to verify kirchhoff’slaws ford.c. circuits engineeering eep151... · sr no. apparatus range...

Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

Department of Electrical Engineering,

Shri Ramdeobaba College of Engineering and Management, Nagpur 440013 INDIA

EXPERIMENT NO: 1 RHS

Aim:To verify Kirchhoff’slaws forD.C. Circuits

Sr No. Apparatus Make Range /rating Make

1 D.C Voltage source

2 Rheostat

3 Ammeters(DC)

4 Voltmeter(DC)

1. Theory:Kirchhoff’s laws

The two laws given by Gustav Robert Kirchhoff (1824-1887) form the fundamental principals used

in writing circuit equations. These laws relate to the topology (i.e. the way the circuit elements are

connected) of the circuit. The laws do not depend on the nature of the elements of the circuit.

A) Kirchhoff’s Current Law (KCL): First law of Kirchoff’s

It is also known as Kirchhoff’s first law.

It states that the algebraic sum of currents meeting at a junction in a circuit is zero. If there are k

no of branches meeting at a junction (also called a node), then

∑ 𝐼𝑗𝑘

𝑗=1=0

Note: this law is just a restatement of principal of conservation of charge. Since charges can not

accumulate at a junction , the amount of charge entering at a instant must be same as the amount of

charge leaving it

Explanation of KCL:

Suppose some conductors are meeting at a point/node “A” as shown in fig 1.a. In some conductors,

currents are incoming to the point “A” while in other conductors, currents are leaving or outgoing

from point “A”.Current entering the node shall be considered to be positive (+) whereas current

leaving the node shall be considered as negative(-)”.

then with respect to fig-1a

I1 + (-I2) + (-I3) + (-I4) + I5 = 0 OR I1 + I5 -I2 -I3 -I4 = 0 OR I1 + I5 = I2 + I3 + I4




i.e.

Incoming or Entering Currents = leaving or Outgoing Currents

OrΣI Entering = ΣI Leaving

For instance, 8A is coming towards a point and 5A plus 3A are leaving that point in fig 1.b,

therefore,

8A = 5A + 3A

8A = 8A.

B)Kirchhoff’s Voltage Law(KVL): Second law of Kirchhoff’s

It states that at any instant the algebraic sum of voltages around a closed loop or circuit is zero.

For a closed loop having k elements = ∑ Vj

𝑘

𝑗=1= 0

In other words it can be defined as in a closed loop/closed path (or circuit) in a network, the

algebraic sum of the IR product is equal to the EMF in that path.

Explanation of KVL:Sign Conventions a) Battery e.m.f.: A rise in voltage should be given a + ve sign and a fall in voltage a –ve sign. Keeping this in

mind, it is clear that as we move from negative terminal of source to positive terminal, there is a rise

in potential, hence voltage E1 should be given a +ve sign. If, on the other hand, we move from +ve

terminal to –ve terminal of voltage source, then there is a fall in potential, hence it is to be

considered as –ve therefore E2 is given a negative sign.

b) Sign of IR Drop: - Whenever we move in the direction of current there is a drop in voltage. Since the current

always flows from point at higher potential to the point at lower potential. Hence Voltage drop in the

current direction is taken as –ve. However, if we go in a direction opposite to that of the current, then

there is a rise in voltage.Hence Voltage drop in the opposite current direction is taken as +ve.

Consider a closed circuit is shown in fig below which contains two emf sources E1and E2. The

overall sum of E.M.F’s of the batteries is indicated by E1-E2. The assumed direction of current is

also shown in the fig.E1 drive the current in such a direction which is supposed to be positive while

E2 interfere in the direction of current (i.e. it is in the opposite direction of the assumed direction of

https://www.electricaltechnology.org/wp-content/uploads/2015/07/Demonstrating-Kirchhoffs-Current-Law-KCL.png




current) hence, it is taken as negative. The voltage drop in this closed circuit depends on the product

of resistance and Current.

In the above fig, I1R1 and I2R2 is positive voltage drop and I3R3 and I4R4 are negative V.D. If we go

around the closed circuit (or each mesh), and multiply the resistance of the conductor and the

flowing current in it, then the sum of the IR is equal to the sum of the applied EMF sources

connected to the circuit.

