JRE SCHOOL OF Engineering
Class Test I – March 2015
Subject Name Electrical Engineering
Roll No. of Student
Date
13th Mar 2015
For EC,IT & EE branch only
Subject Code
Max Marks
Max Duration
Time
NEE -201
30 Marks
1 hour
9:20 am to 10: 20 am
SECTION – A
(5 x 3marks =15)
NOTE: ATTEMPT ALL QUESTIONS
1) What are the internal resistances of ideal dc voltage source & ideal current source?
Draw the characteristics of ideal and practical dc voltage and current sources.
Ans: For ideal voltage source, Rse=0
Ideal voltage source: It is defined as the energy source which gives constant
voltage across its terminals irrespective of the current drawn through its terminals.
The symbol for ideal voltage source is shown in the fig. This is connect to the load as
shown in Fig. At any time the value of voltage at load terminals remains same. This
is indicated by V-I characteristics shown in the Fig.
Practical Voltage Source:
But practically, every voltage source has small internal resistance shown in series
with voltage source and is represented by Rse as shown in the Fig.
For ideal current source, Rsh= ∞
Ideal current source: It is the source which gives constant current at its terminals
irrespective of the voltage appearing across its terminals. The symbol for ideal
current source is shown in the Fig. This is connected to the load as shown in the
Fig. At any time, the value of the current flowing through load I L is same i.e. is
irrespective of voltage appearing across its terminals. This is explained by V-I
characteristics shown in the Fig.
But practically, every current source has high internal resistance, shown in parallel
with current source and it is represented by Rsh. This is shown in the Fig.
Because of Rsh, current through its terminals decreases slightly with increase in
voltage at its terminals.
2) Explain active and passive elements, linear and non-linear elements, uni-lateral and
bi-lateral elements. Give examples.
Ans: Elements which can generate power on their own are called active elements.
Ex. All energy sources.
Elements which cannot generate power on their own but they can absorb power or store
energy are called passive elements. Ex. R, L & C.
Elements which follow ohm’s law, principle of superposition and principle of
homogeneity are called linear elements. Ex. R, L & C.
Elements which do not follow ohm’s law, principle of superposition and principle of
homogeneity are called non-linear elements. Ex. Diode, transistor, op-amp.
Elements in which current flows only in one direction are called uni-lateral elements.
Ex. Diode.
Elements in which current flows in both the directions are called bi-lateral elements.
Ex. R, L& C.
3) Define maximum power transfer theorem. Write the condition for maximum power
transfer.
Ans: Statement : In an active resistive network, maximum power transfer to the load
resistance takes place when the load resistance equals the equivalent resistance of
the network as viewed from the terminals of the load.
Proof of Maximum Power Transfer Theorem
Consider a d.c. source of voltage V volts and having internal resistance of r ohms
connected to a variable load resistance RL as shown in the Fig.
The load current is IL and is given by,
IL 
Vth
Rth  RL
The power consumed by the load resistance R L is,
 Vth
P  I L RL  
 Rth  RL
2
2

 RL

If RL is changed, IL is also going to change and at a particular value of R L, power
transferred to the load is maximum. Let us calculate value of R L for which power
transfer to load is maximum. To satisfy maximum power transfer we can write,
dP
0
dRL
d
dRL
2
 Vth 

 RL  0
 Rth  RL 

RL
d 
 0 …….as voltage is constant

2
dRL  Rth  RL  
Rth  RL 2 d RL   RL d Rth  RL 2  0
dRL
dRL
Vth
2
Rth  RL 2 1  RL 2Rth  RL   0
Rth  RL  2 RL   0
4) Using
RL  Rth
method write equations for
mesh
analysis
the circuit shown in fig.1.
fig.1
Let I1, I2 and I3 are three currents in mesh 1, 2 and 3.
15-2I1-30-1(I1 -I3)=0
30-6I2 -2I2 -10I2 –(I2 - I1 ) =0 [super mesh]
I2 –I3 =2
5) Define RMS and Average value for an alternating current.
Average Value: The average value of an alternating current is that value of steady
direct current which transfers the same charge as the alternating current flowing for
the same time.
Vav the average or the dc content of the voltage across the load for a sinusoidal
voltage is given by
RMS Value: The effective value of an alternating current is that value of steady ,direct
current which produces the same heat as that produced by the alternating current
when passed which produces the same heat as that produced by the alternating current
when passed through the same resistance for the same interval of time.
RMS voltage for a sinusoidal voltage at the load resistance can be calculated as
SECTION – B
(1 x 5marks = 5)
NOTE: ATTEMPT ANY ONE QUESTIONS
6) State superposition theorem? Mention its limitations and applications?
Superposition Theorem:- In any active, linear, bilateral containing two or more than
two energy sources the total response (voltage or current) will be the algebraic sum of
responses produced by each energy source individually. When one energy source is
considered then all remaining energy sources are replaced by their internal resistances.
Ideal voltage source is replaced by short circuit and ideal current source by an open
circuit.
Limitations:(a) Minimum two energy sources must be present in the network.
(b) Network should not contain any non linear and unilateral elements.
This theorem is applicable for linear bilateral networks containing two or more than two
sources.
7) State Thevenin’s theorem.
Thevenin’s theorem:- Any active, linear, bilateral network can be replaced by a
voltage source Vth (Thevenin’s voltage) in series with a resistance R th(Thevenin’s
resistance). Where value of voltage source is the open circuit voltage at the
terminals, where load resistance was connected and value of resistance is
calculated by looking back into the load terminals by neutralising all the sources
present in the circuit.
The voltage VTH is obtained across the terminals A-B with RL removed. Hence VTH is also
called open circuit Thevenin's voltage. The circuit to be used to calculate V TH is shown in
the Fig.for the network considered above. While R eq is the equivalent resistance obtained
as viewed through the terminals A-B with RL removed, voltage sources replaced by short
circuit and current sources by open circuit. This is shown in the Fig.
While obtaining VTH, any of the network simplification techniques can be used.
When the circuit is replaced by Thevenin’s equivalent across the load resistance, then
the load current can be obtain as.
IL 
VTH
RL  Rth
By using this theorem, current through, any branch of the circuit can be obtained,
treating that branch resistance as the load resistance and obtaining Thevenin's
equivalent across the two terminals of that branch.
7) Using delta to star transformation determine the resistance between terminals a-b in
fig.2.
fig.2
SECTION – C
NOTE: ATTEMPT ANY ONE QUESTIONS
(1x10marks= 10)
8. Replace the network of fig.3 to the left of terminal ‘ab’ by its Thevenin’s equivalent
circuit. Hence determine current ia in 10Ω resistor.
fig.3
OR
9) Find Average Value, RMS Value, Form factor (FF) & Peak factor for the voltage
waveform shown in fig.4.
fig.4
Note:
solutions of numerical are given hand written