introduction of active and passive components · 2018-06-11 · experiment no:-1 date ... to...

Basic Electrical & Electronics l S.P.I.T

Introduction of Active and Passive Components


EXPERIMENT No:-1 DATE:- / / 2011

Introduction of Active and Passive Components

Aim :-

a. To introduce active and passive components.

b. Identify the value of resistors using colour codes and wattage according to

size of the resistors.

c. Introduction of different types of capacitors and their values.

d. Identify the pins of transistor and diodes.

Apparaturus & components required :-

i. Digital Multimeter

ii. Backlite sheet mounted with different active and passive components.

Theory: Write theory related with following questions.

1. Define active and passive components.

2. Draw a simple circuit containing 1K resistor and power supply and explain how

to calculate resistor wattage using Ohm’s law.

3. How to identify terminals of Diode and Transistor.

4. Write the ranges given on a given multimeter with appropriate symbol.

Procedure:

A. Find the value of given resistors using color code also verify it using given

multimeter.

B. Identify the terminals in case of Diode and Transistor.

C. Write down the number given on transistor and diode.

D. Find the value of given capacitor also indentify types of capacitor.

E. Measure 230Vrms voltage using given multimeter.

Result:

I. Values of resistors along with the wattages.

II. Values of capacitors with its types.

III. Pin configuration of transistor and diode.

Conclusion:


Introduction of ExpEyes


ExpEyes Junior:



Introduction of ExpEyes

Aim :- 1. Generate and Measure voltages using ExpEyes

2.


1. Battery connected circuit

2. Set of resistor, capacitor and inductor (125mH, 550Ω, 3000 turns, 44SWG)

3. ExpEyes Junior model with software installed on PC

Theory: ExpEYES Junior is interfaced and powered by the USB port of the computer.

For connecting external signals, it has several Input/Output terminals, arranged on both

sides. It can monitor and control the voltages at these terminals. In order to measure other

parameters (like temp erature, pressure etc.), we need to convert them in to electrical

signals by using appropriate sensor elements.

The external voltages connected to expEYES must be within the allowed limits. Inputs

A1 and A2 must be within 5 volts range and Inputs IN1 and IN2 must be in 0 to 5V

range. Exceeding these limits slightly will flash an error message. If the program stops

responding, exit and re-connect the USB to reset the device. Larger voltages wil l result

in permanent damage. To measure higher voltages, scale them down using resistive

potential divider networks.

Specifications:

1. External Input/Outputs terminals

a. Programmable Voltage Source (PVS) : Can be set, from software, to any value in

the 0 to +5V range.

b. Resolution : 12 bits, implies a minimum voltage step of around 1.25 millivolts.

c. Analog Inputs (A1 & A2) :Can measure voltage within the 5 volts range. The

resolution of ADC used is 12 bits. Voltage at these terminals can be displayed as a

function of time, giving the functionality of a low frequency oscilloscope. The

maximum sampling rate is 250,000 per second. Both have an input impedance of

10MΩ.

d. 0-5V Analog inputs (IN1 and IN2)

e. Resistive sensor input (SEN) : for use of sensors like LDR, photo transistor,

thermistor

f. Digital Inputs: (IN1 and IN2) Anything lower than 1V is logic 0 and anything

higher than 2.5V is treated as logic 1.


g. Digital output (OD1): Software controlled 0- 5V Square waves SQR1 and SQR2 :

Maximum swing 0-5V and frequency can be varies from 0.7Hz to 100kHz. Setting

frequency to 0Hz will make the output HIGH and setting it to -1 will make it LOW, in both cases the wave generation is disabled. When the wave generation is disabled, SQR1 and SQR2 can act as digital outputs on channel 8 and 9 respectively

h. Sine wave: Fixed frequency 150Hz, 4V

i. Constant current source: 1mA, software controlled, load resistor should be in the

range of 2K to 4K range.

j. Microphone: Condenser microphone

k. IN->OUT: Inverting amplifier with maximum gain 51

l. Ground terminal: Four ground terminals

Procedure :

a. Measure voltage:

b. Voltage, Current and Resistance:

c. Resistance in series:

d. DC, AC and Power line pick up

e. Measuring resistance of human body

f. Temperature dependent resistor:

g. Measuring conductivity of water:

h. Measuring a capacitance: (Up to 10000 pf)


i. Measuring dielectric constant of a designed capacitance with known value of

Area and Distance between plates.

j. Phase shift in ac circuit (RC or RL circuit)

k. RLC circuit:

l. Transient response of RC circuit (applying dc step input 0-5V)

m. Transient response of RL circuit (applying dc step input 0-5V, inductor )

n. Transient response of RLC circuit

o. Fourier analysis using FTR key

Electromagnetic induction:

Mutual induction:

AC generator:


Electronics

a. Half wave rectifier

b. Full wave rectifier

c. Diode characteristics

d. Common Emitter Transistor Characteristics

e. Frequency of sound

f. Frequency response of Piezo

g. Velocity of sound

h. Capturing a burst of sound

i. Speed of rotation of motor


Superposition Theorem


Circuit diagram:-

Superposition theorem:-

Observation table:-

V1 in volts V2 in volts Current through R3 (mA)

Practical

Current through R3 (mA)

Theoretical



Superposition theorem

Aim :-

To verify Superposition theorem.