The overall equation for the above circuit is:

E1-E2 = i1R1 + i2R2 – i3R3 – i4R4

OR E1-E2 - i1R1 - i2R2 + i3R3 + i4R4 = 0

2. Procedure: 1. Connect the circuit as per circuit diagram

2. Keep all the rheostats at maximum position.

3. Switch on the DC voltage supply, adjust the voltage to any suitable value(e.g.15V,20Vand 30V)

4. Take the reading of all the ammeters.

5. Measure the voltages across all the rheostats.

6. Change the voltage of power supply and repeat step (5) and (6).

7. Keeping the voltage constant, Change the position of rheostats and repeat step 4 and 5.

8. Verify Kirchhoff’s laws.

3. Result & Conclusion:

https://www.electricaltechnology.org/wp-content/uploads/2015/07/Kirchhoffs-Voltage-Law-KVL1.png




4. Discussion Questions:

1. What is the internal resistance of the ideal voltage source and ideal current source?

2. Define a node, a branch and a loop.

3. What are the limitations of Kirchhoff’s law?

4. For a given experimental set up, calculate the total power dissipated in the circuit and total

power supplied .Comment on the result.

5. Find the values of i2, i4and i5 if i1= 3A, i3= 1A and i6=1A

20 ohm

10 ohm

30 ohm

5 ohm

15 ohm

i1

i2

i3

i4i6

i5

6. Find the value of V if V1= 20 V and the value current source is 6 A.

R 2 ohm

++

-

10 ohm 5 ohm

-

V1 V2

V

Your roll no in ohms

100 V




LHS with Pencil

EXPERIMENT NO: 1

Aim:To verify the Kirchhoff’s law for the given DC network

Sr No. Apparatus Make Range /rating Make

1 D.C Voltage source

2 Rheostat

3 Ammeters

4 Voltmeter

1. Circuit diagram:

2. Observation table:

3. Calculations:

i. Verification of KCL and KVL

ii. Error calculation for current and voltage.

Sr.

No.

Source

Voltage

V

Voltage

Across R1

VR1

Voltage

AcrossR2

VR2

Voltage

Across R1

VR3

I1

I2

I3

1

2

3




EXPERIMENT NO :2 RHS

Aim: Verification of Kirchhoff’s laws to AC circuit(RLC series).

Sr No. Apparatus Range /rating Make

1 Single phase dimmer stat

2 Ammeters(AC)

3 Voltmeter(AC)

4 Rheostats

5 Inductor

6 Capacitors(6 No. in series)

7 Multifunction meter

1. Theory: Series RLC Circuit Analysis

Series RLC circuits consist of a resistance, a capacitance and an inductance connected in series

across an alternating supply. The three basic passive components of: Resistance, Inductance, and

Capacitance have very different phase relationships to each other when connected to a sinusoidal

alternating supply.

In a pure ohmic resistor the voltage waveforms are “in-phase” with the current. In a pure inductance

the voltage waveform “leads” the current by 90o, In a pure capacitance the voltage waveform “lags”

the current by 90o, This Phase Difference, Φ depends upon the reactive value of the components

being used.




Series RLC circuits are classed as second-order circuits because they contain two energy storage

elements, an inductance L and a capacitance C. The series RLC circuit above has a single loop with

the current (I)flowing through the loop is same for each circuit element. Then the individual voltage

drops across each circuit element is as under:

i(t) = Imax sin(ωt)

The voltage across a pure resistor, VR is “in-phase” with current

The voltage across a pure inductor, VL “leads” the current by 90o for pure inductor

The voltage across a pure capacitor, VC “lags” the current by 90o for pure capacitor

Therefore, VL and VC are 180o “out-of-phase” and in opposition to each other

As shown in figure below

From phasor diagram we can write the following equations for series R-L-C series circuit.

Where , V= VS= supply voltage.

