Circuit Thoery Lab Manual

8/13/2019 Circuit Thoery Lab Manual

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EE2155 Electrical Circuits Lab Manual

DepartmentofElectrical&ElectronicsEngineeringSUDH RS N ENGINEERING COLLEGE

pudukkottai


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OHMS LAW

CIRCUIT DIAGRAM

TABULATION

S.No

Resistance(R)

()

Current in Amperes

(I)

Voltage across RL(V)

The Exp The Exp


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Ex.No:1

VVEERRIIFFIICCAATTIIOONNOOFFOOHHMMSSLLAAWWAANNDDKKIIRRCCHHOOFFFFSSLLAAWW

AIMTo study and verify the Ohms law and kirchoffs laws by conducting suitable

experiment on the given Electric circuit.

APPARATUS REQUIRED

S.No Apparatus Name Range Type Quantity

1

23

4

Ammeter

Trainer kitMultimeter

Patch Chords

0-100mA

------

---

MC

----Digital

----

3

11

As required

THEORY

Ohms law

At constant temperature, the potential difference (E) across the ends of a conductor is

proportional to the current (I) flowing through it.

i.e. E=IR

Where R is the resistance of the conductor in ohm.

PROCEDURE

1. The resistor values in the circuit are designed for 100 mA, 12V.2. Connections are made as per the circuit diagram.3. Switch on the trainer kit.4. The corresponding voltages & currents are tabulated for variable load resistance (RL)

values.

5. By comparing theoretical and experimental values, the ohms law is verified.


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KIRCHOFFS VOLTAGE LAW

CIRCUIT DIAGRAM

TABULATION

S.No VS (V)V1(V) V2 (V) V3 (V) V= V1+ V2+ V3

The Exp The Exp The Exp The Exp


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KIRCHOFFS VOLTAGE LAW

Kirchoffs voltage law states that In a closed loop Electric circuit the sum of potential

drop is equal to the sum of potential rise.

Along the current direction the polarity of voltage from ve to +ve is considered as

potential rise and +ve to ve is known as potential drop. In the resistor the current entering

terminal is marked as positive and current leaving terminal is marked as negative.

V = V1+ V2 + V3

= IR1+ IR2 + IR3

PROCEDURE

1. Connections are made as per the circuit diagram.2. R1, R2and R3are selected such that the total current in the circuit is within 100mA.3. The voltage across the resistor V1, V2and V3are measured by a Digital

voltmeter/Multimeter for 8V as input voltage and tabulated.

4. Repeat the above procedure for input voltages 10V & 12V.5. By comparing theoretical and experimental values the Kirchoffs voltage law is verified.

KIRCHOFFS CURRENT LAW

Kirchoffs current law states that the sum of current entering into a junction is equal to

the sum of current leaving away from it.

The junction may be interconnection of two or more branches. In the given circuit, the

current entering the node a are I1 and I2and current leaving the node is I3.

PROCEDURE

1. The resistor values in the circuit are designed for 100mA and 12V.2. Connections are made as per the circuit diagram.3. 8V, 10V and 12 V supply is given to the circuit.4. The values of currents are measured by Analog Ammeter and tabulated.5. By comparing theoretical and experimental values Kirchoffs current law is verified.


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RESULT

Thus ohms law and kirchoffs laws were verified with the given resistive circuit.


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THEVENINS THEOREM

CIRCUIT DIAGRAM

TO FIND Rth

TO FIND Vth


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Ex.No:2

VVEERRIIFFIICCAATTIIOONNOOFFTTHHEEVVEENNIINNSSAANNDDNNOORRTTOONNSSTTHHEEOORREEMM

AIM

To verify Thevenins theorem and Nortons theorem by conducting suitable

experiment on the given Electrical circuit.

APPARATUS REQUIRED


1

2

3

4

Ammeter

Trainer kit

Multimeter

Patch Chords

0-100mA

---

---

---

MC

----

Digital

----

1

1

1

As required

THEVENINS THEOREM

A linear, bilateral, lumped network with open output terminals can be reduce to asimple circuit consisting of a single voltage source in series with a resistance .The value

of the voltage source is equal to the open circuit voltage across the open terminals and the

value of resistance is equal to the resistance seen in to the network across the open

terminals.