Resistors:-

Power supply:-

Milliammeter:-

Voltmeter:-

Theory :-

Linear Network:- A network is said to be linear if current is linearly related to voltage

as per Ohm’s law.

Bilateral network:- In a bilateral network, the voltage and current relation is the same

for the current flowing in either direction.

Using the above definition, Superposition theorem can be stated as

In a network containing more than one source, the resulted current in any branch

is the algebraic sum of the currents that would be produced by each source, acting alone,

all other sources of e.m.f. being replaced by their respective internal resistances.

Fig.1.1 Consider 15V source only Fig.1.2 Consider 9V source only


Fig.1.3 Consider 15V & 9V source

Case 1:- According to superposition theorem, each source acts independently. Consider

source V1=15V acting independently. At this time, other sources must be replaced by

internal resistances. But as internal resistances of V2= 9V is not given, i.e., it is assumed

to be zero, V2 must be replaced by short circuit. Using any of the techniques, obtain the

current through resistance R3, i.e. current due to source V1 alone.

Case2:- Now, consider source V2 alone, with V1 replaced by a short circuit, to obtain the

current through R3. The corresponding circuit is shown in Fig.b. Obtain current through

R3 due to V2 alone by using any of the techniques such as mesh analysis, node analysis

and source transformation.

Case 3:- According to superposition theorem, the resultant current through resistance R3

is the sum of the currents through R3 produced by each source acting independently.

Procedure :- 1) Connect the circuit as shown in the figure.

2) Apply voltage V1=15V and remove voltage V2 and short the path.

3) Note down the current reading through R3 due to voltage source

V1=15V.

4) Now remove V1 and replace it by short path. Connect V2=9V and

measure the current through R3 due to V2=9V.

5) Again connect both the supply V1=15V and V2=9V and measure the

current through R3.

6) Find the theoretical reading of current through R3 using superposition

theorem and verify it with the practical reading.

7) Repeat the procedure by changing the voltage V1 and V2.


Calculation:-

Result :-_____________________________________________________________

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Conclusion:-_____________________________________________________________

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3 Φ STAR-DELTA NETWORK


Circuit diagram :-

Star Connection

Fig.2.3 Relation between line and phase voltage

Delta Connection

Fig.2.4 Relation between line current and phase current



3 phase Star Delta network

Aim:-

To study the relationship between phase & line currents and voltages in a

three phase circuit. (Star and delta)

Apparatus required :-

Ameters:-

Voltmeter(DVM):-

Lamp load:-

3 phase autotransformer:-

Theory:-

A 3phase supply contains three times i.e. 3 lines of supply where all three phase

voltages with a phase difference of 120o are available to supply a three phase load.

1) Potential difference between any two lines of supply is line voltage.

2) Line current is the current through any line.

3) Phase voltage is the voltage across any branch of the three phase load.

4) The current through any branch of three phase load is the phase current.

STAR CONNECTED LOAD:-

If the three impedances connected to the power supply, are such that end of each is

connected to a common point called as the neutral point and remaining three ends are

brought for connection to supply terminals R-Y-B, a star connection is formed. Thus in

star connection phase current is always equal to line current.

Fig. 2.1 Star connection

In a star connected load network; IL = I ph & VL = √3 Vph


DELTA CONNECTED LOAD:-

The delta connection is formed by connecting one end of windings to the starting of the

other, in order to form a closed loop, as shown in the figure. The end of the triangle so

formed is taken out for connection using phasor forms current and voltage.

Fig.2.2 Delta connection

In a Delta connection,

VL = Vph & IL = √3 Iph

Procedure:-

1) For a star network, short terminals 1, 3, 5.

2) The phase voltage is measured between points 1& 2.

3) The line voltage is measured between points 1 & 6.

Verify that VL = √3 Vph

4) For Delta, short the following terminals (a) 1 and 6, (b) 2 and 3, (c)4 and 5

5) Measure line voltage between points 1& 2.

6) Note down the currents in ammeter.

7) Verify that IL = √3 Iph .

8) Calculate Power (P)= √3 VL IL cosФ

(Where cosФ = 1, in resistive load)


Observation table:-

I] Star connection for different load

II] Delta connection for different load


Result:-

________________________________________________________________________

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________________________________________________________________________

_______________________________________________

Conclusion:-

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HALF WAVE & FULL WAVE RECTIFIER


CIRCUIT DIAGRAM:-

Fig. 3.1 Half /Full wave rectifier with and without C filter

Fig. 3.2 Measurement of Vr p-p



HALF WAVE & FULL WAVE RECTIFIER

Aim :-

To study half wave and full wave rectifier circuit with and without capacitor filter.