1. Z is the total opposition offered to the flow of current by RLC series circuit and is known as

Impedance of the circuit.

https://circuitglobe.com/wp-content/uploads/2015/10/RLC-SERIES-CKT-EQ1-compressor.jpg

https://circuitglobe.com/wp-content/uploads/2015/10/RLC-SERIES-CKT-EQ2-compressor.jpg




2. XL is the inductive reactance = 2𝜋 f L,3. XC is the capacitive reactance =1

2𝜋𝑓𝐶

Phase Angle(∅)and Power factor:

Phase angle is the angle between voltage and current of the circuit.From phasor diagram, the value

of phase angle will be

Power Factor (Pf): Power factoris defined as the factor with which when apparent power is

multiplied gives the active power. It is given by the cosine of the angle between current and voltage

(phase angle)of the circuit.

∴ 𝑷𝒇 = 𝒄𝒐𝒔(∅)=𝑹

𝒁

The three cases of RLC Series Circuit:

When XL> XC, the phase angle ϕ is positive. The circuit behaves as a RL series circuit in

which the current lags behind the applied voltage and the power factor is lagging.

When XL< XC, the phase angle ϕ is negative, and the circuit acts as a series RC circuit in

which the current leads the voltage by 90 degrees.

When XL = XC, circuit is said to be under resonance. The phase angle ϕ is zero, as a result,

the circuit behaves like a purely resistive circuitand power factor of the circuit is unity.

Impedance Triangle of RLC Series Circuit:

When quantities of the voltage phasor diagram are divided by the common factor ‘I’, a right angle

triangle is obtained,known as impedance triangle. Fig.-1 represents the impedance triangle of theRL

series circuit, when (XL> XC) and fig.-2 represents the impedance triangle of RC series circuit,

https://circuitglobe.com/wp-content/uploads/2015/10/RLC-SERIES-CKT-EQ3jpg-compressor.jpg




when(XC > XL)

Fig-1. XL> XC Fig-2. XL< XC

If the inductive reactance is greater than the capacitive reactance then the circuit reactance is

inductive giving a lagging phase angle. When XL< XC circuit acts as a RC series circuit, and

impedance triangle is obtained as shown in fig-2. When the capacitive reactance is greater than the

inductive reactance the overall circuit reactance acts as a capacitive and the phase angle will be

leading.

Applications of RLC Series Circuit

The following are the application of the RLC circuit

It acts as a variable tuned circuit

It acts as a low pass, high pass, band pass, band stop filters depending upon the type of

frequency.

The Circuit also works as an oscillator

Voltage multiplier and pulse discharge circuit

2. Procedure:

1. Make the connections as per circuit diagram

2. Set the rheostat for maximum resistance.

3. Set the dimmerstat (variac) to zero output & switch on the mains.

4. Adjust the variac so as to apply a suitable voltage (between 100 V to 150 V) to the circuit,

measure the currentI & voltages VR, VL, Vc, VRL , supply voltage Vs , and power factor from

multifunction meter.

5. Take two sets of reading by applying different voltage for maximum position of rheostat.

6. Set the position of rheostat to middle and repeat the steps 4 and 5

7. Make the calculations as shown in calculation table .

Note: Draw phasor diagram as explained in procedure for two sets of readings. one for middle

position and one for maximum position of rheostat.

https://circuitglobe.com/wp-content/uploads/2015/10/IMPEDANCE-TRIANGLE-OF-RLC-SERIES-CIRCUIT-1-compressor.jpg

https://circuitglobe.com/wp-content/uploads/2015/10/IMPEDANCE-TRIANGLE-OF-RLC-SERIES-CIRCUIT-2-compressor.jpg




3. Result & conclusion:


1. Define the following terms w.r.t. alternating quantity.

i. RMS value. ii. Averagevalue . iii. Form factor and peak factor

2. Is KVL & KCL applicable to A.C circuit? State Kirchoff’s laws for AC circuits.

3. Define the term power factor in ac circuit

4. What do you understand by lagging , leading and unity power factor?

4. Define impedance of series RLC circuit? What is it’sunit?

5. Draw phasor diagram for R-L-C series circuit if i. XL = Xc, ii. XL>Xc& iiiXc>XL

6. What do you understand by resonance? What is the condition for series circuit?

7. If a sinusoidal alternating voltage is applied to a resistor (R), inductor (L) &capacitor (C)

separately, draw the wave form of currents (IR ,IL, &Ic ) each w.r.t. voltage as reference.

