NA Lab Manual

LIST OF PRACTICALS 1. Reci Recipr proc ocit ityy Theore Theorem m 2. Maximu Maximum m Power Power Trans Transfer fer Theor Theorem em 3. Star Star Delta Delta Transf Transform ormati ation on 4. Study Study of of RC Series Series AC circ circuit uit 5. Study Study of of RL Seri Series es AC AC circu circuit it 6. To Measure Measure Resonan Resonance ce Frequenc Frequencyy in RLC Serie Seriess Circuit Circuit 7. To Measure Measure Resonance Resonance Frequenc Frequencyy in RLC Series Series Parallel Parallel Circuit Circuit 8. To Observe Observe Damping Damping in in RLC Series Series Circuit Circuit 9. Power Power Fact Factor or Impr Improve ovemen mentt

1

EXPERIMENT # 1 OBJECTIVE: TO VERIFY RECIPROCITY THEOREM APPARATUS:

Ammeter • Power source • Resistors • Connecting wires

1 1 3

•

CIRCUIT DIAGRAM:

T-NETWORK: R1

R3

400 ohm

1000 ohm

E

R2

800 ohm

A

20 V

RECIPROCAL:

A

R1

R3

400 ohm

1000 ohm

R2

800 ohm

E 20 V

2

PIE NETWORK: R2

8 0 0 o hm

E

R1

4 0 0 o h m 1000 1000o ohm

R3 A

20V

RECIPROCAL: R2 800 ohm

A

R1

40 0 oh m

1000 ohm

R3

E 20 V

THEORY:

“ In any linear, bilateral single source network, if source of emf E in any branch produces a current I in any other branch, then the same emf E acting in second branch will produce the same current I in the first branch.” In other words, it simply means that E that E and and I I are are mutually transferable. The ratio E ratio E / I I is known as transfer resistance or impedance (Z) in AC networks. An other way of stating the above theorem is that the receiving point and sending point in a network are interchangeable.

3

When applying reciprocity theorem for a voltage source, following steps must be taken:• Voltage source is replaced by a short circuit in original location • Polarity of source in new location is such that the current direction in that branch remains unchanged. PROCEDURE:

Connect the circuit for T-Network as shown in circuit diagram. Vary voltage from 10 V to 30 V with the difference of 5 V and note the corresponding current I with the help of ammeter. Now interchange the positions of ammeter and power supply to obtain the reciprocal T-Network. Again repeat the same procedure and note current I’. Calculate the percentage error (%E). Connect the same resistances in PIE-Network and repeat the same procedure and calculate currents I, I’ and %E. OBSERVATIONS & CALCULATIONS:

T-NETWORK: Sr. No. 1 2 3 4

E (Volts) 10 15 20 25

I (Ampere)

I’ (Ampere)

%E

E (Volts) 10 15 20 25

I (Ampere)

I’ (Ampere)

%E

PIE-NETWORK: Sr. No. 1 2 3 4

CONCLUSION & COMMENTS: ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________

4

PRECAUTIONS: • • •

While dealing with electric circuits handle the apparatus carefully. Make sure the connections are tight. Observe the readings carefully.

5

EXPERIMENT # 2 OBJECTIVE: TO STUDY MAXIMUM POWER TRANSFER THEOREM APPARATUS:

Power source Resistors • Voltmeter • Connecting wires

1 2 1

• •

CIRCUIT DIAGRAM:

RTH

RL

10 V

SOU R C E

v

VOLTMETER

LOAD

THEORY:

“To obtain maximum external power from a source with an infinite internal resistance. The resistance of the load must be made same as that of the source.” R L = R Th Th

6

The theorem applied to the maximum power, and not to the maximum efficiency. If the resistance resistance of the load is made larger than the source, then efficiency efficiency is higher, since most of the power is generated in the load, but the overall power is lower since the total resistance goes up. If the internal impedance is made larger than the load then most of the power ends up being dissipated dissipated in the source, and although the total power dissipated dissipated is higher, due to a lower circuit resistance, it turns out that the amount dissipated in the load is reduced. The condition of maximum power transfer does not result in maximum efficiency efficiency.. If we define the efficiency η as the ratio of power dissipated by the load to power developed by the source, then

Consider three particular cases: • • •

If If If

, then or , then

, then

The efficiency is only 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity (or the source resistance approaches zero), though the total power level tends towards zero. When the load resistance is zero, all the power is consumed inside the source (the power dissipated in a short circuit is zero) so the efficiency is zero. PROCEDURE:

Connect source voltage, thevenin resistance, load resistance and voltmeter as shown in circuit diagram. Vary the load resistance from 100Ω to 1000Ω in steps of 100Ω . Find VRL, PL, PS, I andη . Note the readings in the table. Now according to theorem to obtain maximum power transfer from source to load, vary load resistance such that it is equal to the thevenin resistance.

