kirchhoff voltage law and current law! and introduction to lumped circuit abstraction..

1. INTRODUCTION AND LUMPED
CIRCUIT ABSTRACTION

2. WHAT IS ENGINEERING?
Purposeful use of science

4. THE ABSTRACTION
 The laws of physics can be viewed as a layer of
abstraction between the experimental data and
the practitioners who want to use specific
phenomena to achieve their goals.
 For example, Newton’s laws of motion are simple
statements that relate the dynamics of rigid
bodies to their masses and external forces.

5. THE CIRCUIT ABSTRACTION
 The abstraction mechanism is very powerful
because it can make the task of building complex
systems tractable. As an example, consider the
force equation:
 F = ma
 This simple force abstraction allows us to
disregard many properties of objects such as
their size, shape, density, and temperature, that
are immaterial to the calculation of the object’s
acceleration.

6. THE LUMPED CIRCUIT
ABSTRACTION

12. EASY WAY

14. THE LUMPED CIRCUIT
ABSTRACTION
 A careful analysis of the physical and electrical
properties of any network element or system is
called lumped circuit abstraction.
 For ex: electric bulb
 V/I = R
 Resistance is the property of interest to us.
 For battery voltage is the property of interest.
 Any electrical element can be treated as black
box through terminals and the behavior of the
element is more important then internal details.
 A lumped circuit element is often used as an
abstract representation of a piece of material
with complicated internal behavior.

15. THE LUMPED CIRCUIT
ABSTRACTION
Notice also that the orientation and shape of the wires
are not relevant to our computation. We could even
twist them or knot them in any way. Assuming for now
that the wires are ideal conductors and offer zero
resistance.
If the wires offer nonzero resistance, we can separate
each wire into an ideal wire connected in series with a
resistor.
lumped circuit abstraction of the
Light bulb circuit

16. LUMPED CIRCUIT ABSTRACTION
 If the battery supplies a constant voltage V and
has zero internal resistance, and if the resistance
of the bulb is R, we can use simple algebra to
compute the current flowing through the bulb as
Q. Draw the lumped circuit of abstraction battery has
a internal resistance r and each connecting wire is also
having resistance of r ohm. Also give Current voltage
relation for the circuit.

17. LUMPED CIRCUIT ABSTRACTION
 Lumped elements in circuits must have a voltage
V and a current I defined for their terminals.
 The circuit comprising a set of lumped elements
must also have a voltage defined between any
pair of points, and a current defined into any
terminal.
 The internal physical phenomena that make an
element to function must interact with external
electrical phenomena only at the electrical
terminals of that element.

18. LUMPED CIRCUIT ABSTRACTION
 Models must be used only in the domain in which
they are applicable. For example, the resistor
model for a light bulb tells us nothing about its
cost or its expected lifetime.

23. THE LUMPED MATTER DISCIPLINE
 The voltage, the current, and the resistance are
defined for an element only under certain
constraints that we collectively call the lumped
matter discipline (LMD).

26. THE LUMPED MATTER DISCIPLINE
 The lumped matter discipline imposes three
constraints on how we choose lumped circuit
elements:
 Choose lumped element boundaries such that the
rate of change of magnetic flux linked with any closed
loop outside an element must be zero for all time. In
other words, choose element boundaries such that
 First constraint allows us to define a unique
voltage across the terminals of an element.

27.  there are no fields related to one element that
can exert influence on the other elements. This
permits the behavior of each element to be
analyzed independently.

28. THE LUMPED MATTER DISCIPLINE
 Choose lumped element boundaries so that there is
no total time varying charge within the element for
all time. In other words, choose element boundaries
such that
 The second constraint results from our desire to
define a unique value for the current entering
and exiting the terminals of the element. A
unique value for the current can be defined if we
do not have charge buildup or depletion inside
the element over time.

29. THE LUMPED MATTER DISCIPLINE
 Operate in the regime in which signal timescales of
interest are much larger than the propagation delay
of electromagnetic waves across the lumped
elements.

30. THE LUMPED MATTER DISCIPLINE
 Consider the following thought experiment.
 Apply a current pulse at one terminal of the filament
at time instant t and observe both the current into
this terminal and the current out of the other
terminal at a time instant t + dt very close to t. If the
filament were long enough, or if dt were small
enough, the finite speed of electromagnetic waves
might result in our measuring different values for the
current in and the current out.

33. LIMITATIONS OF THE LUMPED CIRCUIT
ABSTRACTION
 The abstraction tells us to operate in the regime
in which signal timescales of interest are much
larger than the propagation delay of
electromagnetic waves across the lumped
elements.
 If we are interested in signal speeds that are
comparable to the speed of electromagnetic
waves, then the lumped matter discipline is
violated, and there-fore we cannot use the
lumped circuit abstraction.

