owing illustration. Consider the circuit
given in below Fig. in which it is desired to find the current flowing through the load resistance



removing the load resistance from
Fig. (a),

the circuit shown in
Fig.(b)

is obtained, in which
open circuit volt
age V
th

across the terminals C and D is to be found out.


Current through resistance











Potential difference across



(







)







As the current through,



is zero, the potential difference across CD,


















Fig (c)

s
hows a circuit, in which the source of emf V
1

has been replaced by its internal resistance



and the load resistance remains disconnected.

Resistance of the network between C and D,






















The Thevenin’s equivalent circuit of the gi
ven network has been shown in
Fig.(d).

Hence, the
current flowing through load resistance

























NORTON’S THEORM


This theorem is the dual of thevenin’s

Theorem. It states that current, I. through any load
resistance, R, connected across any two points of a linear, active network can be obtained by
reducing the network across the load terminals by a single current source (Norton equivalent,


)
and a pa
rallel resistance (Norton equivalent resistance,





Then,

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Norton equivalent current


is equal to the current, the current, which flows through the load
terminals when these two terminals are shot circuited. Norton equiva
lent resistance


is equal to
the resistance of the network measured between the load terminals with load disconnected and
the sources replaced with their internal resistances. The Norton equivalent resistance is equal to
the thevenin’s equivalent resist
ance.


N
orton’s theorem will be clearer with the following illustration. Consider the circuit given in Fig
-

(a) in which it is desired to calculate the current flowing through the load resistance R. For
calculating the value of Norton equivalent branch


. Short circuit the load terminals with load
disconnected and the source replaced with their internal resistances. The Norton equivalent
resistance I equal to the Thevenin’s equivalent resistance.


Norton’s theorem will be clearer with the following illus
tration. Consider the circuit given in Fig.
1.23 (a) in which it is desired to calculate the current flowing through the load resistance R. For
calculating the value of Norton equivalent current




short circuit the load circuit the load
resistance R ac
ross the terminals C and D as shown in Fig
-

(b) and solve the network for current
flowing through branch CD,



Fig
-

(c) shows the circuit for calculating Norton’s equivalent
resistance




in which voltage source



is replaced by its internal resi
stance



and the load
resistance remains disconnected. Resistance across the terminal C and D is the Norton’s
equivalent resistance






Fig
-

Illustration of Norton’s theorem

The Norton’s equivalent circuit of the given linear, active network is sho
wn in Fig
-

(d) hence,
the current flowing through the load resistance R is,























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Example on Norton’s Theorem

Solve the circuit shown in Fig. (below) for the current in the branch AB using Norton’s theorem.


Fig

Solution:

As per the Norton’s theorem. First the Norton equivalent current



is to be calculated
by short circuiting the terminals A and B as shown in Fig
-

(a).






Fig. (a)

The voltage source
of 20 V in series with a 2Ω resistance can be replaced by a current source of
10 A in parallel with a resistance of 2Ω as shown in Fig.
-

(b). The two current sources are
supplying the current in the same direction and hence can be replaced by a single curr
ent source
of (10+10) = 20 A. Also, the resistances of 2Ω each are in parallel and can be replaced by a1Ω
resistance in parallel with the current source of 20 A shown in Fig
-

(c).





Fig
-

(b)

Norton equ
ivalent current,









X 20





= 10 A

Next the Norton equivalent resistance


.










Fig
-

(c)


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Is to be calculated by disconnecting the load resistance across terminals A and B and resistances,
across terminal A
and B and replacing the sources with their internal resistances. That is short
circuiting the voltage source and open circuiting the current source as shown in Fig
-

(d).

The two 2Ω resistances are in parallel and the combination is in series with the 1Ω r
esistance, i.e.























Ω






Fig
-

(d)

Thus, the Norton’s equivalent circuit is shown in Fig.
-

(e). Current in the load resistance of 8


is
given by,











X10






= 2A (from A to B)








Fig
-

(e)

Thus, current flowing in the branch AB of 8


resistance is 2Afrom A to B.


MAXIMUM POWER TRANSFER THEOREMS



The maximum power transfer theorem states the following:


A load will receive maximum power from a network when its total resistive value is exactly
equal to the Thévenin resistance of the network applied to the load. That is,

R
L

= R
Th




For loads connected directly to a dc voltage supply, maximum power will be

delivered to the
load when the load resistance is equal to the internal resistance of the source; that is, when:

R
L

= R
int