7

OBSERVATIONS AND CALCULATIONS:

Sr. No. 1 2 3 4 5 6 7 8 9 10

R L

VRL

(Ohms) 10 0 20 0 30 0 40 0 50 0 60 0 70 0 80 0 90 0 100 0

(Volts)

PL = V2RL/ R L (Watts)

PS = EThI (Watts)

I = ETh/( R Th Th + R L ) (Ampere)

η = R L x 100 /

( R Th Th + R L)

GRAPHS:

Draw the following graphs and attach here: • • • • •

PL vs R L PS vs R L VRL vs R L IL vs R L η vs R L

CONCLUSION & COMMENTS: ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ PRECAUTIONS: 8

• • •


EXPERIMENT # 3 OBJECTIVE: TO STUDY STAR DELTA TRANSFORMATION TRANSFORMATION APPARATUS:

Power Source Ammeter Resistors Connecting Wires

1 1 3

CIRCUIT DIAGRAM:

A N3 V Ra

Rc

N1

Rb

N2

N3

A

R3 V R1

R2 N1

9

N2

THEORY:

“Star-delta “Star-delta transformation transformation is a mathematica mathematicall technique technique to simplify the analysis analysis of an electrical electrical network”

Delta Connections

Star Connections

The transformation is used to establish equivalence for networks with 3 terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by tran transfo sform rmin ingg the the impe impedan dance ces. s. For equi equiva vale lence nce,, the the impe impeda dance nce betw between een any any pair pair of terminals must be the same for both networks and hence the current through any pair of nodes must be same for both networks. networks. The equations given here are valid for real as well as complex impedances. Equations for transformation from Δ-Load to Y-load are as follows:

Equations for transformation from Y-load to Δ-load are as follows: follows:

10

PROCEDURE:

Connect the circuit for delta connection as shown in figure 1. Note the current flowing through R a. Repeat the same procedure and note the currents flowing through R b and R c respectively. Note these values in the table. Now connect the circuit for star connection as shown in figure 2. Using same procedure as described above, note the currents flowing through R 1, R 2 and R 3Z 3Z respectively. Note these values in the table. The current through any pair of nodes must be same for both networks. OBSERVATIONS AND CALCULATIONS:

DELTA NETWORK: Sr. No.

Voltage

Ia

I b

Ic

(mA)

(mA)

(mA)

1 2 3

(volts) 5 10 15

Sr. No.

Voltage

I1

I2

I3

(mA)

(mA)

(mA)

1 2 3

(volts) 5 10 15

STAR NETWORK:

CONCLUSION & COMMENTS:

______________________________________________________________________________ ______________________________________________________________________________

11

______________________________________________________________________________ ______________________________________________________________________________

PRECAUTIONS: • • •


12

EXPERIMENT # 4 OBJECTIVE: TO STUDY THE CHARACTERISTICS OF RC SERIES CIRUIT APPARATUS: •

• • • • •

Power Source Capacitor Resistor Square wave generator Oscilloscope Connecting wires

1 1 1 1 1

CIRCUIT DIAGRAM: 1k

SQUARE WAVE

0.1µ F

CRO v

GENERATOR

THEORY:

To understand the behavior of the circuit, suppose the function generator output has been at ground for a long time so the capacitor is uncharged. Then when the function generator changes to it high state, the entire voltage V f will appear across the resistor (because the capacitor is uncharged) and a current Vf /R /R will will flow. But as this charge charge flows it will will accumulate accumulate on the 13

capacitor, a voltage drop V(t) will build up across the capacitor and the amount of current flowin flowingg will will decrease decrease (becaus (becausee the voltage voltage across across the resisto resistor, r, Vf - V(t), V(t), will will decreas decrease). e). Eventually all current will stop flowing when V(t) reaches V f . Mathematically the behavior of the voltage across the leads of the capacitor is given by: V(t) = Vf (1(1- e-t/τ )

while charging,

where τ = RC, the characteristic time constant and e is the base of the natural logarithms, e = 2.71828... 2.71828... . It is called called the characteristic characteristic time because it determines determines (characterizes) (characterizes) how quickly quickly or slowly the RC circuit responds. When the function generator changes back to its low state (ground), the voltage across the capacitor will exponentially decrease as the positive charge on the one plate flows backward through the resistor to neutralize neutralize the negative charge on the the other plate. Mathematically, V(t) = Vf (e-t/τ)

while discharging.