39. NETWORK ANALYSIS
 Network: A network, in the context of electrical
and electronics, is a collection of interconnected
components
 Network analysis: Network analysis is the
process of finding the voltages across, and the
currents through, every component in the
network.

40. CIRCUIT DEFINITIONS
 Node – any point where 2 or more circuit elements are
connected together
 Wires usually have negligible resistance
 Each node has one voltage (w.r.t. ground)
 Branch – a circuit element between two nodes
 Loop – a collection of branches that form a closed path
returning to the same node without going through any
other nodes or branches twice

41. EXAMPLE
 How many nodes, branches & loops?
+
-
Vs Is
R1
R2 R3
+
Vo
-

42. EXAMPLE
 Three nodes
+
-
Vs Is
R1
R2 R3
+
Vo
-

43. EXAMPLE
 5 Branches
+
-
Vs Is
R1
R2 R3
+
Vo
-

44. EXAMPLE
 Three Loops, if starting at node A
+
-
Vs Is
R1
R2 R3
+
Vo
-
A B
C

45. KIRCHOFF’S VOLTAGE LAW (KVL)
 The algebraic sum of voltages around each loop is
zero
 Beginning with one node, add voltages across each
branch in the loop (if you encounter a + sign first)
and subtract voltages (if you encounter a – sign first)
 Σ voltage drops - Σ voltage rises = 0
 Or Σ voltage drops = Σ voltage rises

46. EXAMPLE
 Kirchoff’s Voltage Law around 1st
Loop
+
-
Vs Is
R1
R2 R3
+
Vo
-
A B
C
I2
I1
+
I2R2
-
+ I1R1 -
Assign current variables and directions
Use Ohm’s law to assign voltages and polarities consistent with
passive devices (current enters at the + side)

47. EXAMPLE
 Kirchoff’s Voltage Law around 1st
Loop
+
-
Vs Is
R1
R2 R3
+
Vo
-
A B
C
I2
I1
+
I2R2
-
+ I1R1 -
Starting at node A, add the 1st
voltage drop: -I1R1

48. EXAMPLE
 Kirchoff’s Voltage Law around 1st
Loop
+
-
Vs Is
R1
R2 R3
+
Vo
-
A B
C
I2
I1
+
I2R2
-
+ I1R1 -
Add the voltage drop from B to C through R2: -I1R1 -I2R2

49. EXAMPLE
 Kirchoff’s Voltage Law around 1st
Loop
+
-
Vs Is
R1
R2 R3
+
Vo
-
A B
C
I2
I1
+
I2R2
-
+ I1R1 -
Subtract the voltage rise from C to A through Vs: - I1R1 -I2R2 +Vs = 0
Notice that the sign of each term matches the polarity encountered 1st

50. KIRCHOFF’S CURRENT LAW (KCL)
 The algebraic sum of currents entering a node is
zero
 Add each branch current entering the node and
subtract each branch current leaving the node
 Σ currents in - Σ currents out = 0
 Or Σ currents in = Σ currents out

51. EXAMPLE
 Kirchoff’s Current Law at B
+
-
Vs Is
R1
R2 R3
+
Vo
-
A
B
C
I2
I1
Assign current variables and directions
Add currents in, subtract currents out: I1 – I2 – I3 + Is = 0
I3

52. CIRCUIT ANALYSIS
10 A 8Ω 4Ω
A
B
+
VAB
-
By KVL: - I1∙ 8Ω + I2∙ 4Ω = 0
Solving: I2 = 2 ∙ I1
By KCL: 10A = I1 + I2
Substituting: 10A = I1 + 2 ∙ I1 = 3 ∙ I1
So I1 = 3.33 A and I2 = 6.67 A
And VAB = 26.33 volts
I1 I2
+ +
- -

53. CIRCUIT ANALYSIS
10 A 2.667Ω
A
B
+
VAB
-
By Ohm’s Law: VAB = 10 A ∙ 2.667 Ω
So VAB = 26.67 volts
Replacing two parallel resistors (8 and 4 Ω)
by one equivalent one produces the same
result from the viewpoint of the rest of the
circuit.

55. FIND THE CURRENT FLOWING IN
THE 5 Ω RESISTOR OF THE CIRCUIT
SHOWN

56.  Ans= 0.2353 A

57. SOLUTION
From the left-hand loop: 10 = 2I + 5(I - I )
i.e. 7I - 5I = 10
(1)
From the right-hand loop: 3 = I - 5(I - I )
i.e. -5I + 6I = 3
(2)
5 × equation (1) gives: 35I - 25I = 50
(3)
7 × equation (2) gives: -35I + 42I = 21
(4)
Equation (3) + equation (4) gives: 17I = 71
from which, I = = 4.1765 A
From equation (1) 7I - 5(4.1765) = 10
i.e. 7I = 10 + 20.8825 = 30.8825
and I = = 4.4118 A
Hence, the current in the 5 Ω resistor = I - I = 4.4118 -
4.1765 = 0.2353 A