Electrical Circuits

AN-NAJAH NATIONAL UNIVERSIT

AN-NAJAH NATIONAL UNIVERSITY

FACULTY OF ENGINEERING

ELECTRICAL ENGINEERING DEPARTMENT

ELECTRICAL CIRCUITS LAB

For Electrical students

Eng. Nasim Zayid

2010





Student name:………………………………………………………

Experiment #:………… Experiment Name:……………………………………………. Day :……………. Date: / / Instructor name: …………………………………………


EXPERIMENT # 1: Introduction to Circuits

EXPERIMENT # 2: Ohms Law & Resistors-Series and Parallel Connection

EXPERIMENT # 3: Network Theorems EXPERIMENT # 4: Voltage Source EXPERIMENT # 5: Characteristics in AC EXPERIMENT # 6: Capacitor in the A. C. Circuit EXPERIMENT # 7: Inductors in the A. C. Circuit EXPERIMENT # 8: RLC Series &Parallel EXPERIMENT # 9: Series Resonance EXPERIMENT # 10: Parallel Resonance EXPERIMENT # 11: Three-Phase Alternating Current EXPERIMENT # 12: Using Computer for Analyses



EXPERIMENT # 1

INTRODUCTION TO CIRCUITS Objective:

To practice the use of electrical and electronic engineering labs instruments, such as Digital Multi-meters , Oscilloscopes and the Training Electronic Boards ( prototype boards ) .

Apparatus and components:

1- Digital Multi-meter (DMM). 2- Oscilloscope (some times called “ Cathode Ray Oscilloscope or CRO) 3- Training boards and electronic Components.

Introduction:

Ι- Oscilloscope: An Oscilloscope can be used for monitoring parameters associated with signals

that vary with respect to time . also the same instrument can be used ( although to a lesser extent ) to measure DC signals.

ΙΙ- Digital Multimeter DMM:

This instrument can provide a measure of D.C . signals as well as the r.m.s values of AC signals . Such signals can be that of the output voltage or the currents passing through a particular component in a given circuit .

A DMM can also provides resistance , capacitance and Transistor D.C . amplification factor measurements.

ΙΙΙ- Training Electronic Board

Using this training board and employing suitable electronic components, then many electronic circuits may be realized e.g. Power supplies, A.C generators etc.

See fig-1 [The electronic board].

Experimental procedure: Ι- Using the Digital multi-meter:

Resistance measurement: 1- connect the circuit as shown in fig.2

2- Tabulate you are results as in table-1. 3- How you can read the value of resistance? (Color code)

Voltage measurement: 1- Connect the circuit as shown in fig.3

2-Tabulate you are results as in table-2 (voltage reading raw).

Current measurements:

1- Connect the circuit as shown in fig.4.

2- Tabulate yours results as in table-3 (current reading row).

AN-NAJAH NATIONAL UNIVERSITY FACULTY OF ENGINEERING


EXPERIMENT # 2 Eng.Nasim Zayid Mr.Imad khader

OHM’S LAW AND RESISTORS-SERIES AND PARALLEL CONNECTION

Objective

1- To investigate the various applications of ohm's law. 2- To investigates series and parallel resistive circuits.

Apparatus and components: 1- Training Electronic Board. 2- A DMM. 3- Resistors : 22 Ω , 100 Ω , 150 Ω, 220 Ω , 330 Ω , 470 Ω , 680 Ω, 1k Ω.

Theoretical aspects of Ohm’s law : Ohm's law defines the mathematical relation ships that exist between the Voltage

(v) across a particular electrical component in a circuit and the electrical current (I), passing through it

i.e. V ∝ I or. V = I ∗ R where R is the resistance of the component

This follows that :

RVI = ,

IVR =

If resistors are connected in series, then the same current will follow through them. The value of the current I, can be determined using the following relationships:

RVI

tot

= , where Rtot = R1 +R2 + .. + Rn

Also, the sum of all partial voltages equals the applied voltage. i.e.

V tot = VR1 +VR2 + …… + V Rn

If resistors are connected in parallel, they will all experience the same voltage. i.e. V = VR1 = VR2 = = VRN.

