BITS Pilani
Dubai Campus
INTRODUCTION
Chapter 1
To study the basic circuit elements
and the laws
Passive circuit elements, Voltage and Current sources,
resistors and Ohm’s law, KCL,KVL, Independent and
Dependent sources
Basic elements and Laws
1. Electric circuit or network – collection of electric components. E.g.,
voltage and current sources, resistors, inductors, capacitors etc.
2. Passive elements are those which take energy from the circuit, e.g.,
capacitor stores charge, inductor stores magnetic energy and resistor
dissipates energy. These are passive elements.
3. Active elements are those which supply energy to the circuit e.g.,
voltage and current sources, battery, motor, etc. These are active
elements.
4. Circuit analysis – determining the behavior of a given circuit.
Electric charge is measured in Coulombs (C)
Energy expended on electric charge (joules per coulomb), measured in
Volts (V). Also, a measure of electric potential.
Voltage Sources
- A device that produces a
Terminal 1
voltage or potential difference
of V volts across its terminals
+
-
V volts
- V can be + ve or – ve
Terminal 2
- Following are equivalent ideal voltage sources:
Terminal 1
5 volts
+
Terminal 2
Terminal 1 is at a potential
which is +5v lower than 2
Terminal 1
-5 volts
+
Terminal 2
Terminal 1 is -5 V higher
than that of terminal 2
Voltage Sources
The ideal voltage source is
V
constant and does not change
with time. Such a device is an
3
2
1
t
0
ideal battery.
Constant voltage
Battery symbol:
Terminal 1
Terminal 1
V volts
+
-
+
V volts
Terminal 2
Terminal 2
Generally, the voltage produced by an ideal voltage source will be
a function of time. Then it is represented as v(t).
E.g.,
v(t)=sin(ωt)
Terminal 1
Since voltage is a function of time,
ideal voltage source is represented
as V(t)
V (t) volts
+
Terminal 2
Example 1.1
Suppose the voltage v(t) produced by the ideal voltage
source is described by 𝑉(𝑡) = 5 − 10𝑒 −2𝑡 volts. Determine the
value of this voltage at the instants of time t = 0 s, t = 0.5 s,
t =1 s and t = 1.5s.
Solution
For t = 0 s, V 0 = 5 − 10𝑒 −2×0 = −5V
For t = 0.5 s, V(0.5) = 1.32 V
For t = 1 s, V(1) = 3.65 V
For t = 1.5 s, V(1.5) = 4.5 V
Current Sources
• When there is an electric potential difference (voltage) across
any material, negative charges flows from lower potential to
higher potential.
• Positive charges flows from a higher potential to a lower
potential.
• Flow of charges is denoted as q(t) as it is time dependent.
• Therefore, current
amperes.
and the unit of current is
Current Sources
• Direction of current is conventionally the direction in
which the positive charges flow.
• Arrow indicates the flow of current
• It can also be a function of time, i(t).
Terminal 1
i amps
or i(t)
Terminal 2
Ideal Current Source
A current source places a constraint on the current through it –
there is no constraint on the voltage across a current source.
The voltage across a current source depends on what is
connected to that source. Following are equivalent ideal
current sources.
Terminal 1
-3 amps
Terminal 1
3 amps
Terminal 2
Terminal 2
Constraints on the voltage and
current source:
Voltage Source
There is a constraint on the voltage across its terminal, but there
is no constraint on the current through it
Current depends on what is connected to the source.
Current source
There is a constraint on the current through it terminal, but there
is no constraint on the voltage across it.
Voltage across the current source depends on what is connected to that
source.
Resistors and Ohm’s Law
Consider the following circuit
i(t)
v(t)
+
-
i(t)
+ + + + +
e-
i(t)
•For a current to flow, voltage has to be present
•The resulting current i(t) is always directly proportional to the voltage v(t)
i.e. 𝑣(𝑡) ∝ 𝑖(𝑡)
Or 𝑣(𝑡) = 𝑅 × 𝑖(𝑡)
The proportionality constant R is called resistance and is measured in Ohms
(Ω).
In the above circuit, the material is the resistor.
Resistance
The resistance R is,
𝑣(𝑡) = 𝑅 × 𝑖(𝑡)
&
𝑣(𝑡)
𝑅=
𝑖(𝑡)
𝑖(𝑡) =
𝑣(𝑡)
𝑅
is Ohm’s Law
are also forms of Ohm’s Law
v
R
0
Symbol for R
i
1
A plot of voltage vs current
Resistance (Direction of
current)
If the current is reversed in the circuit below
R
i(t)
v(t)=R i(t)
+ v(t) i1(t)
R
≡
-i1(t)
+ v1(t) -
+ v1(t) -
v1(t)=R [-i1(t)] which is v1(t)=-R [i1(t)]
Alternatively,
i1(t)
R
R
- -v1(t) +
-v1(t)=R i1(t) or v1(t) = -R i1(t)
Drill Ex 1.2
a) For the circuit given below, what value for R will result in
v(t) = -2.5V?