When discharging at t = τ , V(τ ) = Vf /e. /e. That is, V is about one third of the initial voltage (1/e (1/e = 0.368). During the charging process the time constant is the time it takes for the capacitor to charge to about two thirds of its final value. PROCEDURE:

Arrange the apparatus according to the circuit using 0.1µF capacitance and 1kΩ resistance. Set frequ frequenc encyy of square square wave wave genera generato torr at 1000 1000 Hz Obser Observe ve the the oscil oscillo losco scope pe grap graphh show showin ingg chargi charging ng and dischar dischargin gingg charact characteri eristi stics cs of capcit capcitor. or. The rising rising curve curve shows shows chargin chargingg of capacitor and the drooping one is for discharging. Now from the blocks on the screen count the blocks of charging graph and multiply it with the pre-set no. of divsions on voltage mulitiplier. mulitiplier. Find the time for voltage to rise to to 63 % of its maximum value. This This time is the time constant for the circuit compare it with the thoeraticaly thoeraticaly calculated time constant. At t=0 ic is maximum

14

IC

Tim ime

c V

CHARGING

DISCHARGING

TIME

To measure time constant shift the axis to the value of voltage obtained.

15

Time constant value

c V

TIME


No. Of divisions = Voltage = Calculation for charging:

63% of voltage = Calculation for discharging:

36.8% of voltage = Time constants:

Theoratical Time constant = R*C = Observed time constant (from graph) = CONCLUSION & COMMENTS:

16

____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ PRECAUTIONS:

Read graph carefully from oscilloscope. • Make sure the connections are tight. • Observe the readings carefully. •

EXPERIMENT # 5 OBJECTIVE: TO STUDY THE CHARACTERISTICS OF RL SERIES CIRUIT APPARATUS: • •

• • • •

Power Source Inductor Resistor Square wave generator Oscilloscope Connecting wires

1 1 1 1 1

17

CIRCUIT DIAGRAM: CR O V 1k Function Generator

100 mH

THEORY:

To study the response of an RL series circuit, consider an inductor (i.e., a coil with an inductance L) in series with a battery of emf E and a resistor of resistance R. This is known as an RL circuit. There are some similarities between the RL circuit and the RC circuit, and some important differences. First consider what happens with the resistor and the battery. When the switch is closed we have a current; when the switch is opened again we have no current. Now add an inductor to the circuit. When we close the switch now the current tries to jump up to the same value we had with the resistor but the inductor opposes this because a change in current means a change in flux for the coil. If the inductor adds negligible resistance to the circuit the current eventually reaches the same value it had with the resistor but the current follows an exponential curve to get there. An RL Circuit with a Battery:

To see why, apply Kirchoff's loop rule. With the switch closed we have: E - IR - L dI/dt = 0 Solving this for current gives I(t) = Io [1 - e-t/τ ] where Io = E/R is the maximum current and the time constant τ = L/R The potential difference across the resistor has a similar form: VR = E [1 - e-t/τ] The potential difference across the inductor is:

18

VL = E e-t/τ The graph of current as a function of time in the RL circuit has the same form as the graph of the capacitor voltage as a function of time in the RC circuit, while the graph of the inductor voltage as a function of time in the RL circuit has the same form as the graph of current vs. time in the RC circuit. If we take our previous circuit, wait for a while for the current to level off, and then open the switch so the battery is no longer in the circuit, what happens. If the circuit had just a resistor the current would drop immediately to zero. With the inductor the change in current means a change in magnetic flux so the inductor opposes the change. The current does eventually reach zero but it takes some time to get there.