8. For the network shown in fig.1 calculate i. Total impedance ii. Supply current iii. Total Power

consumed v. Power factor and its nature

R= Your Roll No. in

ohms

200 V, 50 Hz supply

L = 1 H C = 2 F

Fig-1




LHS with Pencil

EXPERIMENT NO. 2

Aim: Verification of Kirchhoff’s laws to AC circuit (RLC series ).



2 Ammeters(AC)

3 Voltmeter(AC)

4 Rheostats

5 Inductor



1. Circuit diagram: RLC series circuit

A

Watt meter

P.f. meter

KWH

VR VL VC

VRL

1-ph

230V

AC

Supply




2. Observation table:RLC Series circuit

r-internal resistance of coil =--------Ω

3. Calculations Table : RLC Series circuit

Where,

∗ 𝑍 = √(𝑅 + 𝑟)2 + (𝑋𝐿 − 𝑋𝐶)2

Sr.

No.

Positionofrheostat

Vs

Volt

I

Amp

VR

Volt

VL

Volt

VRL

Volt

Vc

Volt

P.f.ofcircui

t

1

2

Middle position

1

2

Maximum position

Sr No. I

VR R

I

VZ L

L

22 rZX LL

Xc=

𝑉𝐶

𝐼

*Z

Z=𝑉𝑠

𝐼

from

calculation

s

=cos-1

)(Z

rR

from

phasor

diagram

from

p.f.meter

1

2

3

4




4. Sample calculations :

Show sample calculations for all parameters of calculation table

5. Procedure : Series circuit phasor diagram: Reference fig-1 Circuit equations are

i. VS= VRL + VC(phasor sum)

ii. VRL = VR + VL (phasorsum)

1. Choose suitable scale common for all voltages & voltage drops (e.g. 10 Volt = 1cm)

2. Take current I as reference vector & draw VR in phase with I (VR= OA)

3. Draw the arcs AB=VL& OB =VRL intersecting each other at point B

4. Join A with B & O with B.

5. Draw OD=VC at an angle 90 lagging to I.

6. Draw a vertically downwards line from B i.e. BC = VC= OD. Join OC=VS

7. Find the p.f. angle between VS& I.




EXPERIMENT No.: 3 RHS

Aim: Verification of Kirchhoff’s laws to AC circuit (RLC parallel).



2 Ammeters(AC)

3 Voltmeter(AC)

4 Rheostats

5 Inductor



1. Theory :Parallel RLC Circuit Analysis

In parallel circuit instead of the current being common to the circuit components, the applied

voltage is common to all branches /elements connected in parallel, and current through each

branch/element is different.Therefore, Kirchhoff’s current law is applicable to parallel ac circuits.

According to KCL, total current of the circuit is the phasor sum of individual branch currents. The

total impedance, Z of a parallel RLC circuit is calculated similar to that for a DC parallel circuit, the

difference is that for parallel circuit impedance is used instead of resistance.

Consider the parallel RLC circuit below.




In the above parallel RLC circuit, supply voltage, VS is common to all three components whereas the

supply current ISconsisting of three parts. The current flowing through resistor, IR, current flowing

through the inductor, IL and the current through capacitor, IC.The total current IS drawn from the

supply will not be the mathematical sum of the three individual branch currents but is the

phasor sum.RLC parallel circuit can be solved by using phasor /vector method where voltage vector

is considered as reference vector and the three current vectors drawn relative to it at their

corresponding angles.The total current ISis obtained by adding together all the vectors of individual

branches. Angle between V and ISis the circuits phase angle (∅). And cosine of phase angle is power

factor of the circuit.

In this experiment,

The series combination of resistor(R)& inductor(L) is connected in parallel withcapacitor(C)

as shown in the circuit diagram. The a.c.supply voltage of r.m.s. value VS is applied to this parallel

circuit. The source current IT will get divided into two branches as IRL& IC. let IRLbe current passing

through branches consisting of R&L in serieswhile IC be currentpassing through the capacitance

branch, then applying KCL we can write,

IT = IRL + IC (phasor sum)

Also,VR +VL = VRL= Vc = VS (applied voltage)

Here we assume that capacitance is pure,hence IC passing through C will lead VS by 90, while the

inductance is impure, therefore IRL passing through L will lag VS by an angle less than 90

Note that-

VR = IRL*R (drop across resistance)

VL = IRL*XL (drop across inductance)

Vc = IC*Xc (drop across capacitance)

Depending upon the impedance of the two branches the source current(Total current) ITwill lag or

lead the supplyvoltage by some angle which is called as the phase angle.