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TABULATION

S.No RL() IL(A)

The Exe

TO FIND Rth&Vth

S.NoVth(V) Rth()

The Exe The Exe

THEVENINS EQUIVALENT CIRCUIT


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Let us find the value of the current I using Thevenins theorem. First we disconnect RL

and make its terminal open.

The voltage across the open circuit terminals is called as Thevenins voltage Vth.

The value of Rthis found after replacing the voltage source V by short circuit.

Thevenins Equivalent circuit is drawn with Vthand Rth.The disconnected

element is placed at the open circuit terminals of the Thevenins equivalent circuit.

PROCEDURE

1. Connections are given as per the circuit diagram.2. For various values of RL,measure ILvalues.3. To find open circuit voltage Vth, remove RLand measure open circuit voltage

across output terminals.

4. To find Rth,a. Replace the supply sources by their internal resistances.b. Remove load resistor RL.c. Using Multimeter the equivalent resistance across the load terminal was

found.

5. Using the formula IL = Vth/ (Rth+ RL), current value is calculated.


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NORTONS THEOREM

CIRCUIT DIAGRAM

TO FIND RN

TO FIND IN


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NORTONS THEOREM

A linear, bilateral, lumped network with open output terminals can be reduced to

a simple circuit consisting of single current source in parallel with a resistance.

The value of current source is equal to the current passing through the short-circuited

output terminals. The value of the resistance is equal to the resistance seen into the

network across the output terminal.

Let us find the value of current I using Nortons theorem. First we disconnect RLand

short-circuit its terminals.

The current through the short-circuited output terminals is called Nortons current IN. The

RNvalue is found across the open circuit terminals as found in Thevenins theorem, after

short circuiting the voltage source E.


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TABULATION

S.No RL() IL(A)

The Exe

TO FIND RN&IN

S.NoIN(mA) RN()

The Exe The Exe

NORTONS EQUIVALENT CIRCUIT


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Nortons equivalent circuit is drawn with IN and RN.

The disconnected element RLis placed across the open output terminal. Now Nortons

equivalent circuit is,

PROCEDURE

1. Connections are given as per circuit diagram2. For various values of RLmeasure ILvalues.3. To find Nortons short circuit current IN, remove the load resistor RLand connect

ammeter between the terminals, the current through short circuit terminal was

found.

4. To find RN,a. Replace the supply sources by their internal resistances.

b. Remove load resistor RL.

c. Using multimeter the equivalent resistance across the load terminal was found.

5. Using the formula IL = (RN* IN)/ (RN+ RL), current value is calculated.


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RESULT

Thus Thevenins theorem and Nortons theorem were verified both theoretically

and experimentally.


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SUPERPOSITION THEOREM

CIRCUIT DIAGRAM

TO FIND IL

TO FIND IL


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Ex.No:3

VVEERRIIFFIICCAATTIIOONNOOFFSSUUPPEERRPPOOSSIITTIIOONNTTHHEEOORREEMM

AIM

To verify the superposition theorem experimentally.

APPARATUS REQUIRED

S.No

APPARATUS

NAMERange Type Quantity

1

2

3

4

Ammeter

Trainer kit

Multimeter

Patch Chords

0-100mA

---

---

---

MC

----

Digital

----

1

1

1

As required

THEORY

The superposition theorem states that In a linear, lumped element, bilateral

electric circuit that is energized by two or more sources the current in any resistor is equal

to the algebraic sum of the separate current in the resistor when each source acts

separately. While one source is applied, the other sources are replaced by their respective

internal resistance. To replace the other sources by their respective internal resistance,

the voltage sources are short-circuited and the current sources open circuited.

Consider the given electric circuit,


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Tabulation

S.No RRLL() V1 (V)IL(mA)

V2 (V)IL(mA) IL = IL+ IL

The Exe The Exe The Exe


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To find the current through RL, first short circuit the voltage source V2by the internal

resistance.

Let IILL= current through the load while source V1acting alone.

And find current through RL

IILL =

Now short circuit the voltage source V1by the internal resistance and energize the

voltage source V2

Let IILL = current through the load while source V2acting alone.

And find IILL =

IL= IILL+ IILL


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PROCEDURE

1. Connections are made as per the circuit diagram.2. By adjusting the pot, set the load resistor (RL) value.3. Now set the voltage level of V1 & V2 for a constant voltage and measure the

branch current IL.