Apparaturus & Components :- C.R.O.

Center tap transformer:-

Diodes:-

Capacitors:-

Resistors:-

Theory :-

Half wave Rectifier:- When the switch(S1) is open the circuit acts as a half wave

rectifier (HWR). During the positive half cycle of input ac voltage ‘A’ becomes positive

w.r.t ‘B’which makes the diode (D1) forward biased & hence it conducts current. During

the negative half cycle the diode is reversed biased and no current flows in the circuit.

Therefore, current flows through diode only during the half cycle. Hence dc current is

obtained across load.

Full wave Rectifier:- When the switch(S1) is closed the circuit acts as a full

wave rectifier (FWR). During the positive half cycle of the secondary voltage diode

(D1) forward biased & diode (D2) become reverse biased. Therefore D1 conducts and D2

does not conduct. During the negative half cycle D2 is forward biased & D1 is reverse

biased. The load current flows in Both half cycles of ac voltage and in the same direction

through the load resistance. Hence we get the rectified output across the load.

Circuit operation with filter:- In capacitor filter, capacitor is connected in

parallel with load R. During positive half cycle, diode D1 is forward biased & D2 is

reverse biased. Therefore current flows in D1 charging the capacitor to maximum value

say V. Since capacitor and load resistance are in parallel, voltage across capacitor will

appear as output Voltage. Now the capacitor C starts discharging through load resistance

RL. At the same time another diode gets forward biased & starts conducting. The

capacitor C again starts charging and quickly gets charged Through.The forward biased


diode having small forward resistance. The time required by the capacitor to change to

peak value is quite small & diode current again reduces zero.

Theoretical calculation:-

Procedure :-

a. Connect the circuit as shown in the figure.

b. For HWR keep switch “S1” open while for FWR keep the switch

“S1” closed.

c. Observe the waveform for both HWR & FWR with and without filter and

plot the graphs.


Observation table:-

Primary Input Voltage (Vp) = 230 V Secondary Voltage (Vs) =

Without Filter :-

Vm

(volts) Vdc

(volts) Vrms

(volts) Ripple Factor

Frequency Hz

HWR

FWR

With Filter :-

Vr P-P (volts)

Vm (volts)

Vdc = Vm - Vr P-P /

2 (volts)

Vr RMS = (Vr P-P / 2√3)

r = Vr RMS /

Vdc

HWR

FWR


Result :-

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______________________________________________________________

Conclusion:-

______________________________________________________________

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______________________________________________________________


SERIES AND PARALLEL RLC RESONANCE


CIRCUIT DIAGRAM:-

mA

C

Signal

R L

Generator

Fig.4.1 RLC series circuit

mA

Generator

C

LR

Signal

Fig.4.2 RLC parallel circuit



RLC series and parallel circuit

Aim :- To study the Resonance phenomena of Series and parallel RLC circuit.

Apparatus :-

Variable resistors

Inductor

Capacitor

Signal Generator

Milliammeters.

Theory :- Series RLC circuit

In an RLC series circuit impedance is given by Z = √(R2 + (XL - XC))

The frequency at which the reactive part in the impedance becomes zero is called

resonant frequency. So at resonance

XL = XC or Z = R

2 л f0 L = 1/(2 л f0C)

f0 =1/(2 л√LC)

At resonance current is maximum where as impedance ‘Z’ is minimum. Bandwidth

of circuit is given by band of frequencies which lies between the two points at which the current

is 1/√2 times the current at resonant frequency.

Parallel RLC circuit:

In a parallel RLC circuit when the voltage and total current are in phase at a

particular frequency then parallel circuit is said to be at resonance. The frequency at which the

parallel resonance occurs is called resonant frequency. In the circuit , current drawn by inductive

branch is IL = V/ZL where ZL = R + jXL and current drawn by capacitive branch IC = V/XC

where XC = 1/(2 л f C) IL lags voltage ‘V’ by angle ΦL


VC/XC =( V/ZL).(XL/ZL) =( VXL)/ZL2

ZL2 = XL . XC

R2 + (2 л f0 L)

2 =( 2 л f0 L) . (1/2 л f0 C)

(2 л f0 L)2 = (L/C) – R

2

f0 = (1/2 л )√((1/LC)-(R2/L

2))

Now if R is very small as compared to L and C then (R2/L

2) << (1/LC) and f0 = 1/(2 л √LC)

At parallel resonance current is minimum while total impedance of circuit is maximum.

Procedure :-

1) Connect the Series RLC circuit as shown in figure 1.

2) Make sure that the milliammeter is in series with the circuit.

3) Keep the values fixed of R, L and C.

4) Keep varying frequency of the input signal from zero onwards

5) Measure frequency ‘f’ and current ‘I’ in mA.

6) Tabulate readings and draw graphs.

7) Repeat the above procedure for Parallel RLC circuit.


SERIES AND PARALLEL RLC RESONANCE

Observation Table:-

Sr. No. Series Resonance Parallel Resonance

F (KHz) I (mA) F (KHz) I (mA)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.


