Circuit Elements

R, L and C
Resistor
A


 R= ρ l/A Ω l

 A resistor is a passive two-terminal electrical component that
implements electrical resistance as a circuit element. In electronic
circuits, resistors are used to reduce current flow, adjust signal levels,
to divide voltages, bias active elements, and terminate transmission
lines, among other uses
Inductor
 Inductance
 An inductor, also called a coil, choke, or reactor, is a passive two-
terminal electrical component that stores energy in a magnetic
field when electric current flows through it.[1] An inductor
typically consists of an insulated wire wound into a coil.
 e = dλ/dt
 e = Ldi/dt
 e = d(Nφ)/dt
 = Ndφ/dt
 = N dφ/di * di/dt
 = L di/dt
 L= Ndφ/di
 Φ = BA
 H = NI/l
 L =Nφ/I
 =N2 BA/Hl B=μH
 L = N2 μA/l Henry
Capacitor
 Charge separation in a parallel-plate capacitor causes an internal
electric field. A dielectric (orange) reduces the field and
increases the capacitance.
 C=d
Q/V

 E ( z )dz
 Charge
0 density σ = Q/A
 E = σ/ε
 V= = Ed

 V= σ/ε * d
 V=Qd/εA
 C= Q/V = εA/d farad
Sources
Independent Voltage source
 An ideal voltage source is known as an Independent
Voltage Source as its voltage does not depend on
either the value of the current flowing through the
source or its direction but is determined solely by the
value of the source alone.
 So for example, an automobile battery has a 12V
terminal voltage that remains constant as long as the
current through it does not become to high, delivering
power to the car in one direction and absorbing power
in the other direction as it charges.
Dependent Voltage source
contd
 An ideal dependent voltage-controlled voltage source, VCVS,
maintains an output voltage equal to some multiplying constant
(basically an amplification factor) times the controlling voltage
present elsewhere in the circuit.
 As the multiplying constant is, well, a constant, the controlling
voltage, VIN will determine the magnitude of the output voltage,
VOUT.
 In other words, the output voltage “depends” on the value of
input voltage making it a dependent voltage source and in many
ways, an ideal transformer can be thought of as a VCVS device
with the amplification factor being its turns ratio.
 Then the VCVS output voltage is determined by the following
equation: VOUT = μVIN. Note that the multiplying constant μ is
dimensionless as it is purely a scaling factor because
μ = VOUT/VIN, so its units will be volts/volts.
 An ideal dependent current-controlled voltage source,
CCVS, maintains an output voltage equal to some
multiplying constant (rho) times a controlling current
input generated elsewhere within the connected circuit.
Then the output voltage “depends” on the value of the
input current, again making it a dependent voltage source.
 As a controlling current, IIN determines the magnitude of
the output voltage, VOUT times the magnification constant
ρ (rho), this allows us to model a current-controlled voltage
source as a trans-resistance amplifier as the multiplying
constant, ρ gives us the following equation: VOUT = ρIIN.
 This multiplying constant ρ (rho) has the units of Ohm’s
because ρ = VOUT/IIN, and its units will therefore be
volts/amperes.
Independent Current Source
 An ideal current source is called a “constant current source” as it
provides a constant steady state current independent of the load
connected to it producing an I-V characteristic represented by a
straight line. As with voltage sources, the current source can be
either independent (ideal) or dependent (controlled) by a
voltage or current elsewhere in the circuit, which itself can be
constant or time-varying.
 Ideal independent current sources are typically used to solve
circuit theorems and for circuit analysis techniques for circuits
that containing real active elements. The simplest form of a
current source is a resistor in series with a voltage source creating
currents ranging from a few milli-amperes to many hundreds of
amperes. Remember that a zero-value current source is an open
circuit as R = 0.
Ideal current source and Practical
curent source
Dependent Current Source
 An ideal dependent voltage-controlled current source,
VCCS, maintains an output current, IOUT that is
proportional to the controlling input voltage, VIN. In
other words, the output current “depends” on the
value of input voltage making it a dependent current
source.
 Then the VCCS output current is defined by the
following equation: IOUT = αVIN. This multiplying
constant α (alpha) has the SI units of mhos, ℧ (an
inverted Ohms sign) because α = IOUT/VIN, and its
units will therefore be amperes/volt.
OHMS LAW
 Ohm’s principal discovery was that the amount of
electric current through a metal conductor in a circuit
is directly proportional to the voltage impressed across
it, for any given temperature.

