Chapter 2
Basic Laws
•Ohm’s Law
• Nodes, Branches, and Loops.
•Kirchhoff’s Laws
Objectives
we shall discuss some techniques commonly applied in circuit design and
analysis.
These techniques include combining resistors in series or parallel, voltage
division, and current division.
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Georg Simon Ohm (1787–1854), a German physicist, in
1826 experimentally determined the most basic law
relating voltage and current for a resistor.
Born of humble beginnings in Erlangen, Bavaria, Ohm
threw himself into electrical research. His efforts resulted
in his famous law.
Gustav Robert Kirchhoff (1824–1887), a German
physicist, stated two basic laws in 1847 concerning the
relationship between the currents and voltages in an
electrical network. Kirchhoff’s laws, along with Ohm’s
law, form the basis of circuit theory.
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Ohm’s Law
Resistor
It is the algebraic relationship between
voltage and current for a resistor
v=iR
where
R is the resistance in ,
i is the current in A,
v is the voltage in V, with
reference directions as pictured.
If R is given, once you know i, it is easy to
find v and vice-versa.
Since R is never negative, a resistor
always absorbs power…
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Resistors and Ohm’s Law
• Working with very primitive instruments that Ohm
designed and constructed by himself. Ohm
discovered that voltage and current were linearly
related in wires.
I =V R
I
• That means that if you measure voltage across a
wire and plot that against the current through the
wire you get a straight line in the plot.
• Working with very imprecise measurements, Ohm
was able to determine that voltage and current for
any fixed geometrical structure built from conducting
material satisfied a relationship:
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V
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Ohm’s law states that the voltage v across a resistor is directly
proportional to the current i flowing through the resistor.
• V is the voltage across the device, I is the current flowing through
the device, R is a constant.
• R depends upon the material from which the device is constructed
and the geometry of the material.
• Any resistor has a current- voltage
relationship called Ohm’s law:
+
i
v
V=iR
where R is the resistance in
,
−
i is the current in A,
V is the voltage in V, (with reference 5directions as pictured).
• If R is given, once you know i, it is easy to find v and vice-versa.
• Since R is never negative, a resistor always absorbs power.
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The resistance R of an element denotes its ability to resist the flow
of electric current; it is measured in ohms ( ).
1
= 1V/A
Since the value of R can range from zero to infinity, it is important that we
consider the two extreme possible values of R. An element with R = 0 is called a
short circuit,
v = iR = 0
A short circuit is a circuit element
with resistance approaching zero.
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R=
is known as an open circuit,
An open circuit is a circuit element
with resistance approaching infinity.
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Resistance Type
Wire wounded or carbon film
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common variable resistor is known as a potentiometer
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• not all resistors obey Ohm’s law.
• A resistor that obeys Ohm’s law is known as a
linear resistor.
•It has a constant resistance and thus its
current-voltage characteristic is as shown in
figure.
• i-v graph is a straight line passing through
the origin.
•A nonlinear resistor does not obey Ohm’s law.
• Its resistance varies with current.
•Its i-v characteristic is as shown in figure
Example: Varistor
Voltage Dependent
Resistor or VDR
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Conductance G
Conductance is the ability of an element to conduct electric current; it is
measured in mhos ( 1/ ) or Siemens (S).
mho
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Power
•
The power consumed in the
resistor is given by:
P =V ⋅I
According to Ohm' s law
•
•
•
Example: If the power absorbed by Rx
is 20 mW. Find Rx and Vab.
P = 20mW = I (IRx) = 0.002(0.002Rx)
Rx = 5000 Ω
V
I = or V = I ⋅ R
R
Therefore,
V2
P=I R=
R
2
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Example
In the circuit shown, calculate the current i, the conductance G, and the power p.
Solution
Or
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Branches, Nodes, and Loops
Branch
A branch represents a single element such as a voltage source or a resistor.
The circuit has five branches, namely, the 10-V voltage source, the 2-A
current source, and the three resistors.
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Nodes
A node is the point of connection between two or more branches.