Result :-

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______________________________________________________________

Conclusion :-

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THREE PHASE POWER MEASUREMENT

USING TWO WATTMETER METHOD


Circuit Diagram :-

Fig. 5.1 Power measurement using 2 wattmeter method

Observation table:-

Sr.

No.

VR

(V)

VY

(V)

VB

(V)

IR

(A)

IY

(A)

IB

(A)

ΣVI

(W)

W1

(W)

W2

(W)

W1+ W2

(W)

Error

%


EXPERIMENT NO.6 DATE: / / 2011

3 phase power measurement using 2 wattmeter method

Aim

:- To measure power in a 3 phase circuit using two wattmeter method.

Apparatus

:- Two wattmeter:- Ammeters:- Lamp loads Voltmeter(DMM)

Theory :-

If the 3 phase load is balanced or even ubalanced, the power in the load can

be found by two wattmeter method. Let VR, VY & VB be rms value of phase voltages and

IR, IY & IB be the rms value of phase currents. Assume the load to be inductive & hence

phase currents will lag respective phase voltages by some phase angle as in fig. below.

Current through wattmeter W1 = IR. Potential Difference (P.D) across voltage coil of W1

= VRB = VR -VB. Angle between the current through the current coil & the voltage across

the pressure (voltage) coil is (30- ф).

Reading of W1 = IR VRB cos (30 - ф). Similarly, if current through wattmeter W2 = IY

Potential Difference (P.D) across voltage coil of W2 = VYB = VY -VB

From the phasor diagram, phase difference between VYB & IY = 30 + ф

Reading of W2 = IY VYB cos (30 + ф). Since load is balanced therefore VYB = VRB = VYR =

VL (Line voltage) & IR = IY = IB = IL (Line current). W1= VL IL cos (30 - ф),


W2= VL IL cos (30 + ф)

W1+W2 = VL IL cos (30 - ф) + VL IL cos (30 + ф)

= VL IL [cos 30 cos ф + sin 30 sin ф + cos 30 cos ф - sin 30 sin ф]

= VL IL [2 cos 30 cos ф]

=3 VL IL cos ф

W1+W2 = Total power in 3 phase load

Note: For resistive load ф = 0, therefore cos ф = 1

Procedure : 1) Connect the circuit as shown in the circuit diagram.

2) Adjust the 3-ф transformer voltage to 200volts.

3) Vary the lamp load in 3 phases.

4) Measure VR, VY, VB, IR, IY & IB .

5) Take 3 to 4 such readings

Result:- ------------------------------------------------------------------------------------------

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Conclusion:- -------------------------------------------------------------------------------------------

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O.C. & S.C. TEST OF A SINGLE Φ TRANSFORMER


Circuit Diagram for Open circuit test :-

Circuit Diagram for Short circuit test :-



O.C & S.C. test on 1- Φ transformer

Aim :- To carry out open circuit & short circuit test in a single phase transformer and

find percentage efficiency, percentage regulation and equivalent circuit.


Single Φ transformer:-

Variable AC source:-

AC wattmeter:-

Voltmeters:-

Theory :-

A transformer is a device used to change voltages and currents of AC electric

power. In the simplest version it consists of two windings wrapped around a magnetic

core; windings are not electrically connected, but they are coupled by the magnetic field,

as it shown in Figure. When one winding is connected to the AC electric power, the

electric current is generated. This winding is called the primary winding. The primary

current produces the magnetic field and the magnetic flux links the second winding,

called the secondary winding. The AC flux through the secondary winding produces an

AC voltage, so that if some impedance is connected to the terminals, an AC electric

current is supplied.


TRANSFORMER

OPEN CIRCUIT TEST

It determines the iron loss & No load current (IO). One winding is left open and

another

winding of transformers connected to voltage supply. Flux is set up in the core depends

upon the applied voltage & iron loss occurs. IO is very small and Cu loss is negligible in

primary and absent in secondary. The wattmeter readings give us the iron losses. In order

to simplify calculation voltage, current and impedance are either primary or secondary IO

is no load current. RO and XO are resistance and reactance of circuit.

SHORT CIRCUIT TEST

This test determines Cu loss & short circuit current (ISC) which offers the least

impedance (ZSC) & RSC. In this one winding is short circuited and another winding

connected to very low voltage supply. The wattmeter readings give us the copper losses.

The applied voltage is very less so flux set up is very less since iron loss will be

negligible and the current flowing through short circuited winding and hence the rated or

full load current through the wattmeter which represents copper losses.

Procedure :-

A: Open circuit test

1) Make connections as per the circuit diagram of Figure.

2) Adjust the autotransformer to the values specified in the Table and record Io,

V1 and Wi for each voltage step.

3) Set the autotransformer to zero position. Turn the power off.

4) Remove the connections for the circuit.

B: Short circuit test

1) Make connections as per the circuit diagram of Figure. A short wire can be

used for the short circuit connection of the S1 and S2 terminals.