An RL Circuit without a Battery:

To see why, apply Kirchoff's loop rule. With the switch opened we have: IR + L dI/dt = 0 dI/dt = -IR/L Solving this for current gives I(t) = Io e-t/τ The potential difference across the resistor has a similar form: VR = E e-t/τ The potential potential difference difference across the inductor is negative negative because it's acting like a battery battery hooked up in the opposite direction: VL = -E e-t/τ Once again, the graph of current as a function function of time in the RL circuit has the same form as the graph of the capacitor voltage as a function of time in the discharging RC circuit, while the graph of the inductor voltage voltage as a function of time in the RL circuit has the same form as the graph of current vs. time in the discharging RC circuit. PROCEDURE:

Connect the circuit according to the circuit diagram. Set frequency of square wave generator. Observe the oscilloscope graph showing voltage across resistor. Now from the blocks on the screen count the blocks of rising graph and multiply it with the pre-set no. of divsions on voltage

19

mulitiplier. Find the time for voltage to rise to 63 % of its maximum value. This time is the time constant for the circuit compare it with with the thoeraticaly calculated calculated time constant.

VL

Tim ime

20


No. Of divisions = Voltage = Calculation For Rising Graph:

63% of voltage = Calculation For Decaying Graph:

36.8% of voltage = Time constants:

Theoratical Time constant = L/R = Observed time constant (from graph) = CONCLUSION & COMMENTS:

____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ PRECAUTIONS:

Read graph carefully from oscilloscope. • Make sure the connections are tight. • Observe the readings carefully. •

EXPERIMENT # 6 OBJECTIVE: FIND THE RESONANCE FREQUENCY IN RLC SERIES CIRCUIT APPARATUS: • • • • • •

Power Source Function generator Resistor Capacitor Inductor Oscilloscope

1 1 1 1 1 1 21

•

Connecting Wires

CIRCUIT DIAGRAM: CRO CR O L

R A

10Ω

100mH

Function Generator

0.1µF

C

THEORY:

A resonant circuit, also called a tuned circuit consists of an inductor and a capacitor together with a voltage or current source. It is one of the most important circuits used in electronics. For example, a resonant circuit, in one of its many forms, allows us to select a desired radio or television signal from the vast number of signals that are around us at any time. A network is in resonance when the voltage and current at the network input terminals are in phase and the input impedance of the network is purely resistive. The frequency at which the reactances reactances of the inductance inductance and the capacitance capacitance cancel each other is the resonant frequency frequency (or the unity power factor frequency) of this circuit. The voltage across the inductor leads the current through the inductor by 90o 90o, while the voltage across the capacitor lags the current by 90o 90o. The two voltages tend to cancel each other, and actually do cancel at the resonant frequency ω0, as defined by

We then solve for the resonant frequency in radians per second as

The resonant frequency in Hertz is

22

The series impedance, expressed as a phasor, is

where

and

At resonance, Z = R, a real number, so the input current is in phase with the input voltage. The bandwidth of the series circuit is defined as the range of frequencies in which the amplitude of the current is equal to or greater than 1 /√ 1 /√22 times the maximum amplitude. This current produces produces a heating heating effect of half the maximum value, so the frequencies frequencies at the bandwidth bandwidth limits are called the half-power frequencies ω1 and ω2. With a little algebra, the bandwidth is shown to be The maximum value of VR occurs at

It is a plot of the magnitude of output current of a resonance circuit as function of frequency. The response of course starts at zero, reaches a maximum value in the vicinity of the natural resonant frequency, and then drops again to zero as ω becomes infinite. The frequency response is shown in figure. Frequency Response:

23

Frequency response of series RLC circuit

PROCEDURE :

Connect the circuit as shown in circuit diagram. Apply different frequencies to the series RLC circuit using function generator. Calculate the values of capacitive reactance, inductive reactance and total impedance ‘Z’ and note these values in table. Also note the current flowing through the circuit for each frequency applied. The frequency at which the current through the circuit is maximum and the total impedance of the circuit is minimum will be the resonance frequency. OBSERVATIONS AND CALCULATIONS:

Sr. No.

f (Hz)

I (A)

XL=2πfL

1 2 3 4 5 Resonance frequency in Hz = f = Resonance frequency in rad/sec = ω = 2πf =


24

XC=1/(2πfC)

Z=√(R2 +(XL-XC)2)

____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ PRECAUTIONS:

Before applying the voltage to the elements of R, L and C the maximum ratios must be known. • Before switching on the circuit, get it checked by the Instructor. • Switch must be closed only when readings are to be taken. •