The total current equals to the sum of all partial currents flowing in every branch. i.e. Itot = IR1 + IR2 + +Irn

or Rn

V...R

V

R

V

Rtot

V+++=

21

This follows that Rn

...RRRtot

1

2

1

1

11+++=

Voltage Divider Voltage dividers (potentiometer) consist of two series resistors, R1 and R2

connected as shown, in the following diagram. In this case we may write the following equation:

RRR

V

V

221

2

+=

When a load resistor (R3) is connected in parallel with R2, the voltage ratios change, since current branching takes place

RRRR

R ),(eq3232

32 +∗

= -- since R2 in parallel with R3

R ,eq

R ,eqR

VV

32

321

3

+=

Experimental procedure: Ι- Ohm’s law

1- connects the circuit as shown in fig 2.

2- Measure the currents at resistors R = 100, 150, 330 Ω, respectively for voltage

values 0, 2, 4, 6, 8, 10 and12 volts, Tabulate your result as in Table-1. Plot the current against the voltage at each value of resistor. I = f (V) R constant

3- Set the DC supply voltage to 4 V, then complete the row in table - 2 , then change the DC supply to 8 and 12 voltage respectively then fill out the second and third rows in the same table . Plot the current against the resistance.

I = f(R) v constant

R (Ω) 100 150 220 330 470 680 1000 I (mA) at 4 V

I (mA) at 8 V

I (mA) at 12 V

Table - 2

ΙΙ- Series and parallel resistive connection

4- Connect the circuit as shown in fig - 3

5- Measure the voltage VR1, VR2, VR3 and the respective currents IR1, IR2, IR3

Calculate the equivalent resistance of R1, R2, R3, then find the relation between the total resistance and R1, R2, R3. VR1= , VR2= , VR3= IR1= , IR2= , IR3=

6- Connect the circuit as shown in fig - 4.

7- Measure the Voltage VR1, VR2, VR3, and the respective currents IR1, IR2, IR3 and

IRtot Calculate the equivalent resistor of R1, R2, R3, then find the relation ship between the resistances.

VR1= , VR2= , VR3= IR1= , IR2= , IR3= IRtot =


ELECTRICAL ENGINEERING DEPARTMENT Mr.Imad Khader Eng. Nasim Zayed

EXPERIMENT # 3 NETWORK THEOREMS

Objective :

1- To study the effect of more than one voltage source in a network. 2- To find a method of simplifying a network in order to obtain the current in one

particular branch of the network. 3- To study the parameters of the voltage source.

Introduction :

Kirchhoff’s law : Kirchhoff’s current law is stated as follows: The sum of the currents into a junction

(total current in ) is equal to the sum of the currents out of that junction ( total current out ) .

Ι =∑ 0 Kirchhoff’s voltage law is algebraic sum of all the voltages around a closed path is

zero or, in other words, the sum of the voltage drops equals the total source voltage . V =∑ 0

Superposition Theorems:

The superposition method is away to determine currents and voltages in a circuit that multiple source by taking on source at a time , the other sources are replaced by their internal resistance , the ideal voltage source has a zero resistance

Thevenin’s Theorem:

The Thevenin’s equivalent for any resistive circuit consists of an equivalent voltage source ( Vth ) and an equivalent resistance ( Rth ) . Vth is defined to be the open circuit voltage between two specified points in a circuit. Rth is the total resistance appearing between two specified points in a given circuit with all sources replaced by their internal resistance.

I thVthR R

=+

Experimental Procedure Ι - Kirchhoff’s law 1 - Connect the circuit as shown in fig - 1

* Note the polarity of the voltage and current. 2 - Adjust the applied voltage to be12 V. Measure the voltage across R1, R2, R3, R4,

R5, and the currents in each components. Tabulate your results in a table - 1. Apply the Kirchhoff’s law in your results.

3 - From the measured values of current and voltage in each branch calculate the value of the resistance R1 to R5.

4 - Using KCL, KVL, calculate the currents and voltages theoretically. ΙΙ - Superposition Theorem:

1- Connect the circuit as shown in fig - 2.

2- Measure the current in R1 to R5, not both the magnitude and the polarity of each

current and tabulate them. 3- Disconnect the 15 V sources, the circuit shown in fig - 3. Measure and tabulate the

current I’1, I’2, I’3.

4- Connect the circuit as shown in fig - 4, then measure the currents I’’1,I’’2, I’’3, make a table of all the currents. You should have found that the sum of the currents due to individual voltage source is equal to the current resulting when both sources are present in the network.