4. Now short circuit the voltage source V2, and keep the voltage source V1remainsconstant value and measure IL (the current through RL)

5. Now short circuit the voltage source V1, and keep the voltage source V2remainsconstant value and measure IL(the current through RL).

6. Repeat the above procedure for different values of load resistor RLand tabulate it.7. Now the total current through RLis IL= IILL +IILL.8. By comparing the theoretical and experimental values, the super position theorem

is verified.

RESULT

Thus the superposition theorem was verified practically and theoretically.

Then current through RL is IL= IILL+ IILL.


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MAXIMUM POWER TRANSFER THEOREM

CIRCUIT DIAGRAM

TABULATION

Vs= V, Rs=

S.No RL()

IL

(A) Pout

(W)

Actual Calculated Actual Calculated


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Ex.No:4

VVEERRIIFFIICCAATTIIOONNOOFFMMAAXXIIMMUUMMPPOOWWEERRTTRRAANNSSFFEERRTTHHEEOORREEMM

AIM

To verify Maximum power transfer theorem experimentally.

APPARATUS REQUIRED


1

2

3

4

Ammeter

Trainer kit

Multimeter

Patch Chords

0-100mA

---

---

---

MC

---

Digital

----

1

1

1

As required

THEORY

Many circuits basically consist of sources, supplying voltage, current or power to

the load. Sometimes it is necessary to transfer maximum voltage, current or power from

the source to the load.

More voltage is delivered to the load when the load resistance is high as

compared to the resistance of the source. Maximum current is transferred to the load

when the load resistance is small compared to the source resistance.

Thus, Maximum power transfer theorem states that, Maximum power is

delivered from a source to a load when the load resistance is equal to the source

resistance.


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Current in the circuit is I = VS / ( RS+RL ).

Power delivered to the load RLis P = I

2

RL ( )2

2

LS

LS

RR

RV

+

=

To determine the value of RLfor maximum power to be transferred to the load, we

have to set the first derivative of the above equation with respect to RL, i.e., whenL

dR

dP

equals zero.

( )

+=

2

2

LS

LS

LL RR

RV

dR

d

dR

dP

( ) ( ){ }( )

+

++=

4

222

2

LS

LSLLSS

RR

RRRRRV

( ) ( )LSLLS

RRRRR ++22

2 = 0

0222 222

=++LSLLSLS

RRRRRRR

RS= RL

So, maximum power will be transferred to the load when load resistance is equal to

the source resistance.

PROCEDURE

1. The resistor value in the circuit is designed such that RS= RL.2. Connections are made as per the circuit diagram.3. 12 V supply is given to the circuit.4. The current IL is measured for various values of RLby using ammeter.5. By using I = VS/ R, current value is calculated and the output can be calculated as

POUT = I2RL.

6. By comparing the both powers, we can conclude that maximum power will betransferred at RS= RL.


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RESULT

Thus the maximum power transfer theorem was verified.


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RECIPROCITY THEOREM

CIRCUIT DIAGRAM

Circuit 1

Circuit 2


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Ex.No:5

VVEERRIIFFIICCAATTIIOONNOOFFRREECCIIPPRROOCCIITTYYTTHHEEOORREEMM

AIM

To verify Reciprocity theorem experimentally.

APPARATUS REQUIRED

S.NoAPPARATUS

NAMERange Type Quantity

1

2

3

4

Ammeter

Trainer kit

Multimeter

Patch Chords

0-100mA

---

---

---

MC

----

Digital

----

1

1

1

As required

THEORY

The Reciprocity theorem states that In a linear, bilateral network a voltage

source V volts in a branch gives rise to a current I in another branch, the ratio V/ I is

constant when the position of V and I are interchanged.

According to this theorem if a source voltage and ammeter are interchanged, the

magnitude of the current through the ammeter will be same. Consider the given network

fig 1 with two loops 1 and 2 if an ideal voltage source V in loop 1 produces a current I

in loop 2, then interchanging positions as shown in fig - 2, The same ideal source V in

loop 2 produces the same current I in loop 1. The network is said to be reciprocal.

Fig 1 Fig - 2


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TABULATION

Table 1 (For Circuit 1)

S.No Voltage (Vs)Current I1(A)

The Exe

Table 2 (For Circuit 2)

S.No Voltage (Vs)

Current I2(A)

The Exe


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PROCEDURE

1. Connections are made as shown in circuit diagram (1).2. Measure the value of current I1 for various voltage levels and record them in

table1.

