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Electronics Lab Manual
Grade 11 and 12
tdamtew
AdisAbaba,Ethiopia
Table of contents
Laboratory Safety and Conduct Rules for Electronics Laboratory…………………….1
Electro Statics…………………………………………………………………………………………2
Capacitors and Dielectrics…………………………………………………………………………7
Conductivity Resistivity and Resistance……………………………………………………11
The relationship between e.m.f terminal p.d and internal resistance……………..14
Kirchhoff’s Loop Rule……………………………………………………………………...……..17
Kirchhoff’s Current Law…………………..………………………………………………………19
Multimeters………………………………………………………….……………………………….26
The Wheat Stone Bridge……………………….…………………………………………………29
The potentiometer ………………………………….……………………………………………..30
The balanced Bridge……………………………….………………………………………………32
Rheostats…………………………………………….………………………………………………..34
Transformer Turns Ratio………………………….……………………………………………..36
Resistive and alternating currents ……………….…………………………………………..39
Capacitive Circuits and alternating Current ………………………………………………41
Inductive circuits and alternating currents……………………..…………………………43
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Combining an Inductor and a capacitor in a circuit………………..……………………45
Combining resistor, capacitors and Inductors series RCL resonant……..…………47
Reactance in a series R-L series……………………………………………………….……….50
Reactance in a series R-C circuit…………………………………………………….………..53
Reactance in a parallel R-L circuit………………………………………………….…………56
Reactance in a parallel R-C circuit……………………………………………………………58
Laboratory Safety and Conduct Rules for Electronics Laboratory
1. No student is permitted in the laboratory without an instructor.
2. Students may not start an experiment until given permission by the instructor.
Students may not block the aisle in the laboratory with their bags, jackets, notebooks and other articles. Laboratory aisles must be kept uncluttered.
Bare feet and sandals are not acceptable.
No student may invite individuals who are not enrolled in the Electronics laboratory courses to come in the Electronics laboratory class.
No student is permitted to change the configuration of any computer he/she is working on and will only use the computer as instructed to work on physics or astronomy laboratory experiments.
Absolutely no eating or drinking in the laboratory during anytime.
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Every student will clean up his/her work area before leaving. This includes any gum wrapper, paper confetti or eraser crumbs which accumulates during the lab session.
No student will write on or deface any lab desks, computers, or any equipment provided to them during the experiment. They will use all equipment only for the purpose intended.
I have read and understood these rules. I understand that any violation of these rules could lead to dismissal for the lab session and any other appropriate action by the instructor.
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Electro Statics
Objective
1. To apply the principles and laws of electrostatics.
2. How to draw the electric field lines across different charges.
Accessory Required: Pencil, Rubber, Compass, Ruler and drawing paper.
Introduction
Electric field a region of space around a charged object which exerts a force on other object.
Force the capacity to do work or cause physical change
Positive the charge on a body which has a deficiency of electrons.
Electric field lines representing an electric field in a region of space.
Electric field strength. The force per unit positive charge acting on a positive test charge placed in the field.
Vector a quantity specified by its magnitude and direction
Coulomb’s law: Stating that the electrical force between two charged objects is directly proportional to the product of the quantity of change on the objects and inversely proportional to the square of the separation between the two objects.
F=KQ1Q2
R2 =K Q1Q2
R2
F= 14 π EO
Q1Q2
r2
Q1Q2
4π EO r2
Permittivity of Free space: a constant that specifies how strong the electric force is between electric charges in a vacuum .It is an electric constant.
Permeability: Constant on magnetic field on free space.
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Procedure
Complete the following statement concerning charged bodies.
Electric charges of like kind _____________each other, and charges of unlike kind _________each other.
Below is indicated a positively charged body draw lines indicating the field of force body, and show the direction of the field.
Draw lines indicating the field of force existing about the two charged bodies below. Show direction of the field.
A device for detecting electric charges is called ____________.
In the space provided explain what will happen to the leave of the electroscope, if the rod is brought nearer to it, why? Assume the electroscope is positively charged.
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Indicate with lines the field about the two charged bodies below. Show the direction of the lines.
Where would the greatest electrostatic charge accumulate? (Circle the core answer)
B
A D
C
Draw the electrostatic lines of force surrounding
a. Positively charged body
b. Negatively charged
Which of the following statements is correct?
a. Both charges are positive and are attracting each other.b. Both charges are positive and are repelling each other.c. Both charges are negative and attracting each other d. Both charges are negative and are repelling each other.
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+ +
+
-
A B
Determine the force between the two charges when k=1
Q1=+10 esu
Q2=+15 esu
d=5cm
Determine the distance between the charges when
F=128 Dynes of attraction and K=1
Q1=40esu Q2=-80 esu
d
If Q1 = +10 esu, d=5cm, and F=6 Dynes (Repulsion)
Find Q2
Q1=10esu Q2
5cm
Draw the electric field near two parallel plates one charged positively and the other negatively.
