Eee Lab Report 8

7/14/2019 Eee Lab Report 8

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Name of the Experiment:

STUDY OF TRANSIENT BEHAVIOR OF RL CIRCUIT

Course No. EEE 102

Experiment No. 8

Group No. 2

Mursalin Habib

Roll No. 9906114

Department: Electrical and Electronics Engineering

Level 1 Term 1

Session 1999-2000

Partners Roll No. 99061129906113

9906115

9906116

Date of Performance:

Date of Submission:

Bangladesh University of Engineering and

TechnologyObjective:


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To study Transient Response of RL circuit with step input is the objective ofthis experiment. In this experiment we applied a square wave input to an RLcircuit and observed the wave shapes and determine the time constant.

Theory:

The transient response is the temporary response that results from aswitching operation and disappears with time. The steady state response isthat which exists after a long time following any switching operation.

Let us consider an RL circuit shown in figure.

Storage Phase:

When the switch is connected to position 1, applying KVL we can write,

1................

dt

diLRiV +=

If the inductor is initially relaxed, the solution of equation 1 is,

2.....).........1( t

eR

Vi

=

Therefore the voltage across the resistor and inductor are given by

4......

3...).........1(

t

RL

t

R

VeVVV

eVV

==

=

Where RL/= and is called the time constant of the RL circuit. Equation 2,3& 4 are plotted below:


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It is seen from the curves that the voltage across the inductor falls from V tozero voltage exponentially. The current is zero at the start i.e. when the switchis just thrown to position 1, then it increase exponentially and finally reach toV/R amps when the inductor voltage becomes zero.

Decay Phase:

When the switch is connected to position 2 applying KVL we can write

5...........0dt

diLRi +=

The solution of equation 5 is,

6.............t

eR

Vi

=


8.........

7...........

t

C

t

R

VeV

VeV

=

=

Equation 6,7 & 8 are plotted below:

It is seen from the curves that the voltage across the inductor rises from V tozero volts exponentially .The current is maximum at the start i.e. when theswitch is just thrown to position 2, then it decrease exponentially and finallyceases to zero when the inductor voltage becomes zero.

Apparatus:

Resistance 470 Inductance 2.7 mH Oscilloscope & Chords Signal Generator & Chords

Wires Bread Board


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Experimental Setup:


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Procedure:

The circuit was setup as shown in figure 1. 14 kHz square wave was applied from generator The wave shapes were observed at Ch-1 and Ch-2 in dual mode and were

drawn.

Ch-1 & Ch-2 were disconnected & were reconnected as shown in figure 2. The wave shapes at Ch-1 & Ch-2 (INV.) in dual mode were observed &

were drawn.

Question & Answer:

1. Define inductor and inductance. Write the features of an inductor.What does inductance measure?

Inductors are coils of various dimensions designed to introduce specifiedamount of inductance.The ability of a coil to oppose change in current is a measure of theinductance of the coil. It is symbolize by the letter L &, is measured inhenry (H).

Features:

If the current is constant the voltage across the inductors (ideal) is zero.Thus the inductor behaves as a short circuit in the presence of a constantor DC current.

Current cannot change instantaneously in an inductor; that is the currentcannot change by a finite amount in zero time .For and indicator theequation of voltage current relationship.

dt

diLV =

It falls we that this change would require an infinite voltage and infinitevoltages are not possible.

Inductance is a consequence of a conductor linking a magnetic field.Inductance measures the ability of a coil to oppose change in current.

2. Deduce voltage-current relationship for an inductor .Why the currentin an inductor cannot change instantaneously.

The voltage current relation ship for an indicator is,

dt

diLV =


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We know that capacitor stores charge. Like capacitors inductors storevoltage of presents any change of the current in the circuit.

Again if we look at the voltage current relationship equation well see that itis impossible to change that current instantaneously, because if we put O

for dt.That is if dt=0Then =v Which is not possible.This is why the current in an inductor cannot change instantaneously.

3. Define time constant for an RL circuit. What is the significance oftime constant? How time constant can be determined?

