Chapters Rew

8/3/2019 Chapters Rew

1/31

AIM: To plot the frequency response of series resonance circuit and to calculate

the bandwidth.APPARATUS REQUIRED:

S. No. COMPONENTS RANGE QUANTITY

1. Audio Signal Generator (0 - 10)MHz 1

2. Ammeter (0 - 10)mA, mI 1

3. Resistor 47 1

4. DRB - 1

5. DCB - 1

6. DIB - 1

THEORY:

A circuit is said to be a series resonance circuit, if the resistance, inductance

and capacitance are connected in series and it behaves in effect like purelyresistivity circuit. The total current drawn by the circuit is in phase with the appliedvoltage and the power factor will then be unity. Thus, at resonance the equivalentcomplex impedance of the circuit has no imaginary part.

The complex impedance is given as,

Z = R + J(XL - XC),

PRACTICAL CIRCUIT FOR SERIES RESONANCE CIRCUIT:

Expt. No1.(a)FREQUENCY RESPONSE OF SERIES RESONANCE CIRUIT

Date :. Page No..


2/31

MODEL GRAPH:

Imax

Current (mA) Im/f2

f1 f2 f3 Frequency (Hz)

XL = C,

XC = 1/ ,

where XL, XC have directly opposing actions.


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The current is maximum at the resonance frequency frand is equal to

fr= E/R

At (fr) resonance frequency XL=XC and the impedance is purely resistive for

frequencies less than the resonance frequency.

XL< XC

Therefore, it is capacitive

f > fr

XL > XC

and the circuit behaves in effect on RL circuit. Taking logging current is increasedthe maximum current becomes less than frequency and it is not dependent on f.

PROCEDURE:

i. Connections are given as per the circuit diagram.

ii. Varying the frequency and note down the ammeter reading and frequency.

iii. Plot the graph and find out the frequency and bandwidth.

Bandwidth =f2 f1

TABULATION OF SERIES RESONANCE CIRCUIT:

S. No. FREQUENCY(KHz) CURRENT(mA)

Bandwidth = f2 f1

Im =

Im/2 =

RESULT:


4/31

Thus, the frequency response of series resonance circuit was completed andhence Bandwidth =

CIRCUIT DIAGRAM FOR PARALLEL RESONANCE CIRCUIT:

MODEL GRAPH:

Vmax

Voltage (V) Vm/f2


5/31

f1 f2 f3 Frequency f(Hz)

AIM: To plot the frequency response of parallel resonance circuit and to calculatethe bandwidth.

APPARATUS REQUIRED:


1. Audio Signal Generator (0 - 10)MHz 1

2. Ammeter (0 - 10)mA, mI 1

3. Resistor 47 1

4. DRB - 1

5. DCB - 1

6. DIB - 1

THEORY:

A circuit is said to be a series resonance circuit, if the resistance, inductanceand capacitance are connected in series and it behaves in effect like purelyresistivity circuit. The total current drawn by the circuit is in phase with the appliedvoltage. Thus, at resonance the overall power factor is unity.

This is achieved when, XL= XC

TABULATION OF SERIES RESONANCE CIRCUIT:

S. No. FREQUENCY(KHz) CURRENT(mA)

Expt. No1.(b)FREQUENCY RESPONSE OF PARALLEL RESONANCE CIRUIT

Date :. Page No..


6/31

Bandwidth = f2 f1

Im =

Im/2 =

FORMULA:

Bandwidth = f2 f1,

Resonant frequency, fr = 1/(2**(L*C))

PROCEDURE:

i. Connections are given as per the circuit diagram.

ii. Set the input voltage 5V, using signal operator.

iii. Varying the frequency and note down the ammeter reading and frequency.

iv. Plot the graph and find out the frequency and bandwidth.

RESULT:

Thus, the frequency response of parallel resonance circuit was completedand hence Bandwidth =

PRACTICAL CIRCUIT FOR KIRCHHOFFS VOLTAGE LAW:

R 1

1 0 0 k

V

( 0 - 3 0 ) V

V 1

( 0 - 1 5 ) V

V 3

( 0 - 1 5 )

R 2

2 2 0 k

I

( 0 - 5 ) m A

R 3

1 k

V 2

( 0 - 1 5 ) V


7/31

THEORETICAL CIRCUIT FOR KIRCHHOFFS VOLTAGE LAW:

R 4

1 k

R 6

1 k

V

R 5

1 k

AIM:

To verify the Kirchhoffs Voltage law.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 1

2. Ammeter (0 - 15)mA 33. Resistor 100, 1K , 220 1

4. Voltmeter (0 - 15)V 3

Expt. No2.(a) VERIFICATION OF KIRCHOFFS VOLTAGE LAW

Date :. Page No..