Questions Q1- Do the current directions agree with those in the fig? Q2- Can you notice any relation ship between I1, I’1, I”1? Q3- Does the same relation ship hold for I2, I’2, and I”2, also with I3, I’3, and I”3?

ΙΙΙ -Thevenin’s Theorem: 1 - Convert the circuit as shown in fig - 5.

2 - measure the current in the 680 Ωresistors. I680= 3- Remove the 680 Ωresistors, and measure the voltage between terminals X, Y,

Vth= .

4- Remove the source of voltage, the circuit as in fig -6. The resistance of this network

my be found by connecting a voltage source to points X, Y and measure the total current, measure the current for voltage of 2,4,6 and 8 V, then calculate the resistance using ohm’s law, and take the average of the values found (Rth), calculate the current through the 680 Ω Ith.

• Compare the calculated and measured value (step 2) of current.


ELECTRICAL ENGINEERING DEPARTMENT Mr.Imad Khader Eng. Nasim Zayed

EXPERIMENT# 4 Voltage Source

Objectives: 1-To studies the parameters of the voltage source. 2-To study the Series Circuiting of Voltage Sources. 3-To study the Parallel Circuiting of Voltage Sources. Introduction : Voltage Source

The voltage source consists of the initial voltage VOC and internal resistance Ri .If no current flowing in the circuit (IL = 0)

Vt =VOC (Vt :terminal voltage When a current flowing in the circuit :

Vt = VOC - IL Ri RR

VILi

ocL +

= .

A short circuit current ISC is limited by the internal resistance Ri :

SCI ocViR

=

An equivalent voltage source can be represented by a curve as show below

Series Circuiting of Voltage Sources Series circuiting of voltage sources as in the fig-1 (The precondition is that the poles of the voltage source are connected correctly, plus pole of one voltage source to the minus of the next one) gives a higher total voltage:

Vtot=V1+V2

Fig-1

If the poles are reversed the total voltage corresponds to the difference between the initial voltages:

Vtot=V1-V2

The internal resistors of the series-connected voltage sources add to a total internal resistance of:

Ri tot=Ri1+Ri2 A load resistor RL a current IL:

RRRVoI

IILL

21 ++=

Parallel Circuiting of Voltage Sources

By parallel circuiting of several voltage sources the same initial voltage a higher load current IL is achieved.

Fig-2

Matching poles must be connected (Fig-2). If the initial voltages are different, a compensating current Io flows within the voltage sources, which is dependent on the difference voltage of the respective internal resistance’s of individual voltage sources an initial voltage.

RLRiRLRiRiRiRiVoRiVoIL ∗+∗+∗

∗+∗=

21211221

The internal resistors are in parallel, this gives a total internal resistance of:

RiRiRiRiRitot

2121

+∗

=

Experimental Procedure

Ι -Voltage Source : 1- Connect the circuit as shown in fig -3,we add 22Ω resistor as an internal

resistant Ri.

2- Measure the open circuit voltage VOC at points (1, 2), connect a 100 Ω resistor and

measure Vt at points (1,2) and IL, replace 100 Ω by 33 Ω, measure IL and Vt, disconnect RL and connect Ammeter to the points1, 2 then measured ISC.

Plot the graph of equivalent voltage source, showing load lines (100 Ω,33 Ω ).

Q1- What is the maximum power that can be transferred from the source, and under what condition max power transfer occurs? ΙΙ- Series Circuiting of Voltage Sources:

1- Connect the circuit as shown in fig –4 with a power supply unit =2V and a battery.

Fig-4

2-How great is the voltage Vtot:

• Opposite poles connected(+with-) Vtot=

• Matching poles connected (+with+) Vtot=

ΙΙΙ- Parallel Circuiting of Voltage Sources 1-Connect the circuit as shown in fig-5. We add internal resistance Ri=22Ω.

Fig-5

2-Connect two voltages source in parallel and make the flowing measurements at equal and unequal initial voltages. Measurements should be made on both no-load and load operation, as in table-1.

Equal initial voltages VO1=VO2=1.5V No-load

Load=100Ω

Vi1 (V) Vi2 (V) V12 (V) Io (mA) Vi1 (V) Vi2 (V) V12 (V) I1 (mA) I2(mA) IL (mA)

Check the results by calculation.