Applications of Parallel Circuit:

There are many applications of parallel circuit. Some of them are given below

1. In home, industries, factories, we have normally Parallel connection.

2. If we have to give different current to different devices which are attached then we

normallyprefer Parallel Circuit.

3. The electrical wiring to the power points in every household is in the form of Parallel

Circuits ,

4. The computer hardware is designed using Parallel Circuits.




2. Procedure:

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

2. Set the Dimmerstat (variac) to zero output.

3. Set the rheostat to maximum position

4. Switch on the supply.

5. Adjust the variac to give a suitable voltage(120 V to 150 V) to the circuit.

6. Note down the voltagesVR, VL, VC, VRL, currents IT ,IC, IRL& power factor of the circuit.

7. Take two sets of reading for maximum position with two different supply voltages.

8. Switch off the supply and set the position of rheostat to middle.

8. repeat steps 4,5,6 and 7

Note: Draw phasor diagrameachfor one set of reading.

3. Result & conclusion:


1. Define following terms with their units-

i. Admittance ii. Conductance iii. Susceptance

2. Define following terms-i. Active power(P)ii. Reactive power(Q)iii. Apparent power(S)

3. Two branches are connected in parallel. One branch consists of R-L seriescombination while other

branch consists of R-C series combination. Draw the phasor diagram showing all voltages& all

currents.

4. For the network shown in fig. below, calculate i. total impedance of the circuit ii. Total current

iii. Power consumed and v. Power factor and its nature.

200 V, 50 Hz supply

R= Your Roll No. in ohms

L = 0.8

H

C = 30 FR




LHS with Pencil EXPERIMENT NO. 3

Aim:Verification of Kirchhoff’s laws to AC circuit (RLC parallel)


1 Single phase dimmerstat

2 Ammeters(AC)

3 Voltmeter(AC)

4 Rheostats

5 Inductor



1. Circuit diagram: RLC parallel circuit




2. Observation table: RLC parallel circuit:

3. Calculations Table:RLC parallel circuit:-

4. Sample Calculations:-

Sr.

No

Position of

Rheostat

VS

volts

Power

factor

I

Amp

IRL

Amp

IC

Amp

VR

Volt

LV

Volt

VRL

Volt

VC

Volt

1

2

Maximum

3

4

Middle

Sr.

No. RL

R

I

VR

RL

LL

I

VX

C

c

cI

VX

eqZ =Z1| | Z2

𝒁𝒆𝒒=𝑹𝒆𝒒 ± 𝑿𝒆𝒒

from

calculations

=cos-1

)(eq

eq

Z

R

phasor

diagram

p.f.

meter

RL

R

I

VR ,

RL

LL

I

VX ,

C

cc

I

VX

LjXRZ 1 , and jXcZ 2

eqZ =Z1| | Z2 = 21

21

ZZ

ZZ

XcXjR

jXcjXR

L

L

= eqZ ----- ----Polar form

eqeqeq jXRZ -------------------Rectangular form




5. Procedure:Parallel circuitphasor diagram:-

1. Choose suitable scale common for voltages (say1cm=10V)& suitable scale common for all

currents(say,1cm =0.1 A)

2. Take OA =VS = VC = VRL as reference vector.

3. Since VS = VC = VRL =VR + VL, draw the arcs OB= VR from O & AB= VL from A intersecting

each other at point B. Join OB = VR & AB = VL.Show direction of each vectors

4. Draw OC= IRL in phase with VR& OD = IC at an angle 90 leading to VS.

5. Draw OE= I = IC + IRL by vector addition (completing the parallelogram).

6. Find the p.f. angle between VS& I as shown in figure below.




Experiment No: 4 RHS

Aim: To study speed control of D.C. shunts motor by: a) Armature voltage Control method.

b) Field current/flux control method.