8/31

FORMULA USED:

V1 = I*R1, V2 = I*R2, V3 = I*R3,

V =V1 + V2 + V3,

V = I*R1 + I*R2 + I*R3,

Requivalent = R1 + R2 + R3,

I = V/Requivalent

THEORY:

KIRCHHOFFS VOLTAGE LAW:

In any network, the algebraic sum of voltage drop across the circuit elementof the closed path is equal to the algebraic sum of e.m.f in the path. In other words,the algebraic sum of all the branch voltage around any closed path or closed loop isalways zero, i.e., E*V=0.

TABULATION FOR KIRCHHOFFS VOLTAGE LAW:

S.

No.VOLTAGE

(V)

TOTAL

CURRENT

(mA)

THEORETICAL

VOLTAGE DROP

(V)

THEORETICAL

VOLTAGE DROP

(V)

V1

(V)

V2

(V)

V3

(V)

V=(

V1+

V2+

V3)V

V1

(V)

V2

(V)

V3

(V)

V=(

V1+

V2+

V3)V

The law states that if one states at a certain point of the closed path and goestracing nothing all the potential charge in any particular direction till the cutting


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point is reached again it must be cut the same potential with which statics traininga closed path. The law applies equally will to a circuit. By the constant source d.c.

PROCEDURE:

i. Kirchhoffs voltage law connection is made as per circuit diagram.

ii. For various values of supply voltage, note the current voltage drop V1, V2&V3.

iii. Check the value with the theoretically applied voltage.

CALCULATION:

V1 = I*R1,

V2 = I*R2,

V3 = I*R3,

V =V1 + V2 + V3,

RESULT: Thus, the Kirchhoffs voltage law was verified.

PRACTICAL CIRCUIT FOR KIRCHHOFFS CURRENT LAW:

R 1

1 0 o h m

I 1

( 0 - 5 0 ) m A

R 2

1 k

( 0 - 3 0 ) V

I 3

( 0 - 5 0 )I 2

( 0 - 5 0 ) m A


10/31

THEORETCAL CIRCUIT FOR KIRCHHOFFS CURRENT LAW:

R 1

1 k

R 2

1 kV

AIM: To verify the Kirchhoffs Current law.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 1

2. Ammeter (0 - 15)mA 3

3. Resistor 100, 1K , 220 1


FORMULA USED:

I1 = V/R1

I2 = V/R2,

Expt. No2.(b) VERIFICATION OF KIRCHOFFS CURRENT LAW

Date :. Page No..


11/31

I= I1 + I2.

THEORY:

KIRCHHOFFS CURRENT LAW:

The connection of two or more elements create a junction called a node. Thejunction between the two elements is called a simple node. Otherwise, divisiontakes place. Kirchhoffs current law state that the algebraic sum of the currentleaving that is equal to algebraic sum of current entering that node. In the nodeanalysis, KCL is used to analyze the circuit.

TABULATION FOR KIRCHHOFFS VOLTAGE LAW:

S.

No.VOLTAGE

(V)

THEORETICAL

VALUE (mA)

PRACTICAL

VALUE (mA)

I1

(mA)

I2

(mA)

I

(mA)

I1

(mA)

I2

(mA)

I

(mA)

By Kirchhoffs current law,

I0 + I1 + I2 = I3 + I4 + I5.

PROCEDURE:

i. Connection is made as per circuit diagram.

ii. For various values of supply voltage, note the ammeter reading.

iii. Then verify current law,

I = I1 + I2.


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Verify the value with the theoretically applied current.

CALCULATION:

I= I1 + I2.

I1 = V/R1

I2 = V/R2,

RESULT: Thus, the Kirchhoffs current law was verified.

THEORETCAL CIRCUIT

R 2

R 1

( 0 - 3 0 ) VR

R L

R 1

R L

R 1

( 0 - 3 0 ) V

Expt. No3.(a) VERIFICATION OF THEVENIN THEOREM

Date :. Page No..