Unequal initial voltages VO1=3 VO2=1.5V

No-load

Load=100Ω

Vi1 (V) Vi2 (V) V12 (V) Io (mA) Vi1 (V) Vi2 (V) V12 (V) I1 (mA) I2(mA) IL (mA)

Check the results by calculation.

Table-1



ELECTRICAL ENGINEERING DEPARTMENT Mr. Imad Khader Eng .Nasim Zayed

EXPERIMENT # 5 CHARACTERISTICS IN AC

Objective: 1-To study characteristics of the sine wave and square wave voltages ,which are the most forms in electrical and electronic engineering. 2- To practice the use Oscilloscope (some times called “Cathode Ray Oscilloscope or CRO) Introduction Alternating current (AC) changes its direction of flow continuously, in contrast to direct current (DC), which always flows in same direction. The wave (function) of the current or required voltage may adopt different shape.

Fig-2

The functions in time of the sinewave AC voltage during one period as in fig.2 with the values are: T = period duration in S Vp = peak voltage (positive or negative) in V, amplitude peak value. Vpp = peak-to-peak voltage in V. Vrms = root mean square voltage in V. If a sinewave AC voltage is applied to an ohmic resistor, it generates an AC current in it with the same function with the values are: Ip = peak current value (positive or negative) in A. Ipp = peak-to-peak current value in A. Irms = root mean square value of current in A. Other characteristic values and the formulae calculating them are specified below:

Frequency f in Hz: fT

=1 , Radian frequency ω in 1/s, ω= 2.π. f

Wavelength λ in m, λ =vf

and V= propagation speed

Root mean square value Vrms, Irms:

Vrms Vp= ∗12

Irms Ip= ∗12

Momentary values i, v: v Vp t= ∗sinω i Ip t= ∗sinω Active power with sine voltage P in watt

P Irms Vrms= ∗ . P I rms R= ∗2 . P VR

=2

P v i= ∗ -Momentary values, or P Vp Ip=

∗2

Experiment 1- Connect the circuit as in Fig –3

2- Connect the oscilloscope to point A and make suitable sittings in CRO. 3- The sinewave AC voltages shown in the CRO should be draw on simegraph

appear and the sittings. 4- From (CRO) and graph determine:

Vp= Vpp=2Vp= Ip=Vp/R= Vrms= Irms= T = f = ω= λ ( )v = ∗240 106 = Momentary voltage after a third of a period v=

4- Connect the circuit as in Fig- 4

Fig-4

5- Connect the CRO ch1 and ch2 to Y1,Y2 and the ground to C , make the sittings . Draw the displayed curves for i,v. 6- Determine the power from the graph:( r.m.s Values) P Irms Vrms= ∗ = P I rms R= ∗2 =

P VR

=2

=

P Vp Ip=

∗2

=

7-Determine the momentary power for each time? Table-1 P v i= ∗ = [momentary values at times] = 8- Draw the curves for i,v.and P at the same graph paper

ELECTRICAL ENGINEERING DEPARTMENT AN-NAJAH NATIONAL UNIVERSITY

FACULTY OF ENGINEERING Eng.Nasim Zayid Mr.Imad Khader

EXPERIMENT # 6

Capacitor in the A. C.circuit

Objectives

1- To investigate the factors determining the charge and discharge for a capacitor 2 - To investigate what happen when capacitors are connected in series and in parallel 3 - To notice the phase shift between current and voltage . Introduction: In this experiment we wish to see how a capacitor work at AC source . If an AC is applied to a capacitate circuit the current waveform reaches it’s positive peak value 90° . A capacitor charges and discharges flowing an exponential curve, when the voltage has risen to 63 % of the final voltage after 1τ during charging or 37 % of the initial voltage during discharging, τ is the time constant τ = R. C where R in ohm and C in F. The momentary current and voltage are calculated by :

charging case: Vc = Vc ( 1 - e /t τ− )

Ic = C tVR e. − τ

discharging case: Vc = V − te /τ

Ic = − −VR

t

e τ

The resistance to sinusoidal current in a capacitor is called capacitive reactance Xc .The value of the capacitance reactance depends on the capacitance of the capacitor and the frequency. The Xc is calculated by the formula:

fccXC π

=ω

=2

11

If Ic and Vc are known, Xc can be calculated by ohm’s law :

Xc VI

C

C=

If a capacitors are connected in series the total capacitance is equal to:

1 1

1

1

2

1

totalC C C nC= + + +...