EXPERIMENT # 7

25

OBJECT OBJECTIVE IVE:: FIND FIND TH THE E RESON RESONAN ANCE CE FREQUE FREQUENC NCY Y IN RLC SERIE SERIES S PARALLEL CIRCUIT (TANK CIRCUIT) APPARATUS: • • • • • •

Power Source Function generator Resistor Capacitor Inductor Connecting Wires

1 1 1 1 1

CIRCUIT DIAGRAM: A

Function Generator

L=100mH C=0.1µF R=10 ohms

THEORY:

The tank circuit, a common building block in electronic systems, is a parallel resonant circuit compri comprised sed of an induct inductor, or, a capacit capacitor, or, and an option optional al resistor resistor.. Since Since the capacitor capacitor and the inductor both store energy, this type of circuit is referred to as a tank circuit. An LC circuit can store electrical energy vibrating at its resonant frequency. frequency. A capacitor stores energy in the electric field between field between its plates, depending on the voltage across it, and an inductor stores energy in its magnetic field, field, depending on the current through it. If a charged capacitor is connected across an inductor, charge will start to flow through the inductor, building up a magnetic field around it, and reducing the voltage on the capacitor. Eventually all the charge on the capacitor will be gone. However, the current will continue, because inductors resist changes in current, and energy will be extracted from the magnetic field to keep it flowing. The current will begin to charge the capacitor with a voltage of opposite polarity to its original charge. When the magnetic field is completely dissipated the current will stop and the charge will again be stored in the capacitor (with the opposite polarity). Then the cycle will begin again, with the current flowing in the opposite direction through the inductor. The charge flows back and forth between the plates of the capacitor, through the inductor. The energy oscillates back and forth between the capacitor and the inductor until (if not replenished by power from an external circuit) internal resistance makes the oscillations die out. Its action, known mathematically as a

26

harmonic oscillator , is similar to a pendulum swinging back and forth, or water sloshing back and forth in a tank. For this reason the circuit is also called a tank circuit. The oscillations are very fast, typically hundreds to billions of times per second.

A condition of resonance will be experienced in a tank circuit when the reactances of the capacitor and inductor are equal to each other. Because inductive reactance increases with increasing frequency and capacitive reactance decreases with increasing frequency, there will only be one frequency where these two reactances will be equal. What happens at resonance is quite interesting. With capacitive and inductive reactances equal to each other, the total impedance increases to infinity, meaning that the tank circuit draws no current from the AC power source. Suppose the resonance impedances are 100Ω . Now use the parallel impedance formula to see what happens to total Z:

The result approaches a value of infinity as the two parallel impedances get closer to each other. What this means in practical terms is that, the total impedance of a tank circuit is infinite (behaving as an open circuit) at resonance. At f r r, line current is minimum. Total impedance is maximum. In this state circuit is called rejector circuit. • •

Below f r r, circuit is inductive. Above f r r, circuit is capacitive.

Resonance frequency :

To calculate resonance frequency considers the following vector diagram taking voltage as reference.

27

I2

Z1 XL IT R I1sin I1 I 1cos

2

Z 1 =

I 1 =

R + X L

V

V

=

Z 1

2

2

R + X L

I 2

2

V

=

X c

At resonance: I 1 sin sin φ = I 2

V

×

Z 1

X L Z 1

X L Z 1

X L X c ω L ω C L C

= R

= R

ω 2 L2 2 ω

2 ω

=

=

2

=

X c

1

=

2

V

=

X c = Z 1

2

2

2

2

+ ω L 2

2

− R

2

+ ω L

L C

L

−

2

CL 1

LC

−

R

2

2

L 2 R L2

If f r r is the resonance frequency and ω = 2∏ f r r (2Π f r )

2

=

1

LC

28

−

R

2

2

L

2 r

2

4Π f 2

=

f r

=

f r

1

=

−

R 2

LC L2 1 1 R 2 ( − ) 4Π2 LC L2 1

1

2Π LC

−

R 2 L2

Total Current at Resonance : At resonance active component of I1 = IT I T I T

I T L C

= I 1 cos φ =

=

V Z 1

×

R Z 1

VR R

2

+ ω L

= R

2

+ ω L

I T

I T

2

2

2

2

VR =

L C V

=

L CR

PROCEDURE:

Connect the circuit as shown in circuit diagram. Apply different different frequencies to the circuit circuit using function function generator. generator. Calculate Calculate the values of capacitive capacitive reactance, inductive inductive reactance, reactance, IC, IL, total current and total impedance ‘Z’ and note these values in table. The frequency at which the current through the circuit is minimum and the total impedance of the circuit is maximum will be the resonance frequency. OBSERVATIONS AND CALCULATIONS:

Sr. No.

f (KHz)

IC=V/XC (A)

XL=2πfL (Ω )

XC=1/(2πfC) (Ω )

IL-IC (A)

IC-IL (A)

IT (A)

Z=V/IT (Ω )

1 2 3 4 Resonance frequency in Hz = f = Resonance frequency in rad/sec = ω = 2πf = CONCLUSION & COMMENTS:

____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ 29

____________________________ __________________________________________ _____________________________ ________________________ _________ ____________________________ __________________________________________ _____________________________ ________________________ _________ PRECAUTIONS:


EXPERIMENT # 8 OBJECTIVE: TO OBSERVE DAMPING IN RLC SERIES CIRCUIT APPARATUS:

30

• • • • • • •

Power Source Function generator Resistor Capacitor Inductor Oscilloscope Connecting Wires

1 1 1 1 1 1

CIRCUIT DIAGRAM: CRO

R

L C

Function Generator

THEORY:

Resistor-induc Resistor-inductor-ca tor-capacito pacitorr (RLC) circuits are second-order second-order circuits. The response of the RLC circuit can be determined using basic circuit analysis. By KVL, V in

=

V C

V L

+

V R

+

.

(1)

The current i through the inductor is the same as the current through the capacitor. The current i through the capacitor is related to the capacitor voltage by i = C

dV C dt

.

(2)

The voltage drop across the resistor is V R

= iR = RC

dV C

(3)

dt

and the voltage drop across the inductor is given by

31

dV  d V d = L  V L = L  C C  = LC 2C dt dt  dt  dt 2

di

.

(4)

Substituting Equations (3) and (4) into (1) yields 2

LC

d V C 2

dt

+ RC

dV C dt

+ V C = V in .

(5)

Dividing (5) by LC yields d 2V C dt 2

+

R dV C L dt

+

1

=

V C LC

V in LC

.

(6)

The coefficients in (6) are equated to parameters given by α =

R

ω 0 =

2 L

1

.

LC LC

(7 a & b)

The parameter α is the damping coefficient and the parameter ω0 is the undamped resonant frequency. By substitution, 2

d V C 2

dt

+ 2α ⋅

dV C dt

2

2

+ ω 0 ⋅ V C = ω 0 ⋅ V in

.

(8)

The complementary solution of (8) takes the form ( s

2

+

2

2α ⋅ s + ω 0 ) ⋅ K est

(9)

0

=

From which we find the characteristic equation 2

(10)

( s 2 + 2α ⋅ s + ω 0 ) = 0

Solve for the roots, s1 and s2, of the characteristic equation:

s=

R L

2

±

R

2

L 2

-

=

s1, 2

− a1 2a2

4 LC

=-

R 2L

2

±

R

2

4L

-

1 LC

As: α =

R 2L

ωo =

1

then

LC

We define the damping ratio parameter to be 32

s = -α ±

α

2

2

- ω0

±

(

a1 2a2

)2

− a0

a2

ζ =

α ω 0

=

R

C

2

L

(11)

Following are the different types of responses obtained by changing value of ζ Case 1: The Overdamped Response (ζ > 1, α > ω0,

R > R CR CR )

The roots are negative, real and distinct: (a1/2a2)2 > (a0/ a2) The solution is simply

where

xn(t)= A1es1t + A2es2t Case 2: The Critically Damped Response

s1, 2

=

− a1 2a2

±

(

a1 2a2

)2

−

a0 a2

(ζ = 1, α = ω0, R = R CR CR )

The roots are negative, real and indistinct: (a1/2a2)2 = (a0/ a2) So for this case the solution is:

where

xn (t) = (A1 + A2t) est Case 3: The Underdamped Response

s =

− a1

2a 2

(ζ < 1, α < ω0, R < R CR CR )

The roots are distinct and complex: (a1/2a2)2 < (a0/ a2) The solution is simply: xn(t) = est (A1 cos

dt + A2sin

dt )

where s

=

− a1

2a2

and ω d =

Case 4: The Undamped Response (ζ = 0, α = 0, R = 0)