13/31

AIM: To verify Thevenin theorem.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 1

2. Ammeter (0 - 15)mA 1

3. Resistor 2.2, 5.6K 2


FORMULA USED:

IL = Vth/(Rth + RL),

I = V/(R1 + R2),

Rth = (R1*R2)/(R1 + R2),

Vth = I*R2.

where

IL is Load Current

Vth is Thevenin Voltage

R2 is Load Resistance

I2 is Load Current

PRACTICAL CIRCUIT


14/31

( 5 . 6 K ) R 2

( 2 . 2 K ) R 1

V

( 0 - 1 5

V

( 0 - 3 0 ) V

CIRCUIT TO FIND IL

( 2 . 2 K ) R ' t hI( 0 - 1 0 ) m

( 2 . 2 K

V

( 0 - 3 0 ) V

( 5 . 6 K ) R 2

( 2 . 2 K ) R 1I 4 ( 0 - 1 0 )

( 2 . 2 K

V

( 0 - 3 0 ) V

TABULATION FOR THEVENIN THEOREM:


15/31

S.

No.

VOLTAGE

(V)

I = V/

(R1+R2)

(Ma)

THEORETICAL

VALUE

Vth

PRACTICAL

VALUE

Vth

THEORETICAL

VALUE

(Ma)

THEORETICAL

VALUE

(Ma)

1.THEOREM:

Any linear active to passive network consisting of voltage source andcurrent source can be reduced to simple network. The simple network consisting fvoltage source in series with resistance is said to be Thevenin theorem.

THEORY:

This theorem is applicable where it is desire to determine the current (or)voltage across any element. In a network any two terminal by lateral d.c currentcan be replaced by an equivalence circuit consisting of voltage across sources andseries resistor. Any linear bilateral network can be replaced by an equivalentcircuit consisting of a voltage across the open circuiter low terminal and resistanceof source network looking through the open circuiter low terminal. To theThevenin resistance voltage source is resourced by short circuit and currentthrough low resistor can be found out.

PROCEDURE:

i. Connections are made as per the circuit diagram.

ii. Vary the supply voltage and note down the Thevenin voltage Vth.

iii. Connections are made as per the circuit R1

iv. Note the current It across RL.

v. For different power supply voltage note down RL and compare the value of

Vth and IL, which the theoretical value are shown.

CALCULATIONS:

I = V/(R1 + R2),

Vth = I*R2


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Rth = (R1*R2)/(R1 + R2),.

IL = Vth/(Rth + RL),

RESULT: Thus, the Thevenin theorem was verified.

NORTONS THEORETICAL CIRCUIT

( 1 0 0 o h m ) R

( 1 K ) R 2( 2 . 2 K

V

( 0 - 3 0 ) V

Rth

( 1 0 0 o h m )

( 1 K ) R

Rth = (R1*R2)/(R1+R2)

EQUATING CIRCUIT:

R 2

2 2( 1 K ) R 2

I 1

0 A d c


17/31

( 1 0 0 o h m ) R

V

( 0 - 3 0 ) V

( 1 K ) R

AIM: To verify Nortons theorem.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 1

2. Ammeter (0 - 100)mA 1

3. Resistor 2.2K, 100 1

FORMULA USED:

Thevenin Resistance Rth = (R1*R2)/(R1 + R2) ,

Current I = V/R1,

Short circuit current Isc = V/R1,

Load current IL = Isc*Rth/(Rth + RL).

STATEMENT:

Any linear active network with output terminals A, B can be replaced by asingle current source Isc in parallel with simple single impedance Zth.

Expt. No3.(b) VERIFICATION OF NORTONs THEOREM

Date :. Page No..


18/31

THEORY:

PRACTICAL CIRCUIT:

IL

( 1 0 0 K ) R 1

V

( 0 - 3 0 ) V

R 2

2 . 2 k

I 1

( 0 -

ISC

R 2

1 k

( 1 0 0 K ) R 1

V

( 0 - 3 0 ) V

I 1

( 0 - 1

Nortons theorem is a dual of Thevenins theorem. The current to the resistanceconnected to the terminals of the Nortons equivalent circuit must have the sameresistance connected to the original active network.

PROCEDURE:


19/31


ii. Supply voltage is varied in steps and the value of short circuit current isnoted for its step.

iii. For various value of short circuit current, note down the load current.

iv. The theoretical value of the short circuit current and the load current arecalculated and verify with practical value.