 Vα I
 V = IR volts
Series Resistance

Current is same on all

resistances and voltages
are different.

Rtotal = R1+ R2 + R3
Parallel Resistance

The voltage is same

in all the branches.
Current is different
across each resistor
connected in parallel.
Req = 1/R1 + 1/R2 +
1/R3

Req = (R1 R2)/ R1+R2

Problems
 Calculate the equivalent resistance for the circuit
which is connected to 24 V battery and also find the
potential difference across 4 Ω and 6 Ω resistors
connected in series in the circuit.
Solution
 Solution
 Since the resistors are connected in series, the effective
resistance in the circuit
 = 4 Ω + 6 Ω = 10 Ω
 The Current I in the circuit= V/ Req = 24/10 = 2 .4 A
 Voltage across 4Ω resistor
 V1 = IR1 = 2 . 4 A× 4 Ω = 9.6V
 Voltage across 6 Ω resistor
 V2 = IR1 = 2 . 4 A× 6 Ω =14 .4V

Kirchhoffs First Law – The Current
Law, (KCL)

 Kirchhoffs Current Law or KCL, states that the “total

current or charge entering a junction or node is exactly
equal to the charge leaving the node as it has no other
place to go except to leave, as no charge is lost within
the node“. In other words the algebraic sum of ALL the
currents entering and leaving a node must be equal to
zero, I(exiting) + I(entering) = 0. This idea by Kirchhoff is
commonly known as the Conservation of Charge.
Kirchhoffs Laws
 KVL
 Kirchhoffs Voltage Law or KVL, states that “in any
closed loop network, the total voltage around the loop
is equal to the sum of all the voltage drops within the
same loop” which is also equal to zero. In other words
the algebraic sum of all voltages within the loop must
be equal to zero.
Circuit Components
 Circuit – a circuit is a closed loop conducting path in which an
electrical current flows.
 • Path – a single line of connecting elements or sources.
 • Node – a node is a junction, connection or terminal within a
circuit were two or more circuit elements are connected or joined
together giving a connection point between two or more
branches. A node is indicated by a dot.
 • Branch – a branch is a single or group of components such
as resistors or a source which are connected between two nodes.
 • Loop – a loop is a simple closed path in a circuit in which no
circuit element or node is encountered more than once.
 • Mesh – a mesh is a single closed loop series path that does not
contain any other paths. There are no loops inside a mesh.
No1
Find the current flowing in
the 40Ω Resistor, R3
Solution
 The circuit has 3 branches, 2 nodes (A and B) and 2
independent loops.
 Using Kirchhoffs Current Law, KCL the equations are
given as:
 At node A : I1 + I2 = I3
 At node B : I3 = I1 + I2
 Using Kirchhoffs Voltage Law, KVL the equations are
given as:
 Loop 1 is given as : 10 = R1 I1 + R3 I3 = 10I1 + 40I3
 Loop 2 is given as : 20 = R2 I2 + R3 I3 = 20I2 + 40I3
 Loop 3 is given as : 10 – 20 = 10I1 – 20I2
 As I3 is the sum of I1 + I2 we can rewrite the equations as;
 Eq. No 1 : 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2
 Eq. No 2 : 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2
 We now have two “Simultaneous Equations” that can be reduced to
give us the values of I1 and I2
 Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps
 Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps
 As : I3 = I1 + I2
 The current flowing in resistor R3 is given as : -
0.143 + 0.429 = 0.286 Amps
 and the voltage across the resistor R3 is given
as : 0.286 x 40 = 11.44 volts
 The negative sign for I1 means that the direction of current flow
initially chosen was wrong, but never the less still valid. In fact, the 20v
battery is charging the 10v battery.
Mesh Current Analysis Circuit