2) Make sure that the 1-Ф variable autotransformer (VARIAC) is kept in ZERO

Position initially.


3) Turn the power on. Slowly vary the 1-Ф autotransformer (VARIAC) to obtain

the current close enough to the calculated values. Large current variations

should be avoided.

4) Record V1, Isc and Psc for each “% of rated I1current” step calculated.

5) Set the autotransformer to zero position. Turn the power off.

6) Remove the connections for the circuit.

Calculations :-

For Open-Circuit Test

Wi =V1IOcosØO, since, Wi , V1 & IO can be read from the meters,

The no load power factor cosØO = Wi / V1IO can be determine,

Iµ = IOsinØO and IW = IOcosØO

XOC=V1/ Iµ & ROC = V1 / IW

For Short-Circuit Test

ZSC = VSC / ISC

WSC = I2

SC. RSC

. . . RSC = WSC / I

2SC

X SC = √ (ZSC )2 (R SC )

2


Equivalent circuit at no load(open circuit) Equivalent circuit at short

circuit

Observation Table :-

A: Open circuit test

B: Short circuit test


Results :-

ROC = ______________________

XOC = _______________________

R SC = _______________________

X SC = _______________________

ZSC = _______________________

% Regulation:-_________________

%ή for load :-__________________

%ή max:- ____________________

Conclusion:-_____________________________________________________________

______________________________________________________________

______________________________________________________________

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LOAD TEST ON A SINGLE PHASE TRANSFORMER

BY DIRECT LOADING


Circuit Diagram:-

Observations :

V1 = Vs= 220 Volts Constant



LOAD TEST ON A SINGLE PHASE TRANSFORMER

BY DIRECT LOADING

Aim: - To determine the transformer efficiency curve by direct loading of the

transformer.

Apparaturus & components required:-

AC Supply:-

Transformer:-

Ammeters:-

Autotransformer:-

Wattmeter:-

DVM:-

lamp load:-

Theory:-

Transformer is a device used predominantly for two purposes:-.

a) Change the voltage level in AC circuit.

b) Obtain electrical isolation between two circuits.

Construction:

Physically, a basic transformer is made up of a ferromagnetic core and

copper windings on it. Core is made up of stacked Cold Rolled Grain Oriented (CRGO)

steel plates. This material increases the permeability of the flux path, thereby reducing

the leakage flux as well as the hysteresis losses. The stacking of plates helps reduce

the eddy current losses. Each of the plate is called a stamping. Each one is coated with

varnish on either side to insulate it from the adjacent stampings. Also, the stamping

width is varied across the cross section to make it resemble as closely as possible

to a circle. For a given area, circle has the smallest possible perimeter (circumference).

Economically, this makes a big impact in larger transformers where a considerable chunk

of money is spent in copper windings.


Efficiency :

Efficiency of a device is calculated as the ratio of the output power to the input

power. When calculated in percentage, it is known as percentage efficiency. This ratio is

obviously less than 1, but for transformers, it is of the order of 0.95 to 0.99 (at the rated

load). In other words, the transformer is a highly efficient device. A way of determining

the output is to calculate the input and subtract losses from it. The transformer being a

static device, has no moving parts in it. This leaves only two avenues for losses to take

place, the magnetic circuit and the electric circuit. In the magnetic part( core ), the losses

that take place are eddy current losses and hysteresis losses. These losses are independent

of the load and remain constant. The eddy current losses are reduced by using stamped

cores and the hysteresis losses are reduced by using better materials. These losses

together are known as core loss. In addition to this, whenever a current flows in one or

both the windings, there is an associated I2R loss. This loss is as evident completely

dependent on the load known as copper loss.

Regulation :

The output voltage changes with the load current. This variation is termed as

voltage regulation. This takes place because of the drop across the transformer winding

resistances and leakage reactance. The drop across the resistance increases with the load

current. On the other hand, increases in load increases the leakage reactance. The leakage

reactance is nothing but a circuit representation of such a flux. Percentage regulation is

calculated as the ratio of change in voltage from no load to the value to the no load value.

There is a circular former placed on the stacked core, in order to insulate the core

from the windings. The copper windings are wrapped over the former. Most of the

transformers have two windings, though there can be either one or more than two. The

two windings are called primary and secondary. The primary winding is connected to the

AC supply and load is connected to the secondary winding. Although the two windings

are shown on two separate limbs, practically, the windings are not segregated as such.

Both of them are wound on each of the limb, in order to reduce leakage flux. The copper

windings have paper insulator all along their length.


Operation :

The transformer action is based on two mutually coupled coils, with a changing

magnetic flux linking them. Simplistically speaking, primary winding is energized from

an AC voltage source. This voltage causes a current to flow in it. This current produces a

flux in the ferromagnetic core on which it is wound. This flux is also alternating in nature

because of the current. This alternating flux also links with the secondary winding

because they share a common magnetic path. This changing flux induces an emf in the

secondary coil. The magnitude is governed by Farady’s Law and direction by Lenz’s

Law. The voltage across the coil depends on the number of turns in the secondary

winding.