TABULATION FOR NORTONs THEOREM

S. No.

VOLTAGE

(V)

PRACTICA

L VALUE

FOR SHORT

CIRCUIT

Isc(mA)

THEORETICAL

VALUE FOR

SHORT

CIRCUIT

Isc(mA)

VOLTAGE

V=Isc*Rth

PRACTICA

L LOAD

CURRENT

IL (mA)

THEORETICAL

LOAD

CURRENT

IL (mA)

1.

CALCULATIONS:

Short circuit current Isc = V/R1,

Load current IL = Isc*Rth/(Rth + RL),

RESULT:

Thus, the Nortons theorem was verified.

PRACTICAL CIRCUIT:


20/31

V

( 0 - 3

( 2 . 2 K ) R 1

( 5 . 6 K ) R 3V

( 0 - 3 0 ) V

( 1 K ) R 2

THEORETICAL CIRCUIT:

I 1

( 0 - 1 0 ) m A

( 2 . 2 K ) R 1

( 5 . 6 K ) R 3

V

( 0 - 3 0 ) V

( 1 K ) R 2

Expt. No4. VERIFICATION OF SUPERPOSITION THEOREM

Date :. Page No..


21/31

AIM: To verify the superposition theorem.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 2

2. Ammeter (0 - 50)mA 1

3. Resistor 2.2K, 1K , 5.6K 1

FORMULA:

CIRCUIT I

With Vs short circuited

Rth1 = R1 + (R2*R3)/(R2 + R3),

IT1 = V1/RTH1,

I1 = IT1*R2*(R2+R3).

CIRCUIT II

RTH2 = R2+(R1*R3)/(R1+R3),

IT2 = V2/RTH2,

I= = IT2*R1/(R2+R3).

WHEN V1 IS SHORT CIRCUITED:


22/31

I 1

( 0 - 1 0 ) m A

( 2 . 2 K ) R 1

V 2

( 0 - 3 0 ) V( 5 . 6 K ) R 3

( 1 K ) R 2

WITH V1 AND V2:

V 1

( 0 - 3 0 ) V

I 1

( 0 - 1 0 ) m A

( 2 . 2 K ) R 1

V 2

( 0 - 3 0 )( 5 . 6 K ) R 3

( 1 K ) R 2

CIRCUIT III

With V1 and V2


23/31

IT = I1+I2.

THEORY:

With the help of this resistance theorem, we can find the current through or

the voltage across the given element in a linear circuit consisting of two or moresources.

In a linear circuit containing more than one source, the current that flows atany point or voltage that exists between any two points is the algebraic sum ofcurrent or voltage. That would have been reduced by each source taken separatelywith one other replaced or removed.

Total resistance is equal to R1+R2+RL.

IL = (E1-E2)/( R1+R2+RL),

IL = IL+(-IL),

Where

IL is the current to RL because of E1,

IL is current to RL because of E2,

IL is current to RL because of E1 and E2.PROCEDURE:

i. Connections are made as per circuit diagram.

ii. For various values of total current I1 and I2, IL is measured from theammeter reading.

iii. Then verify the theoretical value for different voltage.


24/31

TABULATION FOR SUPERPOSITION THEOREM

S.

No

.

VOLTA

GE

(V)

EXPERIMENT

AL VALUE

I1 (mA)

THEORETIC

AL

VALUE

I1 (mA)

EXPERIMENT

AL VALUE

I2 (mA)

THEORETIC

AL

VALUE

I2 (mA)

VOLTA

GE

(V)

EXPERIMENT

AL VALUE

IT (mA)

THEORETIC

AL

VALUE

IT (mA)

1.

CALCULATIONS:

CIRCUIT - 1

Rth1 = R1 + (R2*R3)/(R2 + R3),

IT1 = V1/RTH1,

I1 = IT1*R2*(R2+R3),

CIRCUIT 2

RTH2 = R2+(R1*R3)/(R1+R3),

IT2 = V2/RTH2,

I= = IT2*R1/(R2+R3),CIRCUIT 3

IT = I1+I2

RESULT:

Thus, the superposition theorem was verified.

THEORETICAL CIRCUIT:


25/31

V 1R L

1 k

R 1

PRACTICAL CIRCUIT:

V

( 0 - 3 0 ) V

D R B I L+

( 2 . 2 K ) R 1

I 3

( 0 - 1 0 )

AIM: To verify the value of load resistance corresponding to power transfer.

APPARATUS REQUIRED:

Expt. No5.(a). VERIFICATION OF MAXIMUM POWER TRANSFER

Date :. Page No..