 Procedure of Mesh Analysis

 Follow these steps while solving any electrical network or circuit using
Mesh analysis.
 Step 1 − Identify the meshes and label the mesh currents in either
clockwise or anti-clockwise direction.
 Step 2 − Observe the amount of current that flows through each
element in terms of mesh currents.
 Step 3 − Write mesh equations to all meshes. Mesh equation is
obtained by applying KVL first and then Ohm’s law.
 Step 4 − Solve the mesh equations obtained in Step 3 in order to get
the mesh currents.
 Now, we can find the current flowing through any element and the
voltage across any element that is present in the given network by using
mesh currents.
 Example
Find the voltage across 30 Ω
resistor using Mesh analysis.
Steps
 Step 1 − There are two meshes in the above circuit.
The mesh currents I1 and I2 are considered in
clockwise direction. These mesh currents are shown in
the following figure.
 Step 2 − The mesh current I1 flows through 20 V
voltage source and 5 Ω resistor. Similarly, the mesh
current I2 flows through 30 Ω resistor and -80 V
voltage source. But, the difference of two mesh
currents, I1 and I2, flows through 10 Ω resistor, since it
is the common branch of two meshes.
 Step 3 − In this case, we will get two mesh equations since
there are two meshes in the given circuit. When we write the
mesh equations, assume the mesh current of that particular
mesh as greater than all other mesh currents of the circuit.
 The mesh equation of first mesh is
 20−5I1−10(I1−I2)=0
 ⇒20−15I1+10I2=0
 ⇒10I2=15I1−20
 Divide the above equation with 5.
 2I2=3I1−4
 Multiply the above equation with 2.
 4I2=6I1−8 Equation 1
 The mesh equation of second mesh is
 −10(I2−I1)−30I2+80=0
 Divide the above equation with 10.
 −(I2−I1)−3I2+8=0
 ⇒−4I2+I1+8=0
 4I2=I1+8 Equation 2
 Finding mesh currents I1 and I2 by solving Equation 1 and
Equation 2.
 The left-hand side terms of Equation 1 and Equation 2 are
the same. Hence, equate the right-hand side terms of
Equation 1 and Equation 2 in order find the value of I1.
 6I1−8=I1+8
 5I1=16
 ⇒I1=165A
 Substitute I1 value in Equation 2.
 4I2=16/5+8
 ⇒4I2=56/5
 ⇒I2=14/5A
 So, we got the mesh
currents I1 and I2 as 16/A and 14/5 A respectively.
 Step 5 − The current flowing through 30 Ω resistor is nothing
but the mesh current I2 and it is equal to 14/5 A. Now, we can
find the voltage across 30 Ω resistor by using Ohm’s law.
 V30Ω=I2R

 Substitute the values of I2 and R in the above equation.

 V30Ω=⟮14/5⟯30
 ⇒V30Ω=84V
 Therefore, the voltage across 30 Ω resistor of the given circuit
is 84 V.
 Loop (Mesh) Analysis

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 Illustrative Example 4
 Let us consider a simple network having only two meshes.
 Although the directions of the mesh currents are arbitrary;
but all must be either clockwise or anticlockwise.
 We shall always choose clockwise mesh currents.
 This results in a certain error-minimizing symmetry.
 Note that by taking mesh currents, the KCL is
automatically satisfied.

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Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 44
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 Resistance Matrix

Mesh current matrix

Source matrix

Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 45

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Applying Crammer’s rule :

The current in 3-ohm resistor is I1 – I2 = 6 – 4 = 2A

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Nodal Voltage Analysis
 Procedure of Nodal Analysis
 Follow these steps while solving any electrical network or circuit using
Nodal analysis.
 Step 1 − Identify the principal nodes and choose one of them
as reference node. We will treat that reference node as the Ground.
 Step 2 − Label the node voltages with respect to Ground from all the
principal nodes except the reference node.
 Step 3 − Write nodal equations at all the principal nodes except the
reference node. Nodal equation is obtained by applying KCL first and
then Ohm’s law.
 Step 4 − Solve the nodal equations obtained in Step 3 in order to get
the node voltages.
 Now, we can find the current flowing through any element and the
voltage across any element that is present in the given network by using
node voltages.
Find the current flowing through 20 Ω resistor of
the following circuit using Nodal analysis.
 Step 1 − There are three principle nodes in the above
circuit. Those are labelled as 1, 2, and 3 in the following
figure. In the above figure, consider node 3 as
reference node (Ground).
 Step 2 − The node voltages, V1 and V2, are labelled in
the following figure.
 In the above figure, V1 is the voltage from node 1 with
respect to ground and V2 is the voltage from node 2
with respect to ground.
 Step 3 − In this case, we will get two nodal
equations, since there are two principal nodes, 1 and
2, other than Ground. When we write the nodal
equations at a node, assume all the currents are
leaving from the node for which the direction of
current is not mentioned and that node’s voltage as
greater than other node voltages in the circuit.
 The nodal equation at node 1 is
 (V1−20)/5+V110+(V1−V2)/10=0
 ⇒2V1−40+V1+V1−V210=0
 ⇒4V1−40−V2=0⇒
 ⇒V2=4V1−40
 Equation 1
 The nodal equation at node 2 is
 −4+V2/20+(V2−V1)/10=0
 ⇒(−80+V2+2V2−2V2)/20=0
 ⇒3V2−2V1=80
 Equation 2
 Step 4 − Finding node voltages, V1 and V2 by solving Equation 1 and
Equation 2.
 Substitute Equation 1 in Equation 2.
 3(4V1−40)−2V1=80
 ⇒12V1−120−2V1=80
 ⇒10V1=200
 ⇒V1=20V