•A node is usually indicated by a dot in a circuit.
• The circuit has three nodes a, b, and c.
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Loop
A loop is any closed path in a circuit.
For example
•The closed path abca containing the 2- resistor is a loop.
•Another loop is the closed path bcb containing the 2- resistor and the current
source.
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Series and Parallel
Two or more elements are in series if they are cascaded or connected sequentially
and consequently carry the same current.
Two or more elements are in parallel if they are connected to the same two nodes
and consequently have the same voltage across them.
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Resistor in Series
•
Suppose two elements are connected with nothing coming off in between.
•
The elements carry the same current.
•
We say these elements are in series.
i1 – i2 = 0
i1 = i2
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•
•
•
•
Consider resistors in series. This means they are attached end-to-end,
with nothing coming off in between.
Each resistor has the same current (labeled i).
Each resistor has voltage iR, given by Ohm’s law.
The total voltage drop across all 3 resistors is19
VTOTAL = i R1 + i R2 + i R3 = i (R1 + R2 + R3)
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•
When we look at all three resistors together as one unit, we see that they
have the same I-V relationship as one resistor, whose value is the sum of
the resistances:
•So we can treat these resistors as just one
equivalent resistance, as long as we are not
interested in the individual voltages.
•Their effect on the rest of the circuit is the
same, whether lumped together or not.
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Resistors in Parallel
•
•
Any set of elements which are directly connected by wire at both ends carry
the same voltage.
We say these elements are in parallel.
Vb – Va = 0
Va = Vb
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Examples
Which of these resistors are in parallel?
R2
R1
R3
R4
R5
None
R6
R8
R7
R7 and R8
R4 and R5
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•
Resistors in parallel carry the same voltage. All of the resistors below have
voltage VR .
•
The current flowing through each resistor could definitely be different. Even
though they have the same voltage, the resistances could be different.
+
R1
i1
R2
i2
R3
i3
i1 = VR / R1
VR
i2 = VR / R2
_
i3 = VR / R3
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•
If we view the three resistors as one unit, with a current iTOTAL going in, and a
voltage VR, this unit has the following I-V relationship:
iTOTAL = i1 + i2 + i3 = VR(1/R1 + 1/R2 + 1/R3)
in other words,
VR = (1/R1 + 1/R2 + 1/R3)-1 iTOTAl
So to the outside world, the parallel resistors look like one:
iTOTAL
iTOTAL
+
+
VR R1
_
i1
VR
R3
R2
i2
i3
24
REQ
_
REQ = (1/R1 + 1/R2 + 1/R3)-1
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Example:
How many branches and nodes does the circuit have? Identify the elements that are
in series and in parallel.
Solution:
Five branches and three nodes are identified.
The 1- and 2- resistors are in parallel.
The 4- resistor and 10-V source are also in parallel.
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Kirchhoff’s Laws
• Kirchhoff’s laws were first introduced in 1847 by the German physicist Gustav Robert
Kirchhoff (1824–1887).
• These laws are formally known as
Kirchhoff’s current law (KCL)
Kirchhoff’s voltage law (KVL).
The I -V relationship for a device tells us how current and voltage are related within
that device.
Kirchhoff’s laws tell us how voltages relate to other voltages in a circuit, and how
currents relate to other currents in a circuit.
KVL: The sum of voltage drops around a closed path must equal zero.
KCL: The sum of currents leaving a node must equal zero.
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Kirchhoff’s current law (KCL) states that the algebraic sum of currents
entering a node (or a closed boundary) is zero.
Mathematically, KCL implies that
where
N is the number of branches connected to the node.
in is the nth current entering (or leaving) the node.
By this law, currents entering a node may be regarded as positive, while
currents leaving the node may be taken as negative or vice versa.
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To Prove KCL,
assume a set of currents ik(t ), k = 1, 2, . . . , flow into a node.
The algebraic sum of currents at the node is
iT (t) = i1(t) + i2(t) + i3(t)+· · ·
Integrating both sides gives
qT (t) = q1(t) + q2(t) + q3(t)+· · ·
Where
qk(t) = ik(t) dt
qT (t) = iT (t) dt .