/

1. Apparatus:

Name of Apparatus Range/Rating Make

Ammeter (DC)

Voltmeter (DC)

DC motor

Rheostats

Tachometer

2. Theory

i. Working principle of DC motor: A DC motor is an electrical machine which converts electrical energy into mechanical

energy. The working of DC motor is based on the principle that when a current carrying conductor

is placed in a magnetic field, it experiences a mechanical force. The direction of the mechanical

force is given by Fleming’s Left-hand Rule and its magnitude is given by F = BIL Newton.

According to Flemings left-hand rule when an electric current passes through a coil placed in a

magnetic field, the magnetic force produces a torque which turns the DC motor to rotate.. Fig-1

represents cross section DC machine

Fig-1 Cross-Section of a DC Machine




ii. Working of DC Motor:

Consider a part of a multipolar DC motor as shown in the figure-2. When the terminals of the motor

are connected to an external source of DC supply:

the field magnets are excited(with field current If) developing alternate North and South poles

the armature conductors carry currents(Ia)

Fig-2 Part of a Multi-polar DC Motor

All conductors under North-pole carry currents in one direction while all the conductors under

South-pole carry currents in the opposite direction. (applying right hand thumb rule)The armature

conductors under N-pole carry currents into the plane of the paper (denoted as ⊗ in the figure). And

the conductors under S-pole carry currents out of the plane of the paper (denoted as ⨀ in the figure).

Since each armature conductor is carrying current and is placed in the magnetic field, a mechanical

force acts on it.

On applying Fleming’s left-hand rule, it is clear that force on each conductor is tending to rotate

the armature in the anticlockwise direction. All these forces add together to produce a driving

torque which sets the armature rotates. When the conductor moves from one side of a brush to the

other, the current in that conductor is reversed. At the same time, it comes under the influence of the

next pole which is of opposite polarity. Consequently, the direction of the force on the conductor

remains the same.By reversing current in each conductor as it passes from one pole to another, it

helps to develop a continuous and unidirectional torque. And thus, motor starts rotating.

iii. Back EMF in DC Motor : When the current carrying conductor placed in a magnetic field, the torque induces on the

conductor. The torque rotates the conductor which cuts the flux of the magnetic field. According to

the Electromagnetic Induction Phenomenon “when the conductor cuts the magnetic field, EMF

induces in the conductor”. The Fleming right-hand rule determines the direction of the induced

EMF.On applying the right-hand rule it is obtained that the direction of the induced emf is

opposite to the applied voltage. Thereby the emf is known as the counter emf or back EMF. The

back emf is developed in series with the applied voltage, but opposite in direction, i.e., the back emf

opposes the current which causes it.

The magnitude of the back emf is given by the same expression shown below.




Where, Eb is the induced emf of the motor known as Back EMF, A is the number of parallel paths

through the armature between the brushes of opposite polarity. P is the number of poles, N is the

speed, Z is the total number of conductors in the armature and ϕ is the useful flux per pole.

The relation between the main supply, back emf and armature current is given as

Eb = V – IaRa.

iv. Significance of Back emf in DC Motor:

1. The back emf opposes the supply voltage. The supply voltage induces the current in the coil

which rotates the armature. The electrical work required by the motor for causing the current against

the back emf is converted into the mechanical energy. And that energy is induced in the armature of

the motor. Thus, we can say that energy conversion in DC motor is possible only because of the

back emf.The mechanical energy induced in the motor is the product of the back emf and the

armature current, i.e., EbIa.

2. The back emf makes the DC motor self-regulating machine, i.e., the back emf develops the

armature current according to the need of the motor. The armature current of the motor is

calculated as, 𝑰𝒂 =𝑽−𝑬𝒃

𝑹𝒂

3. Speed control of DC motor:

The speed of the D.C. shunt motor can be expressed by following mathematical equation,

𝑵 ∝𝑽 − 𝑰𝒂𝑹𝒂

𝝋

Where, V = Applied voltage(DC) in volts.

aI = Total armature current in Amp.

aR = Armature resistance in ohms.