+
25µA
R=?
V(t)
Solution
2.5
𝑅=
= 100,000 = 100𝑘Ω
25 × 10−6
b) For the circuit shown, when R = 4Ω, what voltage will
produce the current i(t) = 3e-2t – 7e-7t A?
i(t)
V(t)
+
-
R
Solution
𝑣(𝑡) = 𝑅×𝑖(𝑡)
= 4 x 3e-2t – 7e-7t
= 12e-2t – 28e-7t volts
i(t)
Current through an ideal
voltage source can be anything
v(t)
+
-
+
v(t)
R
-
i(t)
Voltage across an ideal
current source can be anything
+
i(t)
v(t)
-
R
Short circuit
Consider R = 0
By Ohm’s law
Such a circuit is short circuit.
+
Source
v(t)=0V
-
i(t)
R=0Ω
+
Source
i(t)
v(t)=0V
-
A zero–ohm resistor is equivalent to an ideal voltage source whose
value is zero volts provided the current through it is finite.
Open Circuit
Consider
. Then
Such a circuit is open circuit
+
i(t)=0A
+
v(t)
v(t)
-
-
i(t)= 0V
An infinite resistance is equivalent to an ideal current source
whose value is zero amperes
Kirchoff’s Current Law (KCL)
In an electrical circuit, the connection of two or more
elements is called a node.
KCL: At any node of a circuit, at every instant of time, the
sum of the currents into the node is equal to the sum of the
currents out of the node.
Or
At any node of a circuit, the currents algebraically sum to
zero.
Let us consider the following circuit
i2(t)
vg(t)
i3(t)
Node
i4(t)
is(t)
i1(t)
i5(t)
i1(t)+i4(t)+i5(t)=i2(t)+i3(t)+is(t)
Or i1(t)+i4(t)+i5(t)-i2(t)-i3(t)-is(t)=0
Example: Let us consider a circuit shown below
i3
i1
i2
+
13A
1Ω
2Ω v 2A
3Ω
Let us arbitrarily assign the current and its direction
Apply KCL at either of the two nodes , we get
13-i1+i2-2-i3=0 =➔ i1-i2+i3=11 -------------------(1)
Let the voltage across the 1Ω , 2Ω and 3 Ω resistor be v Volts
By Ohms law
Using (1) we can solve for v
Now i1 , i2 and i3 values can be calculated.
I1=6A , i2=-3A, i3=2A
Consider the following circuit
i6
i3
i1
ig
Region 1
is
i5
Region 2
+
-
i2
i4
Region 3
Let us identify three regions in this
i7
Region 1
Consider Region 1
i6
Applying KCL at n1,
we get i1=i2+i3+is
Applying KCL at n2,
we get i4+i5+is+i3=0
Solving the above 2 equations
we get i2=i1+i4+i5
i1
ig
+
-
i3
Region 1
is n2
n1
i2
i4
i5
i7
Region 2 & 3
Applying KCL at Region 2
i5+i6+i7=0
i6
i3
i1
Applying KCL at Region 3
i2+i7=i4+ig
ig
+
-
i5
is
i2
Region 3
i4
Region 2
i7
RESISTANCES IN PARALLEL
Consider
i
+
Arbitrary
Circuit
v
-
By Ohm’s Law
By KCL
i
n1
R1
i1
+
i2
R2
Arbitrary
Circuit
R=
v
-
n2
n2
𝑖1 =
𝑣
𝑅1
and
𝑖2 =
n1
𝑣
𝑅2
𝑣
𝑣
1
1
𝑖 = 𝑖1 + 𝑖2 =
+
=𝑣
+
𝑅1 𝑅2
𝑅1 𝑅2
Or
𝑖
1
1
1
= =
+
𝑣 𝑅 𝑅1 𝑅2
This is a parallel connection where current is divided
R1 R 2
R1 + R 2
CURRENT DIVISION
Current Division for resistors:
𝑖1 =
i
ARBITRARY
CIRCUIT
v
-
n1
i2
n2
G=
Conductance
𝑎𝑛𝑑
R1
i1
1
𝑚ℎ𝑜
R
𝑅2
𝑖
𝑅1 + 𝑅2
R2
𝑅1
𝑖2 =
𝑖
𝑅1 + 𝑅2
Kirchhoff’s Voltage Law (KVL)
Loop is a complete or closed electrical path with electrical
components and one or more sources with current flow in
it.