The expression

+

= 0)( )(

0 teItitL

R

and

+

= 0,Re)()(

0 ttItvL

R

Include a term of the form .te LR

The coefficient of t namely, R/L determines the rate at which the current orvoltage approaches zero. The reciprocal of this ratio is the time constant ofthe circuit,

R

Ltconstime == tan

Using the time constant concept we write the expression for current,

voltage, power energy as

0),1(2

1

0,

0,Re)(

0,)(

22

0

22

0

0

0

=

=

=

=

+

+

teLIw

teIp

tItv

teIti

t

t

t

t

The time constant is an important parameter for first order circuit somentioning, several of its characteristic is worthwhile. First it is convenientto think of the time elapsed after switching in terms of integral multiples of .Thus one of constant after the inductors has began to release its

energy to the resistor , the current has been reduced to e-1 orapproximately 0.37 of its initial value .

Values of e- for t equal to integral multiple of

t 2 3 4 5 6 7 8 9 10

e-t/3.68x10-1

1.35x10-1

4.98x10-2

1.83x10-2

6.78x10-3

2.48x10-3

9.12x10-4

3.35x10-4

1.23x10-4

4.54x10-5


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When the elapsed time exceeds the time constant the current is less than1% of initial value . Thus we sometimes say that time constant afterswitching has occurred the current and voltage s have ,for most practicalpurposes reached their final value .

For single time constant circuits ( first order circuit ) & with 1% accuracythe phase a long time implies that five or more time constant have elapsed. Thus the existence of current in RL circuit is a momentary event andtherefore is also referred to as the transient response of the circuit .Theresponse that exists a long time after the switching has taken place iscalled the steady response . The phrase a longtime then also means thetime it takes the circuit to reach its steady state value .Any first order circuit is characterized in part by the value of its timeconstant .If we have no method for calculating the time constant of such acircuit, we can determine its value from a plot of circuits natural response.Thats because another important characteristic of the time constant is thatit gives the time required for the current to reach its final value of if the

current continues to change at its interval rate. To illustrate we evaluatedi/dt at 0+ and assume that the current continues to a change at this rate,

1..........)0( 00

II

L

R

dt

di==

+

Now, if I starts as I0 and decreases at a constant rate of I0/ amperes persecond, the expression for i becomes

2..........00 tI

Ii

=

Equation 2 indicates that i. would reach its final value of zero in second.

Fig shows how this graphics interpretation is useful in estimating the timeconstant off a circuit from a plot of its natural response. Such a plot couldbe generated on the oscilloscope measuring output current. Drawing thetangent to the natural response plot at t=0 and reading the value at whichthe tangent intersects the time axis gives the value of .


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4. Describe the storage and decay phase of an RL circuit bothqualitatively and quantitatively.

Storage Phase:

When the switch is connected to position 1 , applying KVL we can write ,

1...............dt

diLRiV +=

If the inductors are initially released , the solution of equation 1 is

2.....).........1( t

eR

Vi

=


4......

3...).........1(

t

RL

t

R

VeVVV

eVV

==

=

when = L/R and is called the time constant of the RL circuit. Equation 2,3and 4 plotted below :

It is seen from the curves that the voltage across the inductor falls from Vto zero volts exponentially. The current is zero at the start i.e. when the

switches is just thrown to position 1 , then it increases exponentially andfinally reach to V/R amps when the inductors voltage becomes zero .

Decay Phase:

When the switch is connected to position 2, applying KVL we can write

5...........0dt

diLRi +=

The solution of equation 5 is


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6.............t

eR

Vi

=

Therefore the voltage across the resistor and inductors are given ,

8.........

7...........

tC

t

R

VeV

VeV

=

=

Equation 6 , 7 & 8 are plotted below

It is seen in the curves that the voltage across the inductor rises from V tozero voltage exponentially. The current is maximum at the start i.e. whenthe switch is just thrown to position 2, then it decreases exponentially andfinally ceases to zero when the inductors voltage becomes zero.

Discussion:

This was our last experiment in L1 T1 of EEE 102. In this experiment weused oscilloscope & signal generator for the second time . At this time we

did not faced that much problem using these apparatus .We did the jobvery skillfully but not quickly enough. It yields a good graph & theobservation experiment was successfully in all sense.

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