= Flux per pole in Weber

N = Speed of motor

https://circuitglobe.com/wp-content/uploads/2015/11/back-emf-in-dc-motor-eq-compressor.jpg




From above expression the speed of DC motor can be varied by using any of following methods.

a) Armature voltage control.

b) Field current control

A) Armature voltage control method:

In this method the external resistance is connected in series with the armature winding & by

controlling the value of this external resistance speed of DC shunt motor can be controlled. The

expression of back emf is given as

AEb

60

ZNP PΦZN

60A

In above equation Z, N, P, A, is constant as they are the design parameters of motor, then

N ∝Eb∅

If flux of motor is constant, then

N Eb ,where Eb = V-Ia*Ra

Thus, keeping field flux constant, speed can be controlled, by controlling the armature voltage.

B) Field current control method:

In D.C. shunt motor flux ( ) is proportional to the current in the field winding (If)of motor. Since

𝑁 ∝𝐸𝑏∅

The speed is inversely proportional to the flux per pole. In field flux control method external

resistance is connected in series with shunt field winding to control the field winding current, which in

turn will control the field flux. If resistance is increased, the current in the shunt field winding will

decrease, therefore, the flux also decreases. Since the speed is inversely proportional to the flux, the

speed will increase. This method is used for obtaining speeds above normal speed.




4. Procedure:

a) Armature voltage control method 1. Connect the circuit as shown in fig.1

2. Set the rheostats in the armature circuit to maximum position and field rheostats in minimum

resistance position.

3. Switch on the D.C. supply and turn on DC motor by using three point starter.

4. Set field rheostat at suitable position by considering suitable value of field current. In this

method value of field current should be kept constant.

5. Measure the motor speed & Armature voltage.

6. Change the armature resistance in steps and repeat step 5

7. Plot the graph between Armature Voltage v/s Speed.

b) Field current(flux) Control Method:

1. Connect the circuit as shown in fig.1

2. Set the rheostats in the armature circuit to maximum position and fieldrheostats in minimum

resistance position.

3. Switch on the D.C. supply and turn on DC motor by using three-point starter.

4. Set armature rheostat at suitable position by considering suitable value of armature voltage. In

this method value of armature voltageshould be kept constant.

5. Measure the motor speed &field current.

6. Change value of field current in steps and repeat step no 5.

7. Plot the graph between field current v/sSpeed.

5. Precautions:

1. While doing connection of instruments, maintain polarity of measuring instruments. 2. Do not turn ON supply to the circuit,without prior permission of instructor.

3. Before turningon supply ensure the correct positions of both the rheostats

4. Before turning on supply conform that starter is at off position.

5. Do not touch the live part of the circuit.

6. While performing the experiment do not closely bend over instruments.




6. Result and conclusion:-

7. Discussion Questions:-

1) List out the types of dc motors? & draw it’scircuit diagram.

2) How the back EMF is generated in DC motor?

3) State the significance of back emf in dc motor.

4) At what condition, Field Control Method will be preferred? Why?

5) At what condition,Armature control method will be preferred? Why?

6) What are the types of starter used for starting DC motor?

7) What is the function of hold on coil & over load coil in three point starter?

8) While starting this experiment, why it is needed to set rheostat in the armature circuit to

maximum position and fieldrheostat in minimum resistance position.




LHS with Pencil

Experiment No:4

Aim: To study speed control of D.C. shunt motor by: a) Armature voltage Control method.

b) Field current control method.

Name of Apparatus Range/Rating Make

Ammeter (DC)

Voltmeter (DC)

DC motor Rheostats Tachometer 1

1. Circuit diagram:

Fig.1 Connection diagram of DC Shunt motor




2. Observation table:-

a) Armature voltage control method. Field current= -----------------------(amp)Constant

Sr. No. Armature Voltage

(volts)

Speed of Motor

(rpm)

1

2

3

4

5

6

7

8

9

10

b) Field current control method.

Armature Voltage =------------------volt(constant)

constant

Sr. No. Field current

(amp)

Speed of Motor

(rpm)

1

2

3

4

5

6

7

8

9

3. Plot the graphs- i) Armature voltage v/s Speed

ii) Field current v/s Speed

to verify kirchhoff’slaws ford.c. circuits engineeering eep151... · sr no. apparatus range...

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