Vctot =VC1 + VC2 + Vcn

If a capacitors are connected in parallel the total capacitance equal to : Ctot = C1 + C2 + + Cn Ictot = IC1 + IC2 + + Icn Experimental procedure : Ι -Charging and discharging process of a capacitor 1- Connect the circuit shown in fig -1 and connect the generator with a positive square wave voltage.

2- Display the input signal (point A) , capacitor signal (point B ) and current signal ( point C ). Draw in same graph paper.

3- From the graph determine the values:

A - Time constant τ.

B- Capacitance C. C -Momentary voltage value Vc after charging time 2ms.

D - Momentary current value Ic after discharging time of 2.5 ms. E - Charge Q .

4 - Confirm the results obtained in step -3 by calculation.

ΙΙ - Phase shift : 5 - In fig –1A set the function generator to Vpp = 3 V sinusoidal,at 1 kHz.

6 - To display the phase shift between voltage and current, connect ch1 of the

CRO to point A(capacitive current), and other ch-2 inverse to point B (capacitor voltage), Draw the curves and determined the phase shift between Vc and Ic .

ΙΙΙ - Capacitive Reactance Xc . 7 - Connect the circuit as shown in fig - 2.

8 - Set the function generator to 4 Vrms at 100 Hz , ... ,1kHz. Tabulate your result in

table -1 then plot the characteristic curve Xc = f ( F ) for 1µ and 0.22µ.

F(KHz) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Vc(rms) 1.0µf 0.22µf Ic(mA) 1.0µf 0.22µf Xc(KΩ) 1.0µf 0.22µf

Table-1 Question

1- What does the curve tell you? 2- Check the Xc of the capacitor 0.22µF at 220 Hz from the graph and calculation?

ΙV -Capacitor in series and parallel 9 - Connect the circuit as shown in fig - 3.

10 - Measure the current in the circuit Voltages across VR1,VC1 ,VC2 ,VC3 and Vctot. Calculate (from measured values) XC1, XC2, XC3 and Xctot , then C1 ,C2 ,C3

and Ctot .Calculate ( from result ) the Ctot and compare between Ctot measured and calculated .

11 -Connect the circuit as shown in fig -4

.Measure the voltage VC1, VC2, VC3, and Vctot And Ictot ,IC1 , IC2 , and IC3 . Calculate (from the measured value) XC1 ,XC2 ,XC3 and Xctot ,then C1 , C2 , C3 and Ctot . Calculate (from result) the Ctot and compare between Ctot measured and calculated. .



ELECTRICAL ENGINEERING DEPARTMENT Eng.Nasim Zayid Mr.Imad Khader

EXPEREMENT # 7 INDUCTORS

Objectives: 1- To describe the characteristics of an inductor under DC and AC conditions. 2- To investigate series and parallel inductive circuits 3- To determine the phase shift that exist between current and voltage in inductive

circuits.

Introduction : In this experiment, an inductors behavior is investigated under DC and AC

conditions. The term inductance (of a particular coil) defines the ability of the coil to induce back emf when subjected to an AC applied voltage , Va ( which causes the flow of an AC drive current , Ia ) . Under such condition , the driving current Ia will be lagging the applied voltage by 90° ( in an ideal case -- ie. Resistance of the coil is neglected ) .

If on the other hand , a DC voltage is applied to the coil ,then the current arising from the DC. Voltage will increases to 63% of its final value in 1τ second , also it will , decrease ( when the DC. Applied Voltage is switched off ) to 37% of its initial value , in 1τ . In both cases , the current requires 5τ second to reach its final destination ( full current or zero current respectively )

RL

=τ [sec.] τ - is the time constant

Where R is the sum of the ohmic resistance of the coil in (Ω ) and the internal resistance of the voltage source .

The instantaneous current and voltage are calculated using:

When switched on IL = ( )VR

te1− − / τ

VL = V te. /− τ

when switched off IL = VR

te−⎛

⎝⎜⎜

⎞

⎠⎟⎟τ

VL= - V. −te / τ A coil has a current limiting effect under AC condition, this arises from the counter

voltage generated in it (induced back emf ). Such effect is directly proportional to the inductive reactance XL of the coil.