The roots are purely imagenary. The solution is simply:

xn(f) = A1cosω dt + A2sinω dt

33

where

s1,2 = ±

−

a0 a2

ao a2

−(

a1 2a2

)

2

PROCEDURE:

Connect the circuit as shown in circuit diagram. Suppose resonance frequency be 5KHz and inductance be 100mH. Find capacitance using following relation: f r

=

1 2Π

LC LC

Calculate R CR CR using following equation: RCR

=2

L C

Now vary R and observe the type of response on oscilloscope oscilloscope.. Following Following types of response will be observed depending on R. • • • •

When R > R CR CR , response will be overdamped. When R = R CR CR , response will be critically damped. When R < R CR CR , response will be underdamped. When R = 0, response will be undamped. In this case observe response across L or C.


fr = L= C= R CR CR = Sr. No.

R (Ω )

Type of Response

1 2 3 4 CONCLUSION & COMMENTS:

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EXPERIMENT # 9 OBJECTIVE: POWER FACTOR IMPROVEMENT APPARATUS: •

Power Source

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• • • • • • •

Resistor Capacitor Inductor Wattmeter Voltmeter Ammeter Connecting Wires

1 1 1 1 1 1

CIRCUIT DIAGRAM: L R C.C

A

A A P.C

C

V

THEORY:

The power factor of an AC electric power system is defined as the ratio of the real power flowing power flowing to the load to the apparent apparent power , and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power Real power is is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power. In an electric power system, a load with low power factor draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor. Linear loads with low power factor (such as induction motors) motors) can be corrected with a passive network of capacitors of capacitors or inductors or inductors.. The devices for correction of power factor may be at a central substation, or spread out over a distribution system, or built into power-consuming equipment. Circuits containing purely resistive heating elements (filament lamps, strip heaters, cooking stoves, etc.) have a power factor of 1.0. Circuits containing inductive or capacitive elements (electric motors, solenoid valves, lamp ballasts, ballasts, and others ) often have a power factor below 1.0.

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AC power flow has the three components: real power (P), measured in watts (W); apparent power (S), measured in volt-amperes (VA); and reactive power (Q), measured in reactive voltamperes (var). The power factor is defined as: . In the case of a perfectly sinusoidal waveform, waveform, P, Q and S can be expressed as vectors that form a vector triangle vector triangle such that:

If φ is the phase the phase angle between the current and voltage, then the power factor is equal to and:

,

Since the units are consistent, the power factor is by definition a dimensionless number between number between 0 and 1. When power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source on each cycle. When the power factor is 1, all the energy supplied by the source is consumed by the load. Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle. It is often desirable to adjust the power factor of a system to near 1.0. This power factor correction (PFC) is achieved by switching in or out banks of inductors or capacitors. For example the inductive effect of motor loads may be offset by locally connected capacitors. When reactive elements supply or absorb reactive power near the load, the apparent power is reduced. Power factor correction:

Power factor correction may be applied by an electrical power transmission utility to improve the stability and efficiency of the transmission network. Correction equipment may be installed by individual electrical customers to reduce the costs charged to them by their electricity supplier. A high power factor is generally desirable in a transmission system to reduce transmission losses and improve voltage regulation at the load. Power factor correction correction brings the power factor of an AC power circuit closer to 1 by supplying supplying reactive power of opposite sign, adding capacitors or inductors which act to cancel the inductive or capacitive effects of the load, respectively. For example, the inductive effect of motor loads

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may be offset by locally connected capacitors. If a load had a capacitive value, inductors (also known as reactors) reactors) are connected to correct the power factor. In the electricity industry, inductors are said to consume reactive power and capacitors are said to supply to supply it, even though the reactive power is actually just moving back and forth on each AC cycle. PROCEDURE:

Connect the circuit as shown in circuit diagram. Close switch 1 and open switch 2. Note the readings in the table. Now close switch 2 and vary the capacitance. Note corresponding readings and observe the effect of capacitor capacitor on power factor. IC increases with capacitance and as a result power factor is improved. OBSERVATIONS AND CALCULATIONS:

Sr. No.

V (Volts)

I (A)

W (Watt)

cosΦ=W/VI

IC (A)

IL (A)

1 2 3 4 5


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NA Lab Manual

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