ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
Now,
V
L
(t) =L
𝑑𝑖(𝑡)
𝑑𝑡
=1*10
-3
*50=0.05V
0=1*10
-3
*0=0V
2=1*10
-3
*(-50) =-0.05V
4The voltage waveform is shown in following figure.
2.
A 0.5uF capacitor has voltage waveform v(t) as shown in following figure, plot i(t) as function of t ?
Solution:
From the given waveform,
For 0Therefore
v(t)=20t
Therefore
i(t)=C
𝑑𝑣(𝑡)
𝑑𝑡
=0.5*10
-6
*20=1*10
-5
A=10uA
For 2Therefore
v(t)=40V
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
Therefore
i(t)=C
𝑑𝑣(𝑡)
𝑑𝑡
=0.5*10
-6
*0=0A
For 40−40
8−4
= −10
Therefore
v(t)=-10t+80
(According to straight line equation i.e. y=mx+c)
Therefore
i(t)=C
𝑑𝑣(𝑡)
𝑑𝑡
=0.5*10
-6
*(-10)=-5uA
The current waveform is shown in following figure
3. A Pure Inductance Of 3mh Carries A Current Of The Waveform Shown In Fig. Sketch The Waveform Of V (t) And
P(t).Determine The Average Value Of Power
Fig
Solution:
i (t)=5t for 0i (t)=10 for 2i (t)=-10t+50 for 4i (t)=-10 for 6
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
i (t)=5t-50 for 8For 0L
(t)=L
𝑑𝑖(𝑡)
𝑑𝑡
=3*10
-3
𝑑(5𝑡)
𝑑𝑡
=15*10
-3
v
For 2L
(t)=L
𝑑𝑖(𝑡)
𝑑𝑡
=3*10
-3
𝑑(10)
𝑑𝑡
=0v
For 4L
(t)=L
𝑑𝑖(𝑡)
𝑑𝑡
=3*10
-3
𝑑(−10𝑡+50)
𝑑𝑡
=-30*10
-3
v
For 6L
(t)=L
𝑑𝑖(𝑡)
𝑑𝑡
=3*10
-3
𝑑(−10)
𝑑𝑡
=0v
For 8L
(t)=L
𝑑𝑖(𝑡)
𝑑𝑡
=3*10
-3
𝑑(5𝑡−50)
𝑑𝑡
=15*10
-3
v
The sketch of v(t) is shown in fig.
Fig.
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
For 0P(t)=v(t)i(t)
=75*10-3t W
For 2P(t)=v(t)i(t)
=0W
For 4P(t)=v(t)i(t)
=-30*(-10t+50)*10
-3
=-0.3W (at t=4)
=0W (at t=5)
=0.3W (at t=6)
For 6P(t)=v(t)i(t)
=0W
For 8P(t)=v(t)i(t)
=15*(5t-50)*10
-3
=-0.15W (at t=8)
=-0.075W (at t=9)
=0W (at t=10)
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
4. Draw the waveforms for current, power for the following elements if a voltage input shown in figure is applied to
these elements.
i.
R=1 OHM
ii.
L=1H
iii.
C=1F
SOLUTION:
From the figure, v (t) is a straight line with slope =
20−0
1−0
=20, For 0Therefore v (t) =20t
i.
R=1 OHM
The voltage and current relation of a resistor is given by, v (t) =R i(t)
i(t) =20t/1=20t
Hence,
When t=0, i(t)=0A
When t=0.5, i(t)=10A
When t=1, i(t)=20A
Therefore the current waveform for the above values of t and i(t) is shown in figure below
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
Power, p(t)=v(t)i(t)
=20t*20t=400t
2
W
Hence,
When t=0, p(t)=v(t)i(t)=0W
When t=0.4, p(t)=v(t)i(t)=64W
When t=0.8, p(t)=v(t)i(t)=256W
When t=, p(t)=v(t)i(t)=400W
Therefore the power wave form for the above values of t and p(t) is shown in below figure.
ii.
L=1 H
The voltage and current relation of a inductor is given by,
i(t)=
1
𝐿
∫
𝑉(𝑡)𝑑𝑡
𝑡
−∞
i(t)=
1
1
[∫
𝑉(𝑡)𝑑𝑡 +
0
−∞
∫ 𝑉(𝑡)𝑑𝑡
𝑡
0
]
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
i(t)=0+
∫ 𝑉(𝑡)𝑑𝑡
𝑡
0
=
∫ 20𝑡𝑑𝑡
𝑡
0
=10t
2
Therefore the current waveform is shown in below figure.
Power, p(t)=v(t)i(t)=20t*10t
2
=200t
3
W
Therefore the power waveform is shown in below figure
iii.
C=1 F
The voltage and current relation of a inductor is given by
i(t)=C
𝑑𝑉(𝑡)
𝑑𝑡
i(t)=1*
𝑑(20𝑡)
𝑑𝑡
=20A
ELECTRICAL CIRCUITS
EEE
MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY
Therefore the current waveform is shown in below figure
Power, p(t)=v(t)i(t)=20t*20=400t W
Therefore the power waveform is shown in below figure.
Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )
Electrical Circuits EEE
UNIT-II
NETWORK ANALYSIS
Introduction
Network Reduction Techniques
Resistive Networks, Inductive Networks and Capacitive Networks
Series, Parallel, and Series Parallel Connections
Star to Delta and Delta to Star Transformations
Mesh Analysis and Super Mesh for DC excitation
Nodal Analysis and Super Node for DC excitation
Network Topology Definitions: Graph, Tree, and Basic Tie-set, Basic Cut-set
Matrices for planar Networks
Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )
Electrical Circuits EEE
Introduction:
A network is a collection of interconnected electrical components. In general, the electrical
networks are made to exchange the energy between different elements .These electrical networks
can be constructed either by using Resistors or Inductors or Capacitors or combination of these
elements. Network analysis is the process of finding the voltage response or the current response
for any element in the network by using the available techniques.
Network Reduction Techniques:
Series Connection of Resistors:
Two or more resistors in a circuit are said to be in series when the current flowing through
all the resistors is the same
.
Consider the circuit in fig(a) ,where two resistors R
1
and R
2
are in series, since the same
current i flows in both of them. Applying Ohm’s law to each of the resistors, we obtain
v
1
=
iR
1
,.......(1)
v
2
=
iR
2
........(2)
If we apply KVL to the loop fig(b) , we have
-v
+
v
1
+
v
2
= 0...........................(3)
v
=
v
1 +
v
2=
i
(
R
1 +
R
2)
i
=
v/ (R
1 +
R
2)
......................(4)
v
=
i R
eq................(5)
Implying that the two resistors can be replaced by an equivalent resistor; that is,
R
eq =
R
1 +
R
2
Note
: The equivalent resistance of any number of resistors connected in series is the sum of
the individual resistances.
If "n"resistors are in series, Req=R1+R2+......Rn
If "n"resistors of same value are in series, Req=nR
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