3. Change the voltage source position as shown in circuit diagram (2).4. Measure the value of current I2for various voltage levels as in step 1 and record

them in table2.

5. By comparing theoretical and experimental values, the reciprocity theorem isverified.

RESULT

Thus the Reciprocity theorem was verified both theoretically and experimentally.


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Ex.No:6

FREQUENCY RESPONSE OF SERIES AND PARALLEL RESONANCE

AIMTo analyze the characteristics of RLC series and parallel resonance circuits

experimentally.

APPARATUS REQUIRED


1

2

3

4

5

6

Resistor

Capacitor

Inductor

Functional Generator

Multimeter

Connecting Probes

500

0.1F

70 mH

--------

--------

--------

------

disc

Decade box

-------

Digital

-------

1

1

1

1

1

As required

THEORY

SERIES RESONANCE

Consider a circuit consisting or resistor R, inductor L and capacitor C all

connected in series. A fixed voltage, variable frequency sinusoidal voltage is applied

across this series combination. The following observations are based on experimental

study of the behavior of the above circuit:

1. As the supply frequency is varied, the current drawn by the circuit phase varieswith applied frequency. Also, the phase difference between the applied voltage

and the current drawn varies with frequency. The phase difference is a function of

the frequency.2. The current is maximum at a particular frequency, called resonant frequency fr,

and hence the impedance offered by the circuit is minimum at minimum fr. The

voltage applied and the current drawn, are in phase with each other at fr. The load

appears purely resistive at resonant frequency.


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3.( )LC

fr2

1=

i. for f < fr - I leads V,ii. for f > fr - I lags V

4. At f= fr, VR=VS.By measuring the voltage drop across R (VR), for each applied frequency, the

corresponding value of IScan be calculated, by using IS =V/R.

A Graph is drawn between ISand f on semi log graph sheet. The variation in

the phase difference between the applied voltage and current drawn by the circuit is a

function of frequency.

PARALLELRESONANCE

In parallel resonant circuit, an inductor and a capacitor are connected in

parallel to each other, with respect to the supply source. The current through the

inductance has the value VS / XLand lags VSby 900. Thus, two currents are out of phase

with each other.

If XC < XL, then IC > IL and the circuit acts capacitive, if XC > XLthen the

circuit in inductive. When XC = XLthe inductive and capacitive currents are equal and the

circuit is said to be in resonant condition.

At resonance, the resonant frequency for the parallel resonant circuit is thesame as that for the series resonant circuit. The frequency at which resonant occurs in a

parallel LC circuit is sometimes called as the anti resonant frequency to distinguish it

from the resonant frequency of the series LC circuit.

For a parallel resonant circuit, at resonance condition,

i. XL = XC and IL = IC.ii.

( )LCfr

2

1=

iii.P

S

Z

VI = (minimum and in phase with applied voltage)

iv. ZPis maximum and resistive.


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TABULATION

PARALLEL RESONANCE

S.No Frequency F (Hz) Current I (A)


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PROCEDURE

SERIESRESONANCE

1. Connect the RLC components as shown in diagram.2.

Connect the sine signal from a functional generator to the sockets marked as Input

3. Select digital multi meter with AC milli ampere range and connect it across thesocket marked as output.

4. Set the input amplitude to a pre-defined level say RMS volt in DMM 2V (P-P)and this amplitude is said to be maintained constant throughout the experiment.

5. Repeat the same procedure for different frequency settings and the correspondingreadings are noted down in the tabular column.

6. Use the readings to draw the resonance curve.

PARALLELRESONANCE

1. Connect the RLC components as shown in circuit diagram.2. Repeat the above procedure from steps 2 to 6.

RESULT

1. Thus the Resonance curves for RLC series and parallel resonance circuits aredrawn.

2. Resonant frequencya) For series fr= ------------ Hzb) For parallel fr= ------------ Hz


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MESH CURRENT ANALYSIS

CIRCUIT DIAGRAM


41/57

Ex.No:7

VVEERRIIFFIICCAATTIIOONNOOFFMESH AND NODAL ANALYSIS

AIM

To verify Mesh and Nodal analysis by conducting suitable experiment on the

given Electrical circuit.