26/31


1. RPS (0 - 30)V 1

2. Ammeter (0 - 10)mA 1

3. Resistor 2.2K 1

4. DRD (0-100)K 1

FORMULA:

Power = [I*I]RL

RL is the value of resistance for which maximum power is delivered.

THEORY:

Maximum power will be delivered from the voltage source to the loadresistance. This source to terminal resistance of network. Consider a voltage will

be delivered from a voltage source from a internal resistance RG, connected to aload resistance. When the power transfer to the load is maximum RL = RG. At this

solution the total resistance RL+RG=2 RL and load current IL = RG/2RL.Power distributed to RL = (IL*I2)RL

MODEL GRAPH :

Power (mW)

Pmax

Rmax Resistance


27/31

TABULATION FOR MAXIMUM POWER TRANSFER

S.

No.

RESISTANCE

()

CURRENT

(mA)

POWER P = (I*I)RL

(mW)

1.

(VG*VG)KRL.

Under the condition, the efficiency is 50%, since only half of the total power

generated is less in dissipation within the source.PROCEDURE:


ii. Supply voltage is fixed at 10V.

iii. Load resistance, RL is varied.

iv. Current through RL is noted.

v. Step 2 is repeaed for various of RL.

vi. Power dissipated in PLP = (I*I)RL, is calculated.

vii. The Power of resistance RL from which maximum power delivered is foundfrom the graph plotted resistance on X-axis and power on Y-axis.

RESULT:

Thus, the maximum power transfer theorem was verified.

a). RECIPROCITY THEOREM:


28/31

R 2 R 4

V

R 1 R 3 R 5

To find Output Current when Source is connected at AA input R1

VR 2 R 4

R 1 R 3 R 5

PRACTICAL CIRCUIT:

CIRCUIT I

I 2

( 0 - 3 0( 1 K ) R 2V

( 0 - 3 0 ) V

( 1 0 0 K ) R 4

( 1 0 0 K ) R 1 ( 1 0 0 K ) R 3( 2 2 0 K ) R


29/31

AIM: To verify the value of load resistance corresponding to reciprocity theorem.

APPARATUS REQUIRED:


1. RPS (0 - 30)V 1

2. Ammeter (0 - 15)mA 1

3. Resistor 100, 220K, 1K 1

FORMULA:

Applying measurement analysis to the values given

Ia = Ib = I3 = 3/

- determinant of resistance matrix

3 determinant of resistance matrix whose third column is replaced

by voltage matrix

RECIPROCITY THEOREM STATEMENTS:

In linear, bilateral network the voltage source, b volts in a branch gives acurrent in another branch. If Vis applied in second branch, the current in the first

branch in the B2. This V/I is called transfer resistance or Impedance.

CIRCUIT II:

Expt. No5.(b). VERIFICATION OF RECIPROCITY THEOREM

Date :. Page No..


30/31

( 1 K ) R 2 ( 1 0 0 K ) RI 3

( 0 - 5 0 ) m A

( 1 0 0 K ) R 1

V

( 0 - 3

( 1 0 0 K ) R 3( 2 2 0 K )

TABULATION FOR RECIPROCITY THEOREM:

S.

No.VOLTAGE

(V)

THEORETICAL VALUE PRACTICAL VALUE

IA (mA) IB (mA) IA (mA) IB (mA)

THEORY:

The reciprocity theorem states that a linear bilateral network the voltage

source V volts in a branch gives rise to a current I, in another branch. The ratio V/I,is a constant then the position of V and I are interchange.

It is found that on changing the voltage source from branch 1 to 2. Thecurrent I in branch 2 appears in branch one as shown in the figure. The circuit AAdenote the input terminals and BB denotes the output terminals. The applicationof voltage B cross AA produces current I at BB. The position of source andresponses are in interchange then by continuity voltage across BB. The resultantcurrent I will be at the terminals AA. According to Reciprocity theorem, the ratio

of input to responses in both cases. This theorem is also valid theorem for networkconnect the single current source.

PROCEDURE:



31/31

ii. The source voltage is connected as input side at AA.

iii. Source B is varied and the corresponding values of current Ia, is noted.

iv. Connect the circuit as per circuit- II.

v. The source voltage B is connected at output side BB.

vi. The source voltage B is varied in same steps and is 3 corresponding is noted.

vii. According to this theorem, Ia and Ib should be same.

viii. Theoretical values in Ia and Ib are calculated and varying with practicalvalues.

CALCULATION:

RESULT:

Thus, the reciprocity theorem was verified.

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