The value of XL depends on the inductance of a coil and frequency of the applied AC source, as shown in the following expression:

XL = 2 π f L for a sinusoidal signal

If IL and VL are known , then XL can be calculated using ohm’s law as follows :

ILVLXL =

When inductors are connected in parallel then the total inductance will be determined using the following equation:

1 1

1

1

2

1

totL L L nL= + + +....

Also when inductors are connected in series then the total inductance will be determined using the following equation :

Ltot = L1 + L2 + + Ln

Experimental Procedure: 1- Construct the circuit as shown in fig -1 and set the function generator to :

positive square wave , amplitude of voltage 4 V and at 1 kHz .

2- Plot the input signal at point A . Inductance voltage signal at point B. Inductance current signal at point C . Determine (from graph) τ. Then calculate the inductance L. 3- Calculate the values of τ and L and compare the calculated values with those

obtained from 2. 4- To determine the phase shift between the current IL and the voltage VL. Change

the positive square wave to sin wave with amplitude 8 Vp-p, then connect ch1 at point B (voltage monitor), and ch2 at point C (current monitor) Plot the graph and determine the phase shift.

5- To determine the inductive reactance XL, connect the circuit shown in fig -2

6-Set the function generator to 4Vrms at 1kHz , tabulate your result in table-1, then

plot the characteristic curve XL = f (F) for 40mH and 200mH .

F (kHz ) 1 2 3 4 5 6 VL 40 mH

200 mH IL 40mH

200mH XL 40mH

200mH Table-1

Question • What can you deduce from the curve ? • Check the value of XL , from graph , for L= 200mH at 3 kHz and compare it

with calculated value ? 7- Connect the circuit as shown in fig - 3.

8- Measure the current I L1, I L2, V L1, V L2 and V Ltot .

Calculate (from measured values ) X L1 ,X L2 , X Ltot then L1 , L2 and L tot . and subsequently , Calculate L tot , compare between the measured and calculated value for Ltot .

9-Connect the circuit as shown in fig - 4, measure the voltage V L1, V L2, VLtot and ILtot , IL1 and IL2 . Calculate (from the measured values) XL1,X L2 and X Ltot ,then L 1, L2 ,Ltot. Calculate (from result) the Ltot and compare between Ltot



ELECTRICAL ENGINEERING DEPARTMENT Eng: Nasim Zayid Mr:Imad Khader

EXPERIMENT #8

RLC Series & Parallel Circuits Objective of experiment 1 - To investigate the impedance of RLC circuit. 2 - To determine the distribution of the applied voltage V and current I in the R, L and C elements. 3 - To determine the phase angle between the voltage and the current for each element in the RLC circuit. Theoretical aspects : Series circuits If a sinusoidal AC voltage is applied across a series circuit, containing resistor, capacitor and coil, the same current would flow through all the components. In the case of the resistor, the voltage across the resistor, VR will be in phase with the current, IR passing through it. However, the voltages across the capacitor and across the coil experience a phase shift with their respective currents. The Apparent voltage V (the voltage applied to the circuit) may be defined as follows: V V V VcR L= + −2 2( ) or V Z I= ∗ where Z is called the apparent resistance Z ( or more commonly known as the Impedance ): The Impedance Z may be calculated as follows :

Z R X XcL= + −2 2( ) XL =w∗ L, Xc = 1w C∗

IVZ =

The phase angle ϕ : (ϕ being the angle between the applied voltage & current for the circuit )

tanϕ =−

=−V Vc

VX Xc

RL

R

L

Parallel circuits In a parallel RLC circuit, the voltages across all the components are the same. and the total current , I , is divided into active current IR, capacitor current IC, and coil current IL. A phase shift occurs between each of the currents IL, IC, and the total current I ( it is clear that no phase shift exist between IR and I ) , due to the reactance’s XL of the coil and XC of the capacitor. The current IC precedes (leads) IR ( and therefore I ) constantly by 90° -- assuming that the capacitor contains no resistance , While the current IL lags the active current IR constantly by 90° -- also assuming that the coil is pure inductive , i.e contains no resistance .The Currents IC opposes IL (180° angle phase ) and thus tends to equalize each other depending on their magnitude . The apparent current I ( the total current supplied to the circuit ) can be calculated using the following equation :

I I (Ic I )2

R

2L= + −

Also the apparent conductance Y of the circuit may be obtained from the following equation:

Y B Bc L= + −2 2G ( ) Bc w C= ∗ BL

L=

∗1

ω G

R=

1

Tan of the phase angle ϕ :

tanϕ =−

=−Ic I

IBc B

GL

R

L

Experimental procedure Ι - Series circuit 1 -- Connect the circuit as shown in fig -1, and set the function generator at the

following voltage: Vrms = 3V (sinusoidal), f = 1kHz .

