Electrical elements are conceptual abstractions representing idealized electrical components, such as resistors, capacitors, and inductors, used in the analysis of electrical networks. All electrical networks can be analyzed as multiple electrical elements interconnected by wires.

1. ELEMENTS OF ELECTRICAL ENGINEERING
DC CIRCUITS
GUJRAT TECHNOLOGICAL
Mahatma Gandhi Institute Of Technical
Education And Research Center

2. PRESENTED BY :
Vaibhav Tandel (160330109xxx)
BRANCH: ELECTRICAL ENGINEERING
SUBJECT: ELEMENTS OF ELECTRICAL
ENGINEERING

3. CIRCUIT ELEMENTS
1-3
• Resistors
• Capacitors
• Inductors
2

4. RESISTORS
1-4
Resist the flow of current
Energy (voltage) is required to
force current thru them
this “voltage energy” is
transformed to heat
analogous to a valve like a faucet
Ohms’ Law: v = iR
2

5. CAPACITORS
1-5
Stores up and releases charge
(energy) depending on the
change in voltage
i = C dv/dt
i is the “displacement current”
cuz the dielectric it does not
conduct current
2

6. INDUCTORS ( )
1-6
Stores up and releases
(voltage) energy
depending on the change
in current
v = L di/dt
2

7. Circuit Elements: Ideal independent voltage source
An ideal dependent voltage source is characterized as having a
constant voltage across its terminals, regardless of the load
connected to the terminals.
The ideal voltage source can supply any amount of current.
Furthermore, the ideal independent voltage source can supply any
amount of power.
The standard symbols of the ideal independent voltage source are
shown below.
+
_v(t) E
Sometimes
used
Most often
used
16

8. Circuit Elements: Ideal independent current sources
An ideal independent current source is characterized as
providing a constant value of current, regardless of the load.
If the current source is truly ideal, it can provided any value
of voltage and any amount of power.
The standard symbol used for the ideal independent current
source is shown below.
i(t)
17
1 meg 1 amp
+
-
V
V = ?

9. Circuit Elements: Dependent voltage source
A dependent voltage source is characterized by depending on
a voltage or current somewhere else in the circuit. The symbol
For the current source is shown below. Note the diamond shape.
A circuit containing a dependent voltage source is shown below.
10  20 
30 
12 +_5 V
Iy
10Iy
A circuit with a current
controlled dependent
voltage source.
19

10. Circuit Elements: Dependent current source
A dependent current source is characterized by depending on
a voltage or current somewhere else in the circuit. The symbol
for a dependent current source is shown as follows:
A circuit containing a dependent current source is shown below.
10  20 
30 
12 +
5 V
+
_
vx
4vx_
A circuit with a voltage
controlled dependent
current source
20

11. RESISTORS IN SERIES AND
PARALLEL
As with capacitors, resistors are often in series
and parallel configurations in circuits
Series
Parallel
The question then is what is the equivalent resistance

12. RESISTORS IN SERIES
Since these resistors are in series, we have the same current in all three
resistors
IIII  321
We also have that the sum of the potential differences across the three resistors
must be the same as the potential difference between points a and b
ybxyaxab VVVV 

13. RESISTORS IN SERIES
Then using
321 ;; RIVRIVRIV ybxyax 
We have that  321 RRRIVab 
Now the equivalent resistor, R, will also have the same potential difference
across it as Vab, and it will also have the same current I
RIVab 
Equating these last two results, we then have that

i
iRRRRR 321
The equivalent resistance for a sequence of resistors in series is just the sum
of the individual resistances

14. RESISTORS IN PARALLEL
Here we have that the voltage across each resistor has to be the same (work
done in going from a to b is independent of the path, independent of which
resistor you go through)
abVVVV  321

15. RESISTORS IN PARALLEL
We now deal with currents through the resistors
At point a the current splits up into three distinct currents
We have that the sum of theses three currents must add to the value coming
into this point
321 IIII 
We also have that
3
3
2
2
1
1 ;;
R
V
I
R
V
I
R
V
I ababab 
The equivalent resistor, R, will have also have the current I going
through it

16. RESISTORS IN PARALLEL
Using
R
V
I ab
and combining with the previous equations, we then have
321 R
V
R
V
R
V
R
V abababab 
or

i iRRRRR
11111
321
The inverse of the effective resistance is given by the sum of the inverses of the
individual resistances

17. SOLVING RESISTOR NETWORKS
• Make a drawing of the resistor network
• Determine whether the resistors are in
series or parallel or some combination
• Determine what is being asked
• Equivalent resistance
• Potential difference across a particular resistance
• Current through a particular resistor

18. SOLVING RESISTOR NETWORKS
• Solve simplest parts of the network first
• Then redraw network using the just calculated
effective resistance
• Repeat calculating effective resistances until
only one effective resistance is left

19. SOLVING RESISTOR NETWORKS
Given the following circuit
What is the equivalent resistance and what is the current through each
resistor
We see that we have two resistors in parallel with each other and the effective
resistance of these two is in series with the remaining resistor