Procedure:

1) Connect the circuit as shown in the diagram.

2) Increase the input voltage to the transformer rated primary voltage.

3) Vary the lamp load in step from no load to the full load value.

4) Take the readings of the input wattmeter (W), load current (IL) and load

voltage (V2). (Minimum 10 readings).

5) Calculate efficiency and regulation.

6) Plot percentage efficiency and regulation v/s load current.

Calculations:

100%

100Re%

1

22

)0(2

2)0(2

xW

IVEfficiency

xV

VVgulation


Results :-

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________________________________________________________________________

________________________________________________________________________

Conclusion :

________________________________________________________________________

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MAXIMUM POWER TRANSFER THEOREM


Circuit Diagram:-

Observation Table:-

V1 = ___________ (volts) ; V2 = _______________ (volts)

For maximum power transfer,

RL (observed) ( Ω) = _______________ (from the graph)


EXPERIMENT NO:-09 DATE:- / / 2011

MAXIMUM POWER TRANSFER THEOREM

Aim: To experimentally verify the maximum power transfer theorem.

Apparatus:

DC Voltage source:-

Ammeter:-

DVM:-

Resistor:-

Breadboard.

Theory:

The power in load resistance is given by I2R, where I is the current trough the

resistance R. Also any complex circuit can be reduced to its equivalent Thevenin’s

circuit. The maximum power transfer theorem states that the power transferred to the load

is maximum when the value of the load resistance is equal to the internal resistance of the

source.In other words, the power dissipated in any resistor in a given resistive network

will be maximum when the rest of the network is replaced by its Thevenin’s equivalent

and the value of the load resistance is equal to the Thevenin’s resistance.

The general Expression for the power in the load resistor is

L

LTH

THLLL R

RR

VRIP *

)(

)(2

22

To find the value of RL for which PL is maximum

TH

THTH

THTH

THMAX

THL

LTH

LTHLTH

L

L

LTHLTH

LTH

L

L

R

VR

RR

VP

RR

RR

RRRV

dR

dP

RRRR

RV

dR

dP

4

)(*

)(

)(

0)(

2

0)(

1

)(

2

2

2

2

3

2

23

2


Hence the maximum power is transferred to the load when the load

resistance value is equal to the internal resistance of the Thevenized

circuit.

Procedure:-

1) Connect the voltage sources and the milliammeter as shown in the circuit.

2) Keep the voltages such that the maximum current does not go beyond the

range of the milliammeter ( i.e. resistance minimum).

3) Vary the load resistance from minimum to maximum.

4) Note the corresponding load current (IL) and voltage across the load VL.

5) Calculate the value of load resistance by taking the ratio of VL to IL.

6) Calculate the power dissipated in the load resistance as VL IL.

7) Plot the graph of P L v/s R L.

Calculations:-


Result :

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Conclusion :

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Verification of Kirchhoff’s voltage and current law using given

network


Circuit Diagram:-

Observation Table:-

V1(Volts) V2(Volts) V3(Volts) V4(Volts) V5(Volts)

I1(mA) I2(mA) I3(mA) I4(mA) I5(mA)



Verification of Kirchhoff’s voltage and current law using given network

Aim:- To verify the Kirchhoff’s current and voltage law.

Apparatus:- Powersupply:-

Ammeter:-

DMM

Bread board

Theory:

The following are the important terms should know before defining Kirchoff’s law.

1] Active element:- A active element is one that supplies electrical energy to the circuit.

2] Passive element:- A passive element is one that receives electrical energy, then

converts it into heat (resistance) or stores in electrical field

(capacitance) or magnetic field (inductor). For example resistance

3] Node:- A node of network is an equi potential surface at which two or more circuit

elements are joined.

4] Junction :- A junction is that point in a network where three on more circuit elements

are joined. Remember, all junctions are the nodes but all the nodes are not

junctions.

5] Branch:- A branch is the part of a network lying between two junction points.

6] Loop :- A loop is any closed path of a network.

7] Mesh:- A mesh is the most elementary form of a loop and cannot be further divided

into other loops.

Using the above definition we can define Kirchoff’s law. Kirchoff’s law is divided into

two parts one is related with current and other related with voltage and E.M.F.

A] Kirchoff’s current law:- It states that the algebraic sum of currents meeting at a

junction or node in an electric circuit is zero.

B] Kirchoff’s Voltage law:- It states that in any closed circuit or mesh, the algebraic sum

of the electromotive forces and the voltage drops is equal to zero.


Procedure:-

A] Kirchoff’s current law:-

1] Connect the circuit as shown in the figure for Kirchoff’s current law.

2] Keep the voltage source V1 and V2 at 5V and 10V respectively.

3] Measure the current through each branch.

4] Verify theresult using Kirchoff’s current equation for the respective node.

5] Take four more readings by changing the supply (V1 and V2) at some

reasonable voltage.