 Substitute V1 = 20 V in Equation1.
 V2=4(20)−40
 ⇒V2=40V
 So, we got the node voltages V1 and V2 as 20 V and 40 V respectively.
 Step 5 − The voltage across 20 Ω resistor is nothing but the node
voltage V2 and it is equal to 40 V. Now, we can find the current flowing
through 20 Ω resistor by using Ohm’s law.
 I20Ω=V2/R
 Substitute the values of V2 and R in the above equation.
 I20Ω=40/20
 ⇒I20Ω=2A
 Therefore, the current flowing through 20 Ω resistor of given circuit
is 2 A.

Example 14
 Using nodal analysis, determine the current through the 2-
Ω resistor in the network given.

Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 55

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• Solution : It is much simpler to write the
KCL equations, if the conductance (and not
the resistances) of the branches are given.

Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 56

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• It has 3 nodes. So, we have to write KCL
equations for only 2 nodes.
• We just equate the total current leaving the
node through several conductances to the
total source-current entering the node.
• At node 1,

• At node 2,

Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 57

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• Writing the above equations in matrix form,

0.7  0.2 V1  3 

 0.2 1.2  V   2
  2   

Solving for V1, using Calculator, we get

Click
V1  5 V
Finally, the current in the 2-Ω resistor,
V1 5
I    2.5 A
2 2
Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 58
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Nodal Analysis
 The above examples suggests that it is
possible to write the nodal analysis
equations just by inspection of the network.
 Such technique is possible if the network
has only independent current sources.
 All passive elements are shown as
conductances, in siemens (S).

Monday, May 23, 2022 Ch. 3 Network Analysis- Part II 59

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  In case a network contains a practical
voltage source, first convert it into an
equivalent practical current source.
 Write the Conductance Matrix, Node-
Voltage Matrix and the Node-Current
Source Matrix, in the same way as in the
Mesh Analysis.