KVL: In any loop in any circuit, at every instant of time,
the sum of voltages having one polarity equals the sum of
the voltages having the opposite polarity.
Or
The sum of the voltage drops around a closed loop is
equal to the sum of the voltage sources of that loop
Example
Consider the following simple circuit
+V - vR1 - vR2 = 0
R1=10
+V = iR1 + iR2
+
_
V = i(R1 + R2)
i = V/(R1 + R2)
i = 5/(10+20)
=0.16 A
V= 5v
+
_
+
R2= 20
_
Consider the following loop
a
+
v6
-
6Ω
1Ω
b
+ v1 -
3Ω
v7
4Ω
f
- v8 +
5Ω
- v5 +
c
- v2 +
+
2A
2Ω
e
- v4 +
+
d
3V
Consider the following loop
- V1+ v2+ v 3- v 8+ v 6=0
OR V1- v2 - v 3+ v 8- v 6=0
a
+ V 2- 3 – v 4+ v 7 = 0
The elements between
nodes b and e is an
Open circuit.
+
v6
-
6Ω
1Ω
2Ω
b
+ v1 -
+
v7
2A
- e
f
- v8 +
5Ω
- v5 +
c
- v2 +
3Ω
4Ω
- v4 +
+
d
3V
Drill Ex 1.5
For the circuit shown below, to what value of V such that the
resulting current is i=2A ?, what are the corresponding value of
v1, v2, and v3?
i
2Ω
?V
+ v1 +
v2
-
10V
6Ω
+ v3 -
4Ω
Drill Ex 1.5
For the circuit shown below, to what value of V such that the
resulting current is i=2A ?, what are the corresponding value of
v1, v2, and v3?
i
𝑣1 + 𝑣 + 𝑣2 − 𝑣3 − 10 = 0
∴ 𝑣1 = 2𝑖 = 4𝑉
𝑣2 = 4𝑖 = 8𝑉
𝑣3 = −6𝑖 = −12𝑉
+
v2
-
10V
6Ω
+ v3 -
4 + 8 + 12 + 𝑣 = 10
Thus
𝑣 = −14𝑉
?V
+ v1 -
2𝑖 + 4𝑖 − (−6𝑖) + 𝑣 = 10
𝑖 = 2𝐴
2Ω
4Ω
Resistors connected in series
We can generalize it as for m resistors in series
R=R1+R2+R3+….Rm
i
+
ARBITRARY
CIRCUIT
n1
+ v1 -
v
-
R1
+
v2
-
R2
n2
i
n1
+
ARBITRARY
CIRCUIT
v
-
R = R1 + R 2
n2
VOLTAGE DIVISION
Voltage v is divided among R1 and R2, as v1 and v2
respectively.
i
ARBITRORY
CIRCUIT
+
v
-
n1
R1
+ v1 -
n2
+
v2
-
R2
𝑅1
𝑣
𝑅1 + 𝑅2
𝑅2
𝑣2 =
𝑣
𝑅1 + 𝑅2
𝑣1 =
Drill Ex 1.6
Given following circuit, find the current i and the resistance
loading the source i
9Ω
5Ω
28V
+
v1
-
4Ω
+
v2
-
3Ω
Drill Ex 1.6
Given following circuit, find the current i and the resistance
loading the source i
9Ω
5Ω
28V
i
28V
+
v1
-
i
5Ω
+
v1
-
+
v2
-
4Ω
4Ω
+
12Ω
v2
-
28V
i = 3.5 A and source resistance =8Ω
3Ω
5Ω
+
v1
-
3Ω
Sources can be combined as follows
Independent and Dependent
Sources
Independent Source: This is a source, in which the current
and voltages are independent of the behavior of circuit
to which the sources belong.
Dependent Source: This is an ideal source, whose value
depends upon some variable in the circuit to which the
source belongs.
vs
+
_
is
Independent source
Drill Ex 1.10
In the circuit find v1, i1 & i2
i1
2A
i2
+
3Ω v
-
2i1
2Ω
Drill Ex 1.10
In the circuit find v1, i1 & i2
By KCL, i1 + i2 +2 = 2 i1
– i1 + i2 = –2
i1
2A
3Ω v
-
By Ohm’s Law
i1 = v / 3 & i2 = v / 2
Thus
−𝑣 𝑣
+ = −2
3
2
Or
 v = – 12 V, Thus i1 = – 4 A
And i2 = – 12 / 2 = – 6 A
i2
+
−2𝑣 + 3𝑣 = −12
2i1
2Ω
Thank you