B] Kirchoff’s voltage law:-

1] Connect the circuit as shown in the figure for Kirchoff’s current law.

2] Keep the voltage source V1 and V2 at 5V and 10V respectively.

3] Considering the different loops and assuming the direction of current find the

theoretical current passing through each branch. If the theoretical reading is

negative change the assumed direction of current.

4] Measure and verify the reading of voltage and current using kirchoff’s law.

Calculations:-

Result :

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________________________

Conclusion :

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COMMON EMITTER CHARACTERISTCS OF A

TRANSISTOR


Circuit Diagram:-



COMMON EMITTER CHARACTERISTCS OF

TRANSISTOR

Aim :- To plot CE characteristics of Bipolar Junction Transistor (BJT BC547)

Apparatus:- Powersupply:-

Ammeter:-

DVM

BJT (547)

Resistances

connecting wires

Theory :- In Common Emitter (C.E) configuration, input is applied between the

base – emitter and output is taken across collector – emitter. Here emitter

of the transistor is common to both input and output, hence the name

common emitter configuration. The most important characteristics of

transistor in any configuration are input and output characteristics.

A. Input Characteristics :-

It is the curve between input current IB and input voltage VBE at constant

collector emitter voltage VCE. The input characteristic resembles a forward

biased diode curve. After cut in voltage the IB increases rapidly with small

increase in VBE. It means that dynamic input resistance is small in CE

configuration. It is the ratio of change in VBE to the resulting change in

base current at constant collector emitter voltage. It is given by ΔVBE / ΔIB

B. Output Characteristics:-

This characteristic shows relation between collector current IC and

collector voltage for various values of base current. The change in

collector emitter voltage causes small change in the collector current for

the constant base current, which defines the dynamic resistance and is

given as ΔVCE / ΔIC at constant IB. The output characteristic of common


emitter configuration consists of three regions: Active, Saturation and

Cut-off.

Active region: In this region base-emitter junction is forward biased and base-

collector junction is reversed biased. The curves are approximately

horizontal in this region.

Saturation region: In this region both the junctions are forward biased.

Cut-off: In this region, both the junctions are reverse biased. When the base

current is made equal to zero, the collector current is reverse

leakage current ICEO.

The region below IB = 0 is the called the cutoff region.

Procedure :

A. Input Characteristics

1. Make the circuit connection as shown in the circuit diagram.

2. Set the voltage VCE = 5 V and vary IB with the help of VBB and

measure VBE.

3. Set the voltage VCE = 10 V and vary IB with the help of VBB and

measure VBE.

4. Plot graph of IB v/s VBE.

B. Output Characteristics

1. Keep IB constant say 20 μA, vary VCE and note down the

collector current IC.

2. Now keep IB = 40 μA, vary VCE and note down the collector

Current IC.

3. Plot graph of IB v/s VCE.


Observation Table:

A. INPUT CHARACTERISTICS

Sr.

No.

For VCE = 5 V For VCE = 10 V

IB (μA) VBE (V) IB (μA) VBE (V)

B. OUTPUT CHARACTERISTICS

Sr.

No.

For IB = 20μa For IB = 30μA

VCE (V) IC (mA) VCE (V) IC (mA)


Result :

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Conclusion :

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V-I characteristics of Diode


Circuit Diagram:-

A] Forward Biased:-

B] Reversed Biased:-



V-I Characteristics of Diode

Aim :- To plot V-I characteristics of Diode.

Apparatus:-

Power supply

Ammeters

Voltmeter

Resistances

Theory:-

Semiconductors, like Silicon or Germanium, are elements having

resistivity that in intermediate between a conductor and an insulator. They inherently

have four electrons in the valence band which helps them to form covalent bonds with

four neighboring silicon atoms. Hence, at absolute zero, the material behaves like an

insulator. At room temperature, few of these electrons absorb enough energy to break

away from the nucleus and serve as conduction electrons. The conduction properties

can also be easily changed by changing the doping (adding different elements to) the

semiconductor. Addition of a pentavalent impurity such as Phosphorus, N – type dopant

, gives an additional electron after the four silicon bonds are satisfied. Similarly, a

trivalent impurity such as Boron, P-type dopant, creates an absence of electron, a hole.

The entire semiconductor material is a single crystal, with one region dopes to be P-type

, with excess holes , and the adjacent region to be N- type , with excess electrons . This

creates a metallurgical junction between the p and n regions. The contact to the p region

is called the anode and that of the n region is called cathode.

Equilibrium P – N junction :-

A large density gradient in both hole and electron concentrations occur at this junction.

Initially, then, there is a diffusion of holes from the p region to the n region and diffusion

of electrons from n region to the p region. The flow of holes from p region uncovers

negatively charged acceptor ions, and the flow of electrons uncovers positively charged

donor ions. This action creates a charge separation which sets up an electric field

oriented in the direction from the positive to the negative charge. This sets up an electric


field in such a direction as to oppose the movement of electrons and holes eventually.