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AC Circuits
 he circuit that is excited using alternating source is
called an AC Circuit. The alternating current (AC) is
used for domestic and industrial purposes. In an AC
circuit, the value of the magnitude and the direction of
current and voltages is not constant, it changes at a
regular interval of time.
 It travels as a sinusoidal wave completing one cycle as half
positive and half negative cycle and is a function of time (t) or
angle (θ=wt).
 In DC Circuit, the opposition to the flow of current is the only
resistance of the circuit whereas the opposition to the flow of
current in the AC circuit is because of resistance (R), Inductive
Reactance (XL=2πfL) and capacitive reactance (XC = 1/2 πfC) of
the circuit.
 In AC Circuit, the current and voltages are represented
by magnitude and direction. The alternating quantity may or
may not be in phase with each other depending upon the various
parameters of the circuit like resistance, inductance, and
capacitance. The sinusoidal alternating quantities are voltage
and current which varies according to the sine of angle θ.
 For the generation of electric power, all over the world
the sinusoidal voltage and current are selected because
of the following reasons are given below.
 The sinusoidal voltage and current produce low iron
and copper losses in the transformer and rotating
electrical machines, which in turns improves the
efficiency of the AC machines.
 They offer less interference to the nearby
communication system.
 They produce less disturbance in the electrical circuit
 Alternating Voltage and Current in an AC Circuit
 The voltage that changes its polarity and magnitude at
regular interval of time is called an alternating
voltage. Similarly, the direction of the current is
changed and the magnitude of current changes with
time it is called alternating current.
 When an alternating voltage source is connected
across a load resistance as shown in the figure below,
the current through it flows in one direction and then
in the opposite direction when the polarity is reversed.
The waveform of the
alternating voltage with
respect to the time and
the current flowing
through the resistance
(R) in the circuit is
shown below
 There are various types of AC circuit such as AC circuit
containing only
 resistance (R),
 AC circuit containing only capacitance (C),
 AC circuit containing only inductance (L),
 the combination of RL Circuit,
 AC circuit containing resistance and capacitance (RC),
 AC circuit containing inductance and capacitance (LC)
and resistance inductance and capacitance (RLC) AC
circuit.
 The various terms which are frequently used in an AC circuit are
as follows
 Amplitude
 The maximum positive or negative value attained by an
alternating quantity in one complete cycle is called Amplitude or
peak value or maximum value. The maximum value of voltage
and current is represented by Em or Vm and Im respectively.
 Alternation
 One-half cycle is termed as alternation. An alternation span is of
180 degrees electrical.
 Cycle
 When one set of positive and negative values completes by an
alternating quantity or it goes through 360 degrees electrical, it
is said to have one complete Cycle.
 Instantaneous Value
 The value of voltage or current at any instant of time is called an
instantaneous value. It is denoted by (i or e).
 Frequency
 The number of cycles made per second by an alternating quantity is
called frequency. It is measured in cycle per second (c/s) or hertz (Hz)
and is denoted by (f).
 Time Period
 The time taken in seconds by a voltage or a current to complete one
cycle is called Time Period. It is denoted by (T).
 Wave Form
 The shape obtained by plotting the instantaneous values of an
alternating quantity such as voltage and current along the y-axis and
the time (t) or angle (θ=wt) along the x-axis is called a waveform.
Peak Value
Definition: The maximum value
attained by an alternating quantity
during one cycle is called its Peak
value. It is also known as the
maximum value or amplitude or
crest value.
Average Value
Definition: The average of all the
instantaneous values of an
alternating voltage and currents
over one complete cycle is
called Average Value.
If we consider symmetrical waves
like sinusoidal current or voltage
waveform, the positive half cycle
will be exactly equal to the
negative half cycle. Therefore, the
average value over a complete cycle
will be zero.
The work is done by both, positive
and negative cycle and hence the
average value is determined
without considering the signs.
Definition: That steady current
which, when flows through a
resistor of known resistance for a
given period of time than as a
result the same quantity of heat is
produced by the alternating
current when flows through the
same resistor for the same period
of time is called R.M.S or effective
value of the alternating current.
In other words, the R.M.S value is
defined as the square root of
means of squares of instantaneous
values.
Let I be the alternating current
flowing through a resistor R for
time t seconds, which produces the
same amount of heat as produced
by the direct current (Ieff ). The base
of one alteration is divided into n
equal parts so that each interval is
of t/n seconds as shown in the
figure below.
Let i1, i2, i3,………..in be the mid
ordinates
Then the heat produced is
Since Ieff is considered as
the effective value of this
current, then the total heat
produced by this current
will be
=
Ieff2 Rt/J
 RMS value of sine wave is Vrms = Vm/√2
 Peak Factor
 Definition: Peak Factor is defined as the ratio of
maximum value to the R.M.S value of an alternating
quantity. The alternating quantities can be voltage or
current. The maximum value is the peak value or
the crest value or the amplitude of the voltage or
current.
 he root mean square value is the amount of heat produced by the
alternating current will be same when the direct supply of current is
passed through the same resistance in the same given time.
 Mathematically it is expressed as:
Em Im
 Peak Factor = or
 Erms Irms
Where,
Im and Em are the maximum value of the current and the voltage
respectively, and Ir.m.s and Er.m.s are the roots mean square value of the
alternating current and the voltage respectively.
 For the current varying sinusoidally, the peak factor is given as:

The value of Peak Factor is 1.4142

 Definition: The ratio of the root mean square value to
the average value of an alternating quantity (current or
voltage) is called Form Factor. The average of all the
instantaneous values of current and voltage over one
complete cycle is known as the average value of the
alternating quantities.
Resistors in an AC Circuit
Consider a circuit
consisting of an AC source
and a resistor.
The AC source is
symbolized by
ΔvR = DVmax= Vmax sin wt
ΔvR is the instantaneous
voltage across the resistor.

Section 33.2
Resistors in an AC Circuit, cont.
The instantaneous current in the resistor is
Dv R DVmax
iR   sin ωt  I max sin ωt
R R

The instantaneous voltage across the resistor is also

given as
ΔvR = Imax R sin ωt

Section 33.2
Resistors in an AC Circuit, final
The graph shows the current
through and the voltage across the
resistor.
The current and the voltage reach
their maximum values at the same
time.
The current and the voltage are
said to be in phase.
For a sinusoidal applied voltage,
the current in a resistor is always in
phase with the voltage across the
resistor.
The direction of the current has
no effect on the behavior of the
resistor.
Resistors behave essentially the
same way in both DC and AC
circuits.