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.
-
-
-
-
-
+
+
+
+
+
Draw the electric field between two oppositely charged parallel plates is uniform .This means the field lines are equal spaced between the plates.
200 v
100 v
There are three other consequences of Gauss’s law and Columbus law concerning the distribution of charges on a charged conductor.
1. The net electric charge of a conductor resides entirely on its surface. This is due to the repulsion of like charges the charges are pushed as far a part as possible and so spread out on the surface.
2. The electric field inside the conductor is zero
If there were to be any charge inside the object then this would cause there to be a net force acting on some of the charges and they would accelerate .This is no the case the charge remain static.
The electric field at the surface of the conductor is perpendicular to that surface .If the field were to act at an angle then there would be a horizontal component of the force. This would again cause the charges to move around rather than remaining static.
Note: In a uniform field the lines of equi potential are equidistant parallel lines. The diagram below shows the lines of equi potential between two oppositely charged parallel plates. The lines of equi potential between parallel plates are parallel lines, However, these lines curve as the field weakens as you move outside the plates.
Draw the equipotential lines between the two charges?
200V
100v
0v
9
+++++++++++++++++
-- -------------------------
Note: If the electric field strength is increased then the lines of equipotential move closer together.
Draw the equipotential lines
300v
50v
0v
10
Capacitors and Dielectrics
Objective
List the factors that affect capacitance
What three physical factors affect capacitance. What is meant by a voltage rating of a capacitor. List five dielectrics used in the construction of capacitors.
Introduction
The words capacitor and condenser are used interchangeably. Inductors store energy in their magnetic fields, while capacitor store energy in their electric fields, whenever two bodies having opposite charges on them are brought close to one another, an electric field is produced the strength of this field is determined by the amount of charge on each body and the bodies physical size the material between the plates of capacitor is called the Dielectric.
Capacitor: an electrical device characterized by its capacity to store an electric charge.
Capacitance: an electrical phenomenon where by an electric charge is stored.
The amount of capacitance of a capacitor depends up on.
The area of the plates The distance between the plates And the type of dielectric material that is used
c= AKD ………………………………………………. Eqn
where C=the amount of capacitance ,A=,K=Dielectric constant
and D=distance between the plates.
Type of Dielectric Material
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Vacuum Dielectric Constant 1
Air Dielectric constant 1
Paper, Paraffined Dielectric constant 2.2
Paper bees waxed 3.1
Glass 4.2
Castor oil 4.7
Porcelain 5.5
After you observe the charge of the capacitor
Turn of the source voltage Now the capacitor starts discharge and the meter will read in the reverse
direction.
II. Connect the circuit below
Observe the wave across the capacitor using channel 1 of the oscilloscope.
Put the 2nd channel of the oscilloscope across resistor Compare the wave shape across the capacitor and across the resistor .What
do you observe.
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Remove the generator input and connect it very quickely as shown
What happens to the direction of current.
Mica Dielectric constant
It requires less force to distort the electron orbits in mica than it does to distort the electron orbits in air the greater the dielectric constant the greater the capacitance capacitance can be calculated the equation.
c=QV ………………………………………………………………………..Eqn
Where C=Capacitance, Q=Charge and V=Voltage.
Capacitors have two resistances.
One resistance is the resistance of the conducting materials which opposes the current on charge and discharge.
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The other resistance is the resistance of the dielectric which is so great that the current that does leak through is negligible.
Note the current leads the voltage by 90 in a purely capacitive.
Breaking down when the voltage applied across a capacitor is too high the dielectric ceases to act as an insulator and the charge starts to spark across the plates.
Equipment Required Accessory Required
Signal generator Capacitors Galvanometer Connecting lead Multimeter Resistors Dry cell
Procedure: connect the circuit below.
Break down Voltage
Q V for a capacitor .When ploting a graph of Q against V the gradient of the line is equal to the capacitance. However, this can not continue to increase indefinitely. Eventually the p.d across the plates will become too high .Charge will begin to spark across from one plate to the other when this happens the capacitor is said to be breaking down it ceases to store charge and begins to conduct electricity.
14
Breaking down when the voltage applied across a capacitor is too high the dielectric ceases to act as an insulator and the charge starts to spark across the plates.
Q(c)
Capacitor
V(v)
Conductivity Resistivity and Resistance
Objective: To determine the resistance of different materials.
Equipment required
Multimeter
Accessory: Copper, Silver, Aluminuim.
Procedure: Collect different types of conducting material cut it all with equal length.
The diameter should be of equal size. Measure the resistance of each material using ohm meter. Compare them all
Electric Current
An electric current is a flow of charge if you compare an electric current with water, a small current is like a trickle passing through a pipe a really large current is like a river in flood.
Coulomb: If charge flows at a rate of one ampere and continues to flow like that for a second, then the total amount of charge that has passed is one coulomb.