But the law of conservation of electric charge requires that the algebraic sum of
electric charges at the node must not change; that is, the node stores no net charge.
Thus
qT (t) = 0
iT (t) = 0, confirming the validity of KCL.
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Example
Apply Kirchhoff'
s law
i1 + (−i2) + i3 + i4 + (−i5) = 0
since currents i1, i3, and i4 are entering the node,
while currents i2 and i5 are leaving it.
By rearranging the terms, we get
i1 + i3 + i4 = i2 + i5
The sum of the currents entering a node is equal to the sum
of the currents leaving the node.
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KCL Equations
In order to satisfy KCL, what is the value of i?
KCL says:
24 A + -10 A + (-)-4 A + -i =0
18 A – i = 0
24 µA
-4 µA
i = 18 A
10 µA
i
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Example
Apply Kirchhoff'
s law
The combined or equivalent
current source can be found by
applying KCL to node a.
IT
IT = I1 + I3 - I2
I1
IT + I2 = I1 + I3
I2
I3
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Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of all voltages
around a closed path (or loop) is zero.
Expressed mathematically, KVL states that
Where
M is the number of voltages in the loop (or the number of branches in the loop)
vm is the mth voltage.
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Writing KVL Equations
+ v2 −
1
+
va
−
b
−
a
v3
+
What does KVL
say about the
voltages along
these 3 paths?
2
+
vb
-
c
+
vc
−
3
Path 1:
− va + v 2 + vb = 0
Path 2:
− vb − v3 + vc = 0
Path 3:
− va + v2 − v3 + vc = 0
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Example
Apply Kirchhoff’s KVL
Solution
Suppose we start with the voltage source and
go clockwise around the loop.
The voltages would be −v1,+v2,+v3,−v4, and
+v5.
Thus, KVL yields
−v1 + v2 + v3 − v4 + v5 = 0
v2 + v3 + v5 = v1 + v4
Sum of voltage drops = Sum of voltage rises
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Example
Apply Kirchhoff’s KVL
−Vab + V1 + V2 − V3 = 0
Vab = V1 + V2 − V3
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Example
For the circuit shown, find voltages v1 and v2
Solution:
To find v1 and v2, we apply Ohm’s law and
Kirchhoff’s voltage law.
Assume that current i flows through the loop
From Ohm’s law,
v1 = 2i, v2 = −3i ………………………………(1)
Applying KVL around the loop gives
−20 + v1 − v2 = 0 ………………………………(2)
Substituting (1) into (2),
−20 + 2i + 3i =0
5i = 20 i = 4 A
v1 = 8 V
v2 = −12 V
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Example
Find the currents and voltages in the circuit shown
ohm
KVL
KCL
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Solution
By applying
-Ohm’s law
-Kirchhoff’s laws.
v1 = 8i1
v2 = 3i2,
v3 = 6i3
At node a, KCL gives
i1 − i2 − i3 = 0 … … … … … … … … … … … … … … … ..… … … … .… … … .1
Applying KVL to loop 1
−30 + v1 + v2 = 0
−30 + 8i1 + 3i2 = 0
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… … … … … … … … … ...2
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Applying KVL to loop 2,
−v2 + v3 = 0 v3 = v2
6i3 = 3i2
… … … … .3
Substituting Eqs. (2) and (3) into (1) gives
i2 = 2 A
i3 = 1 A
i1 = 3 A
v1 = 24 V , v2 = 6 V , v3 = 6 V
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Example
For the circuit shown find: (a) v1 and v2, (b) the power dissipated in the 3-k and
20-k resistors, and (c) the power supplied by the current source.
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Example
Find v using KVL and
KCL
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Solution
+
-
A
-
B
+
+
-
KCL at A gives
IAB = 3 -1 = 2A
KCL at B gives
IB = 2 + 2A = 4 A
KVL around the blue path gives
(-18) - (1* 6) + (2 * 3) + (4 * 4) – V = 0
V=-2V
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