Section 33.2
Phasor Diagram
To simplify the analysis of AC circuits,
a graphical constructor called a phasor
diagram can be used.
A phasor is a vector whose length is
proportional to the maximum value of
the variable it represents.
The vector rotates counterclockwise
at an angular speed equal to the
angular frequency associated with the
variable.
The projection of the phasor onto the
vertical axis represents the
instantaneous value of the quantity it
represents.

Section 33.2
A Phasor is Like a Graph
An alternating voltage can be presented in different
representations.
One graphical representation is using rectangular coordinates.
 The voltage is on the vertical axis.
 Time is on the horizontal axis.
The phase space in which the phasor is drawn is similar to polar
coordinate graph paper.
 The radial coordinate represents the amplitude of the voltage.
 The angular coordinate is the phase angle.
 The vertical axis coordinate of the tip of the phasor represents the
instantaneous value of the voltage.
 The horizontal coordinate does not represent anything.
Alternating currents can also be represented by phasors.
Section 33.2
rms Current and Voltage
The average current in one cycle is zero.
Resistors experience a temperature increase which depends
on the magnitude of the current, but not the direction of the
current.
The power is related to the square of the current.
The rms current is the average of importance in an AC
circuit.
 rms stands for root mean square
Imax
Irms   0.707 Imax
2
Alternating voltages can also be discussed in terms of rms
values. DVmax
DVrms   0.707 DVmax
2
Section 33.2
Power
The rate at which electrical energy is delivered to a
resistor in the circuit is given by
 P=i2R
 i is the instantaneous current.
 The heating effect produced by an AC current with a
maximum value of Imax is not the same as that of a DC current
of the same value.
Pav The2
Irms R
maximum current occurs for a small amount of time.
 The average power delivered to a resistor that carries an
alternating current is

Section 33.2
Notes About rms Values
rms values are used when discussing alternating
currents and voltages because
 AC ammeters and voltmeters are designed to read rms
values.
 Many of the equations that will be used have the same
form as their DC counterparts.

Section 33.2
Inductors in an AC Circuit
Kirchhoff’s loop rule can
be applied and gives:

Dv  Dv L  0 , or
di
Dv  L 0
dt
di
Dv  L  DVmax sin ωt
dt

Section 33.3
Current in an Inductor
The equation obtained from Kirchhoff's loop rule can be
solved for the current
DVmax DVmax
iL 
L  sin ωt dt  
ωL
cos ωt

DVmax  π DVmax
iL  sin  ωt   I max 
ωL  2 ωL

This shows that the instantaneous current iL in the

inductor and the instantaneous voltage ΔvL across the
inductor are out of phase by (p/2) rad = 90o.

Section 33.3
Phase Relationship of Inductors
in an AC Circuit
The current is a
maximum when the
voltage across the inductor
is zero.
 The current is
momentarily not
changing
For a sinusoidal applied
voltage, the current in an
inductor always lags
behind the voltage across
the inductor by 90° (π/2).
Section 33.3
Phasor Diagram for an Inductor
The phasors are at 90o
with respect to each other.
This represents the phase
difference between the
current and voltage.
Specifically, the current
lags behind the voltage by
90o.

Section 33.3
Inductive Reactance
The factor ωL has the same units as resistance and is
related to current and voltage in the same way as
resistance.
Because ωL depends on the frequency, it reacts
differently, in terms of offering resistance to current, for
different frequencies.
The factor is the inductive reactance and is given by:
 XL = ωL

Section 33.3
Inductive Reactance, cont.
Current can be expressed in terms of the inductive
reactance:
DVmax DVrms
Imax  or Irms 
XL XL

As the frequency increases, the inductive reactance

increases
 This is consistent with Faraday’s Law:
 The larger the rate of change of the current in the inductor,
the larger the back emf, giving an increase in the reactance
and a decrease in the current.

Section 33.3
Voltage Across the Inductor
The instantaneous voltage across the inductor is

di
Dv L  L
dt
 DVmax sin ωt
  Imax X L sin ωt

Section 33.3
Capacitors in an AC Circuit
The circuit contains a
capacitor and an AC
source.
Kirchhoff’s loop rule
gives:
Δv + Δvc = 0 and so
Δv = ΔvC = ΔVmax sin ωt
 Δvc is the instantaneous
voltage across the
capacitor.