The region surrounding the junction which contains immobile charges is called the

“space charge” or “depletion” region. The electric field creates a potential difference

across the region, which is called the built-in potential barrier. This is about 0.7 V for a Si

diode at room temperature.

Reverse Biased P-N junction :

A voltage source with its positive terminal connected to the n region and negative

terminal connected to the p region reverse biases the P-N junction. This increased

electric filed holds back the holes in the p region and electrons in the n region and

hence, there is no current flow. The electric field and the width of the space-charge

region increases. There is also a decrease in junction capacitance associated due to

increase in the width. Thus, the reverse bias region is characterized by neglible current

(due to minority carriers) even on the application of a very high voltage across the

terminals, the limit being decided by reverse breakdown voltage of the diode.

Forward Biased P-N junction :

Application of a positive voltage to the p region and negative voltage to the n region

creates an additional electric field in the space charge region. But this time the field

opposes the space – charge E-field. This disturbs the balance between diffusion and E-

field force. Hence majority carriers from the p region diffuse over to the n side and

electrons from n side move over to the p side of the junction. This process continues as

long as the voltage is applied. Thus, in the forward bias mode, the diode carries a large

current.

Procedure:-

1) Connect the diode in the forward bias mode.

2) Connect a current limiting resistor in series with the diode.

3) Make the connection for milliammeter. (Check the polarity before making the

connections.)

4) Slowly increase the voltage applied, and measure the current (I) through the

diode and the voltage across the diode (VD). Take more than 10 readings.

5) Plot the graph of VD v/s I.

6) Calculate the static and dynamic resistance of the diode.


Result :

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Conclusion :

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Transistor as a Switch


Circuit Diagram:



To verify the application of a transistor as a switch

Aim :- To switch on a 6V bulb using a transistor.

Components and instruments required:

1. Transistor BC547

2. Resistor

3. Power supply

Theory:- Write based on the following questions.

1. Explain transistor operation.

2. State the concentration and area of collector, base and emitter.(Highly

doped/lightly doped/ large area)

3. Write equations which explains operation of transistor as a switch.

Procedure:

Result :

Conclusion:


How to read Resistor Color Codes

First the code

Black Brown Red Orange Yellow Green Blue Violet Grey White

0 1 2 3 4 5 6 7 8 9

The mnemonic

B. B. ROY Great Britain Very Good Wife

Black is also easy to remember as zero because of the nothingness common

to both.

(Please don't add or change the mnemonic - it will only get reverted -admin)


How to read the code

First find the tolerance band, it will typically be gold ( 5%) and

sometimes silver (10%).

Starting from the other end, identify the first band - write down the

number associated with that color; in this case Blue is 6.

Now 'read' the next color, here it is red so write down a '2' next to the

six. (you should have '62' so far.)

Now read the third or 'multiplier exponent' band and write down that

as the number of zeros.

In this example it is two so we get '6200' or '6,200'. If the 'multiplier

exponent' band is Black (for zero) don't write any zeros down.

If the 'multiplier exponent' band is Gold move the decimal point one to

the left. If the 'multiplier exponent' band is Silver move the decimal point

two places to the left. If the resistor has one more band past the tolerance

band it is a quality band.

Read the number as the '% Failure rate per 1000 hour' This is rated

assuming full wattage being applied to the resistors. (To get better failure

rates, resistors are typically specified to have twice the needed wattage

dissipation that the circuit produces). Some resistors use this band for

temco information. 1% resistors have three bands to read digits to the left

of the multiplier. They have a different temperature coefficient in order to

provide the 1% tolerance.

At 1% the temperature coefficient starts to become an important

factor. at +/-200 ppm a change in temperature of 25 Deg C causes a

value change of up to 1%


BS 1852 Coding for resistor values

BS 1852(British Standard 1852). The letter R is used for Ohms and K for

Kohms M for Megohms and placed where the decimal point would go.

At the end is a letter that represents tolerance Where M=20%, K=10%, J=5%,

G=2%, and F=1% D=.5% C=.25 B=.1%

BS 1852 coding examples

R33 0.33 ohms

2R2 2.2 ohms

470R 470 Ohms

1K2 1.2K ohms

22K 22K ohms

22K2 22.2K ohms

4M7 4.7M ohms

5K6G 5.6K ohms 2%

33KK 33k Ohms 10%

47K3F 47.3 K Ohms 1%


Question Bank

Sub:- Basic Electrical & Electronics

Sem:- I

A] DC CIRCUITS

Qu.1 Explain ideal voltage Source and ideal current source.

Qu2. Explain the concept of source transformation with suitable example.

Qu.3 Explain kirchoff’s voltage and current law with suitable example.

Qu4.Explain maxwell’s mesh current method with suitable example.

Qu.5 Explain the steps to find Thevenin’s resistance and current with suitable example.

Qu.6 State and explain maximum power transfer theorem.

Qu.7 Derive the relation to express three star connected resistance into equivalent delta.

B]

introduction of active and passive components · 2018-06-11 · experiment no:-1 date ... to...

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