APPARATUS REQUIRED


1

2

3

4

Ammeter

Trainer kit

Multimeter

Patch Chords

0-100mA

---

---

---

MC

----

Digital

----

3

1

1

As required

MESH CURRENT ANALYSIS

In the loop current method, we need to write and solve equations for as many

currents as the number of independent loops. The number of independent loops is equal

to (b-n+1). This is based on kirchoffs voltage law applicable to only planar networks.

Here KCL is applied automatically. In this method the loop current is an independent

variable. The number of loop current equations is equal to number of independent loops.

Consider the networks shown below. For convenience numerical values are not

given.


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TABULATION

V1= V,V2 = V

S.NO RL()I1(A) I2(A)

I3=I1I2

(A)

Exe The Exe The Exe The


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NODAL VOLTAGE ANALYSIS

CIRCUIT DIAGRAM


45/57

In general the above matrix form is expressed as below:

R11 R12 I1 V1

R21 R22 I2 = V2

Here , V1= the algebraic sum of the emf in the loop 1.

V2= the algebraic sum of the emf in the loop 2

An emf is assigned positive if the loop current leaves at positive terminal of the

source.

PROCEDURE

1. Connections are made as per the circuit diagram.2. Set the load resistor value RLto constant value.3. Note down the branch current I1, I2& I3values by using ammeter.4. Repeat the above procedure for different values of load resistor RL.5. On comparing the theoretical and experimental values Mesh current analysis was

verified.

NODAL VOLTAGE ANALYSIS

Consider the network shown below. The voltages V1and V2can be calculated by

using Nodal Voltage analysis.

In the above circuit three nodes are there. Consider the third node (node 3) as the

reference node. To calculate the node voltages the matrix form can be written as follows.


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TABULATION

VA = V,VB = V

S.NO RL()V1(V) V2(V)

Exe The Exe The


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RA-1

+ RB-1

+ RC-1

-RC-1

V1 = Va/RA

-RC-1

RC-1

+ RD-1

+ RE-1

V2 Vb/RE

GA+ GB + GC

-GC

V1 = I1

-GC GC+ GD+ GE V2 I2

Here, Conductance G = R-1

By solving the above matrix, V1, V2 can be found.

PROCEDURE

1. Connections are made as per the circuit diagram.2. Set the load resistor value RLto constant value.3. Note down the node voltage V1 & V2values by using voltmeter.4. Repeat the above procedure for different values of load resistor RL.5. On comparing the theoretical and experimental values Nodal voltage analysis was

verified.

RESULT

Thus Mesh and Nodal analysis were verified by theoretically and experimentally.


48/57


49/57

Ex.No 8

TTRRAANNSSIIEENNTTAANNAALLYYSSIISSOOFFRRLLAANNDDRRCCCCIIRRCCUUIITTSSAIM

To study the transient response of RL and RC circuit using series D.C and A.C

circuit trainer.

APPARATUS REQUIRED


1.

2.

3.

4.

Series A.C and D.C circuit

trainer kit

D.C ammeter

A.C Ammeter

Patch chords (or) wires

-

(0-200) mA

(0-200 )mA

-

-

-

-

-

1

1

1

As required

THEORY

RESPONSE OF AN R-LCIRCUIT

Consider the RL circuit shown in fig.1 with switch S and applied DC voltage

V. When the switch S closed, we can determine the complete solution to apply the

kirchoffs voltage law to the circuit.


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dt

diLRiV += -------------------------- (!)

Divide equation 1 by L

L

Vi

L

R

dt

di+= -------------------------- (2)

The above equation is linear differential equation; it should refer like this,

kPxdt

dx=+

Whose solution is

ptptptcedtkeex

+= ---------------------- (3)

Where c is arbitrary constant. In a similar way to write the current equation,

( ) ( ) ( )dte

L

Vecei

t

LRt

LRt

LR

+=

( )

R

Vcei

tL

R

+=

---------------------------- (4)

To determine the constant c to apply the initial condition at t=0 (i.e.) just before closing

the switch and the current in the inductor is zero, so at t=0, I=0 to substitute in equation 4,

we get,

R

VC = ------------------------- (5)

Substitute equation 5 in 4,

+= t

LR

RV

RVi exp

= tL

R

R

Vi exp1 ------------------------- (6)

In the above equation, steady state part V/R and the transient part (V/R) e-(R/L) t


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Error!