Section 33.4
Capacitors in an AC Circuit, cont.
The charge is q = CΔVmax sin ωt
The
i 
instantaneous
dq current is given by
 ωC DV cos ωt
C max
dt
 π
or iC  ωC DVmax sin  ωt  
 2

The current is p/2 rad = 90o out of phase with the

voltage

Section 33.4
More About Capacitors in an AC
Circuit
The current reaches its
maximum value one
quarter of a cycle sooner
than the voltage reaches its
maximum value.
The current leads the
voltage by 90o.

Section 33.4
Phasor Diagram for Capacitor
The phasor diagram
shows that for a
sinusoidally applied
voltage, the current always
leads the voltage across a
capacitor by 90o.

Section 33.4
Capacitive Reactance
The maximum current in the circuit occurs at cos ωt = 1
which gives
DVmax
Imax  ωC DVmax 
(1 / ωC )

The impeding effect of a capacitor on the current in an

AC circuit is called the capacitive reactance and is
given by
1 DVmax
XC  which gives Imax 
ωC XC

Section 33.4
Voltage Across a Capacitor
The instantaneous voltage across the capacitor can be
written as ΔvC = ΔVmax sin ωt = Imax XC sin ωt.
As the frequency of the voltage source increases, the
capacitive reactance decreases and the maximum
current increases.
As the frequency approaches zero, XC approaches
infinity and the current approaches zero.
 This would act like a DC voltage and the capacitor would
act as an open circuit.

Section 33.4
The RLC Series Circuit
The resistor, inductor,
and capacitor can be
combined in a circuit.
The current and the
voltage in the circuit vary
sinusoidally with time.

Section 33.5
The RLC Series Circuit, cont.
The instantaneous voltage would be given by Δv =
ΔVmax sin ωt.
The instantaneous current would be given by i = Imax
sin (ωt - φ).
 φ is the phase angle between the current and the
applied voltage.
Since the elements are in series, the current at all points
in the circuit has the same amplitude and phase.

Section 33.5
i and v Phase Relationships –
Graphical View
The instantaneous voltage across the
resistor is in phase with the current.
The instantaneous voltage across the
inductor leads the current by 90°.
The instantaneous voltage across the
capacitor lags the current by 90°.

Section 33.5
i and v Phase Relationships –
Equations
The instantaneous voltage across each of the three
circuit elements can be expressed as

Dv R  Imax R sin ωt  DVR sin ωt

 π
Dv L  Imax X L sin  ωt    DVL cos ωt
 2
 π
DvC  Imax XC sin  ωt    DVC cos ωt
 2

Section 33.5
More About Voltage in RLC Circuits
ΔVR is the maximum voltage across the resistor and
ΔVR = ImaxR.
ΔVL is the maximum voltage across the inductor and
ΔVL = ImaxXL.
ΔVC is the maximum voltage across the capacitor and
ΔVC = ImaxXC.
The sum of these voltages must equal the voltage from
the AC source.
Because of the different phase relationships with the
current, they cannot be added directly.
Section 33.5
Phasor Diagrams

To account for the different phases of the voltage drops,

vector techniques are used.
Remember the phasors are rotating vectors
The phasors for the individual elements are shown.

Section 33.5
Resulting Phasor Diagram
The individual phasor diagrams can
be combined.
Here a single phasor Imax is used to
represent the current in each element.
 In series, the current is the same
in each element.

Section 33.5
Vector Addition of the Phasor
Diagram
Vector addition is used to combine
the voltage phasors.
ΔVL and ΔVC are in opposite
directions, so they can be combined.
Their resultant is perpendicular to
ΔVR.
The resultant of all the individual
voltages across the individual elements
is Δvmax.
 This resultant makes an angle of
φ with the current phasor Imax.

Section 33.5
Total Voltage in RLC Circuits
From the vector diagram, ΔVmax can be calculated

 
2
DVmax  DVR2  DVL  DVC

 ( Imax R )2   Imax X L  Imax X C 

2

DVmax  Imax R 2   X L  X C 
2

Section 33.5
Impedance
The current in an RLC circuit is
DVmax DVmax
Imax  
R 2   X L  XC  Z
2

Z is called the impedance of the circuit and it plays the

role of resistance in the circuit, where

Z  R 2   X L  XC 
2

 Impedance has units of ohms

Section 33.5
Phase Angle
The right triangle in the phasor diagram can be used to
find the phase angle, φ.