2 -- With the DMM measure VL(AB), VC(BC) and VR(CD) . • Using the appropriate calculation and the respective vector diagram ,determine the above voltages as well as the phase angle , ϕ , between the total voltage supplied to the circuit and the total current. 3 -Connect the oscilloscope’s ch-1 to point C, ch-2 to point A and connect point D to ground , and draw the displayed voltage waveforms and determines the phase angle ϕ.

ΙΙ − Parallel circuits:

4 -Connect the circuit as shown in fig -2, and set the function generator to voltage: Vrms = 3V (sinusoidal), f = 1kHz.

5 -- With the DMMs measure I, IR, IC, IL and deduce from calculations the total current I and phase angle ϕ. 6 -Using the related calculations, constructs the victor diagrams.


ELECTRICAL ENGINEERING DEPARTMENT Mr: Imad Khader Eng: Nasim Zayed

EXPERIMENT # 9 SERIES RESONANCE

Objective:

To investigate the series resonance in RLC circuit around the point of minimum impedance resonance.

Theory:

For series resonance, we can notice that the impedance of the circuit is changing with frequency.

Z = R + jωL + 1J Cω

Z= ( )2 21 1 1R L C L C

R+ − − −

ω ωω ω/ tan

/

The frequency at which the capacitive reactance is equal to be inductive reactance, and they cancel each others is called the resonant frequency of the circuit. The circuit impedance at this frequency is just the resistance of the circuit.

Xcfc

=1

2π= 1

ωc ; XL=2πfL = ωL, at resonance f = f0 , ω = ωo , Xc = XL

12πfc

=2πf L ------- fo = LCπ2

1 ----- ωo= 1

LC

At resonance ωoL= 1ω oc

and Io = ViR

VL =ωoLIo = ω o L VR

= V LR

Vo

Vc = Ioc

VcR

LR

Vo o

o

ω ωω

= = = VL from which

ωω

oLR oCR

Q= =1

Where Q is the quality factor, a dimensionless ratio

Experimental procedure: Series resonance 1- Connect up the circuit fig1, then connect one of the ac voltmeters between point

P&S.

2-For a suitable generator output value, vary the frequency of the generator from

about 100 Hz to 1kHz, and notice the variation in current and voltage, then find the resonant frequency.

Now we are going to draw Vo, I ,VL and VC with frequency. Determine the output of the function generator, which must be constant for the whole range of frequency.

3- construct a table like fig-2 containing of freq. against each of V, I, VC, VL and take readings for changing the frequency from 100Hz to 1000Hz by 50 Hz step.

NOTE: Around the resonant frequency take several readings.

F(Hz) Vo I Vc VL 50 …..

To 1000 Fig-2

On a sheet of single cycle logarithmic graph paper, draw curves of I, VL, and VC

against frequency. 4- Repeat above steps for L=200mH.

Questions 1- From the curves plotted with R = 100Ω, what is the resonant frequency? 2-What are the values of I, VL, VC at res. frequency? 3- what is the quality factor (Q) theoretically and practically? 4- What is the bandwidth?


ELECTRICAL ENGINEERING DEPARTMENT Mr:Imad Khader Eng:Nasim Zayid

EXPERIMENT # 10

Parallel Resonance Objective: To investigate the parallel RLC circuit around the point of minimum impedance resonance. Theory: Ideally, parallel resonance occurs when Xc=XL, the frequency at which resonance occurs is called the resonance frequency. When Xc=XL, the two branch currents Ic and IL are equal in magnitude and they are always 180° out of phase, the two currents Ic and IL cancel and the total current is zero Fig –1.

For an ideal parallel resonant circuit, the frequency at which resonance occurs is determined by the same formula as in series resonant circuits:

of LC=

12π

In practice we have not only L and C, but also other components which cause energy loss. The most obvious one is the winding resistance of the inductor, but there are others in practice. The effect of the winding resistance is to rotate the phasor IL slightly. Consequently the currents IL and Ic cannot quite cancel each other, and small residual current flows in the terminals.