In Fig. 2 when switch S is closed, the response reaches the steady state

value after some time interval. Thus the transient period is defined as the time taken for

the current to reach the final value.

DC RESPONSE OF AN R-CCIRCUIT

Consider the RC circuit shown in fig. 3. with switch S and applied DC voltage

V. When switch S closed, we can determine the complete solution to apply Kirchoffs

voltage law to the circuit.

+= idtcRiV

1 ---------------------- (7)


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Differentiate the above equation,

c

i

dt

diRO += ----------------------- (8)

Divide the equation 8 by R

01

=+ iRdt

di ----------------------- (9)

The above equation is a linear differential function with only complementary

solution. The solution for this equation is

RctCei /

= ---------------------- (10)

When switch S is closed at t=0, the capacitor never allow sudden changes in

voltage, it will act as a short circuit. At t=0,i=V/R. substitute this current to equation 4.

C=V/R

The current equation becomes,

RCte

R

Vi /

= ------------------------- (11)

InFig. 4 After 4TC the curve reaches the 99 percent of it final value. In that

solution equation 5, the quantity 1/RC is time constant and is denoted by


53/57

PROCEDURE

D.CSERIES R-L

1. Switch ON the CRO.2. Connect 230v supply to the trainer.3. See the waveform in the CRO, it should be exponential manner.

D.CSERIES R-C

1. Switch ON the CRO.2. Connect 230v supply to the trainer.3. See the waveform in the CRO, it should be inverse exponential manner.

RESULT

Thus the Transient response of RL and RC circuits are studied and the output

waveforms are seen using CRO.


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Circuit Diagram:

CONNECTION DIAGRAM:


55/57

Ex.No. 9

MEASUREMENT OF SELF INDUCTANCE OF A COIL

AIM:

To study the measurements of self inductance of coil using Hays bridge method.

Theory:-

The hays bridge is the modification of the Maxwell Bridge. This bridge uses a resistance

in series with the standard capacitor. The bridge has four resistive arms in which the arms one is

consists of the resistor R1, L1.The arm 2 is consists of the variable resistance R3.The low value

of the resistance is obtain by the low resistive arms of the bridge. The value of R4and C4is the

standard value of the capacitor and resistance. By using the known non-inductive resistance R1,R2, R3& R4and standard capacitor C4value, the unknown value of inductor is calculated. The

unknown value of inductance and Quality factor of the coil is obtained by formula.

L1= (R2R3C4) / (1 + 2R4

2C4

2)

Quality factor (Q) = (1 / 2R4

2C4

2)

Basic AC bridges consist of four arms, source excitation and a balanced detector(G).

Commonly used detectors for AC bridges are:

(1) Head phones

(2) Vibration galvanometers

(3) Tunable amplifier detectors

Vibration galvanometer is extremely useful at power and low audio frequency ranges. Vibration

galvanometers are manufactured to work at various frequencies ranging from 5 KHZ to 1 KHZ.

But one most commonly used between 200HZ.

Advantage-1) This Bridge gives very simple expression for unknown for High Q coil.2) This bridge also gives a simple expression for Q factor.

Disadvantage-1) The hays bridge is suited for the measurement of the High Q inductor.

2) It is used to find the inductor having the Q value of the smaller then 10.


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Model Calculation:

By considering the Hay Bridge is in balanced condition,

R1 = 6.25

R2 = 10

R3 = 5

R4 = 8

f = 50 HZ

C4= 156 F

= 2f = 314 rad/sec

The unknown self inductance value of coil is calculated by using formulaL1 = (R2R3C4) / (1 +

2R4

2C4

2)

= (10515610^-6

) / (1 + (314)2(8)2(15610^-6

)2)

L1 = 6.8 mH

The Quality factor of the coil is calculated by using formula

Quality factor (Q) = (1 / 2R4

2C4

2)

= (1 / ((314)2(8)

2(15610^

-6)

2))

Q = 6.5


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Procedure:

1) Study the given circuit.

2) Make all connections to complete the bridge.

3) Set the bridge is in balanced condition by adjusting R3.

4) Note value of R1, R2, R3, and R4 from the circuit diagram.

5) Measure value of L & Q by calculating using given formula.

Result:-

The measurements of self inductance of coil using Hays bridge were studied.

Circuit Thoery Lab Manual

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