 X  XC 
φ  tan1  L 
 R 
The phase angle can be positive or negative and
determines the nature of the circuit.

Section 33.5
Determining the Nature of the
Circuit
If f is positive
 XL> XC (which occurs at high frequencies)
 The current lags the applied voltage.
 The circuit is more inductive than capacitive.
If f is negative
 XL< XC (which occurs at low frequencies)
 The current leads the applied voltage.
 The circuit is more capacitive than inductive.
If f is zero
 XL= XC
 The circuit is purely resistive.
Section 33.5
Vidhyadeep Institute Of Management & Technology, Anita, Kim
Electronics & Communication Department

Guided by Mr. Vicky Paperwala

1. Fuse

2. Circuit breaker

1. MCB

2. MCCB

3. ELCB
A type of low
resistance resistor to
provide overcurrent
protection.

Prevents short-circuiting,
overloading, mismatched
loads or device failure.
The size and construction of
the element is determined so
that the heat produced for a
normal current does not cause
the element to attain a high
temperature.
 FUSE

LOW HIGH
VOLTAGE VOLTAGE
FUSE FUSE

REWIREABLE CARTRIDGE
This kind of fuse is most
commonly used in the case
of domestic wiring and
small scale usage.

The main advantage of this type of

fuse is that it is easy to install and
also replace without risking any
electrical injury. But there are
certain shortcomings associated
with it too.
The level of sub division in this
case includes:
1.D type
2.Link Type
Link type further divided into
1.Knife blade type HRC fuse
2.Botled type HRC link fuse
High-voltage fuses are used to
protect instrument transformers
used for electricity metering, or for
small power transformers where
the expense of a circuit breaker is
not warranted.
A circuit breaker is a manually or
automatically operated electric switch
designed to protect an electrical circuit
from damage caused by overload or short
circuit.
A safety device used in electrical
installations with high earth
impedance to prevent shock.

A device used to directly detect

currents leaking to earth from an
installation and cut the power and
mainly used in TT earthing
systems.

There are two types of ELCBs:

1. Voltage Earth Leakage circuit breaker
(voltage-ELCB)
2. Current Earth Leakage Current Earth
Leakage Circuit Breaker (Current-ELCB).
A voltage-operated ELCB detects a rise in
potential between the protected
interconnected metalwork (equipment
frames, conduits, enclosures) and a distant
isolated earth reference electrode.

Disadvantages of the voltage-operated

ELCB are the requirement for a second
connection, and the possibility that any
additional connection to earth on the
protected system can disable the detector.
An electrical wiring device that disconnects
a circuit whenever it detects that the
electric current is not balanced between
the energized conductor and the return
neutral conductor.
ELCBs have one advantage over RCDs:
they are less sensitive to fault conditions,
and therefore have fewer nuisance trips.

1.As with RCDs, electrically leaky

appliances such as some water heater,
washing machine may cause the ELCB to
trip.
2.ELCBs introduce additional resistance
and an additional point of failure into the
earthing system
It automatically switches off the electrical circuit during abnormal
condition of the network means in over load condition as well as
faulty condition.

MCB is much more sensitive to over current than fuse.

 Miniature Circuit Breaker
Working Principle

There are two

arrangement of
operation of
miniature circuit
breaker.
1.One due to
thermal effect of
over current
2.electromagnetic
effect of over
current.
1. It automatically switches off the electrical
circuit during abnormal condition of the
network means in over load condition as
well as faulty condition.

2. Handling MCB is more electrically safe

than fuse.

3. Another advantage is, as the switch

operating knob comes at its off position
during tripping, the faulty zone of the
electrical circuit can easily be identified.
SR. PARTICULAR FUSE CIRCUIT
N BREAKER
O
1. FUNCTION PERFORMS DETECTION PERFORMS
& INTERRUPTION INTERRUPTON
FUNCTION
2. OPERATION INHERENTLY REQUIRES
COMPLETELY ELABORATE
AUTOMATIC EQUIPMENT
3. BREAKING CAPACITY SMALL VERY LARGE

4. OPERATING TIME VERY SAMLL VERY LARGE

5. REPLACEMENT REQUIRES DOES NOT

REQUIRE
DOMESTIC WIRING