20. SOLVING RESISTOR NETWORKS
Current through this effective resistor is given by
Amps
R
V
I
eff
3
6
18

The current through the resistors in the intermediate circuit of Step 1 is
also 3 Amps with the voltage drop across the individual resistors being
given by
VoltsV
VoltsV
623
;1243
2
4





21. Step 1: Combine the two resistors that
are in parallel
yielding
Step 2: Combine the two resistors that are in series
 624 effR yielding
SOLVING RESISTOR NETWORKS


2;
2
1
3
1
6
11
 eff
eff
R
R

22. To find the current through the resistors of the parallel section of the initial circuit,
we use the fact that both resistors have the same voltage drop – 6 Volts
Amps
Volts
I
Amp
Volts
I
2
3
6
;1
6
6
3
6








SOLVING RESISTOR NETWORKS

23. KIRCHOFF’S RULES
• Not all circuits are reducible
There is no way to reduce the four resistors to
one effective resistance or to combine the three
voltage sources to one voltage source

24. KIRCHOFF’S RULES
First some terminology
A junction, also called a node
or branch point, is is a point
where three or more
conductors meet
A loop is any closed
conducting path

25. KIRCHOFF’S RULES
Kirchoff’s Rules are basically two statements
1. The algebraic sum of the currents into any
junction is zero   0I
A sign convention:
0321  III
A current heading towards a junction, is considered to be positive,
A current heading away from a junction, is considered to be negative
Be aware that all the junction equations for a
circuit may not be independent of each other

26. KIRCHOFF’S RULES
2. The algebraic sum of the potential differences in any loop including those
associated with emfs and those of resistive elements must equal zero
  0V
Procedures to apply this rule:
Pick a direction for the current in each branch
If you picked the wrong direction, the current will come out negative

27. KIRCHOFF’S RULES
Pick a direction for traversing a loop – this direction must be the same for
all loops
Note that there is a third loop along the outside branches
As with the junction equations not all the loop equations
will be independent of each other.

28. KIRCHOFF’S RULES
Starting at any point on the loop add the emfs and IR terms
An IR term is negative if we traverse it in the same sense as the current that
is going through it, otherwise it is positive
An emf is considered to be positive if we go in the direction - to +,
otherwise it is negative
Need to have as many independent
equations as there are unknowns

29. KIRCHOFF’S RULES
For loop I we have 031432111  RIRIRI
For loop II we have 0324332  RIRI
Junction equation at a gives us 0321  III
We now have three equations for the three unknown currents

30. KIRCHOFF’S RULES
Assume that the batteries are: 1 = 19 V; 2 = 6 V; 3 = 2 V
and the resistors are: R1 = 6; R2 = 4; R3 = 4; R4 = 1
you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A
The minus sign on I2 indicates that the current is in fact in the opposite
direction to that shown on the diagram
Complete details can be found here

31. STAR-DELTA TRANSFORMATION

32. EQUIVALENCE
•Equivalence can be found on
the basis that the resistance
between any pair of terminals in
the two circuits have to be the
same, when the third terminal is
left open.

33. • First take delta connection: between A and C, there are
two parallel paths, one having a resistance of R2 and
other having a resistance of ( R1+R3)
Hence resistance between terminal A and C is
= R2.(R1+R3)/[R2+( R1+R3)]

34. • Now take the star connection
The resistance between the same terminal A and C is (RA+RC)
Since terminal resistance have to be same so we must have
(RA+RC) = R2.(R1+R3)/[R2+( R1+R3)] (1)
Similarly for terminals A and B, B and C, we can have the following
expression
(RA+RB) = R3.(R1+R2)/[R3+( R1+R2)] (2)
(RB+RC) = R1.(R2+R3)/[R1+( R2+R3)] (3)

35. DELTA to STAR
Now subtracting 2 from 1 and adding the result to 3, we will get the following values
for R1,R2 and R3.
How to remember?
Resistance of each arm of star is given by the product of the
resistance of the two delta sides that meet at its ends divided by
the sum of the three delta resistance

36. STAR to DELTA
Multiplying 1 and 2, 2 and 3 , 3 and 1 and adding them together and
simplifying, we will have the following result.
How to remember: The equivalent delta resistance between any two point is
given by the product of resistance taken two at a time divided by the opposite
resistance in the star configuration.

37. PROBLEM
• A delta-section of resistors is given in figure. Convert
this into an equivalent star-section.
.5.1;0.1;3:.  CBA RRRAns

38. PROBLEM
The figure shows a
network. The
number on each
branch represents
the value of
resistance in ohms.
Find the resistance
between the points E
and F.

39. SOLUTION

40. Ans. : 5.6 Ω

41. PROBLEM
• Find the current drawn from the 5 volt battery in the
network shown in figure.

42. SOLUTION :

46. Ans. : 0.974 A

47. Note :
• During the network reduction or
simplification process, some points
in the original network are lost.
•Care must be taken during this
process that no point of ultimate
relevance is lost.

48. THANK YOU