Experimental procedure:

1- Connect the circuit as shown in fig-2, and then connect two AC Ammeters for Ic and IL.

2- From a function generator output vary the frequency at 100 to 1kHz,In table-1 write your results, and then find the resonant frequency. Output of function generator must be constant.

Frequency (Hz) Ic (mA) IL (mA) IR (mA)

Table-1 3- Connect the circuit as shown in fig- 3, then vary the frequency of function

generator slowly from 100 Hz to 1kHz, and notice the variation of V1 and V2 on the osill. (CRO), tabulate your results in a table-2.

Frq. (Hz) V1 V2 100 ----- ----- 1000

Table-2 4- Find the resonant frequency, and calculate the total current (I) at this moment and the impedance of the RLC circuit. Questions: 1- Compare between series and parallel resonance circuits? 2- Compare between the theoretical and practical values of resonant frequencies?



EXPERIMENT # 11 Mr:Imad Khader Eng:Nasim Zayid

Three-Phase Alternating Current

INTRODUCTION Three-phase current systems result when several phases shifted voltages are connected together. The most common three-phase generator which is usually used in the public power supply network supplies three sinusoidal AC voltages which are linked to each other and phase shifted by 120° (fig-1).

Fig- 1

Two basic circuit types are common in the coils of the three-phase current generator and on the load side when connecting motors, for example, the star circuit and the delta circuit as in fig-2 and fig-3.

fig-2 fig-3

In the star circuit three conductor voltages (380V) are available between the conductors L1, L2 and L3 as will as three phase voltages (220V) between the conductors L1 ,L2 and zero conductor (N). In the delta circuit three voltages are available between the conductors L1, L2 and L3 with 380V each. The three necessary sinusoidal AC voltages are not taken directly from the mains, they are generated electronically by phase shifters. For risk less experimentation the voltage 220V is limited to 12V. The conductor voltage VL and phase voltage V phases are linked with factor 3 , also called the linkage factor. In the public power supply network therefore, at a phase voltage of 220V there is a conductor voltage of:

V V VL ph= = =3 173 220 380* . . *

EXPERIMENTAL PROCEDURE: Ι- Potential gradient in three-phase current systems 1-Display the phase voltages of a three-phase current system on the oscilloscope draw the displayed voltage curves in a diagram and determine the angle of phase shift between the individual voltages. 5- Measure the phase and conductor voltages with a multimeter and establishes the

linkage factor. Q. What is the peak value of phase and conductor voltage? ΙΙ- Load in star circuit: 1- Connect the circuit as shown in fig - 4. Set each of R1=R2=R3=100Ω (Symmetrical load).

Fig-4

2- Measure the currents IL and IN and the voltages VL and Vph. (r.m.s values) with a mulltimeter, tabulate your result in table-1. 3- Set each of R1=100Ω, R2=220Ω, R3=470Ω (Unsymmetrical load) and measure the currents and the voltages, tabulate your result in table -1.

Star circuit Load Symmetrical Unsymmetrical Conductor IL1 currents IL2 IL,IN,Iph IL3

IN Conductor VL1, L2 voltages VL2,L3

VL VL3,L1 Phase VL1,N

voltages VL2,N Vph VL3,N

Table-1

4- Disconnect IL1 and measure IN for Symmetrical load 5- Disconnect IL1 and IN and measure VL1 VL2 for Symmetrical load 6- Disconnect IN for for Unsymmetrical load and and measure voltages as in table ΙΙΙ-Load in delta circuit: 1-Connect the circuit as shown in fig -5 2-Measure the phase currents and conductor currents as well as the conductor voltages in a three-phase AC mains on a delta circuited when R1=R2=R3=100Ω (Symmetrical load) and tabulate your result in table -2. 3- Set each of R1=100Ω, R2=220Ω, R3=470Ω (Unsymmetrical load) and measure the currents and voltages, tabulate your result in table-2.

Fig-5

Delta circuit

Load

Symmetrical Unsymmetrical Conductor IL1 currents IL2

IL IL3

Phase IR1 currents IR2

Iph IR3 VL1,L2

VL=Vph VL2,L3 VL3,L1

Table-2

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