The formula linking amperes, coulombs and seconds is: Q=It
Where Q stands for the quantity of charge which passes when a current I flows for a time t.
15
The units are
Q/Coulombs (C)
I/ Amperes (A)
t/ Seconds (S)
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Conductivity Resistivity and Resistance
Conductivity: Away of measuring a materials ability to allow an electric current to flow.
Resistivity: A measure of how much a material resistor the flow of an electric current.
Resistance: a property of a material that controls the amount of current that flows through it. Ohm the unit of resistance.
The resistance of a metal wire at a given temperature is determined by three factors.
The length L in meters – The resistance is proportional to L,so if the length doubles so does the resistance.
Its area of cross section A is m2 the resistance is inversely proportional to A, So a wire twice the cross sectional area will have only half the resistance.
The resistivity (in ohm meters) of the material from which the wire is made .A material with higher resistivity will have a higher resistance
Resistivity and resistance are thus related by the equation R= PL
A
Eg. What will the resistance of a copper cable be if it has a cross –sectional area of cm2 and a length of 2 km? the resistivity of copper is 2x 108 Ω m .
Becareful over the units.
L= 2km =2x103
A= 1CM2 = 1X10-4 M2 (Since there are 100x100 square centimeter in a square meter).
R(Ω) P (Ωm) L(m) A(m2)
Use R=PL
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A
Putting the values, we get
R= 2X10 -8 X 2X10 3 = 0.4 Ω
1X 10-4
Drift Velocity
Even when no current flows through a piece of copper, the free electrons are moving rapidly about. Their speed is about 106 m/s or 3000 times the speed of sound in air. How ever, since they are moving at random, there is no net flow of electrons in any particular direction and so there is no current.
When an electric field in the form of a voltage is applied the electrons gain an additional velocity so that there is a net flow along the wire .This extra Velocity is called their drift velocity.
Drift velocity the average velocity that an electron reaches when an electric field is applied across a conductor.
Current Density
It is a vector quantity which means it has both magnitude and direction. Its magnitude is the current per cross sectional area. Its units are A/m2 and it is given the symbol J.
There is a common approximation to the current density which assumes that the current is proportional to the electric field E that produces it .The relationship is
J= QE
Where J is the current density, is the electrical conductivity of the material and E is the electric field
Eg. Find the approximate current density when an electric field of 5 V/m is applied to a copper conductor .The conductivity is 59.6 x 106 s/m
J(A/M2 (S/M) E(V/M)
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? 59.6X106 S Use J=E
= 5.96 X106X5
=2.98X106 A/M2
How does a source of e.m.f produce ap.d?
Electrical circuits transfers energy from batteries to the other components .The chemicals in the battery are a store of energy. When the circuit is complete the energy from the battery pushes the current around the circuit and transfer the energy to the component which can then work. The energy or push that the battery gives to the circuit is called the voltage or electromotive force (e.m.f) of the battery. It is measured in volts, which have the symbol V.
The higher the voltage is the greater the amount of energy that can be transferred .Voltage is a measure of the difference in electrical energy between two parts of a circuit .Because the energy is transferred by the component there must be more entering the component than there is leaving the component voltage is sometimes called potential difference (P.d)Potential difference measures the difference in the amount of energy the current is carrying either side of the component.This voltage drop across the component tells us how much energy the component is transferring.
Potential difference is defined as energy per unit charge the unit of potential difference is the volt (V) using the definition we can define the volt as joules per coulomb.
IV= 1J/C
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The relationship between e.m.f terminal p.d and internal resistance.
Objective
To determine voltage, current power internal resistance of the battery and main resistance of the circuit component
Equipment: Multimeter, 6 VDC, or 9 VDC dry cell battery patching board.
Accessory parts Connecting wires different values of two resistors.
Introduction
Suppose you short-circuit a battery .This means that you join its two terminals by a circuit that effectively has no resistance, a short piece of very thick copper wire, for instance the battery has an e.m.f V but the circuit apparently has no resistance R.What happens then? Does the current increase without limit? The thing we are forgetting is that the battery has to pump the charge round the whole circuit, and that includes the bit within the battery as well as the outside circuit .The internal resistance varies a lot between the different sorts of batteries, but that is what finally sets a limit to the current they can supply.
A 1.5 V torch battery typically has an internal resistance of up to an ohm. This means that ever if you short –circuit the battery there is still that ohm of resistance present. The biggest current it can deliver is given by I= 1.5 = 1.5 A
There is a formula that relates e.m.f (E) terminal p.d (V) and the internal resistance of the source (r): E=v+Ir
Procedure
1. Measure and record the resistance of each resistor.
2. Measure the battery voltage with no load.
3. Connect the circuit shown below.
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4. Measure the voltage drop across ER1,and ER2.
5. Add ER1, and ER2 (ER1+ER2)
6. Compare the voltage on item 5 with the source voltage
7. E-ER1+ER2= Er
8. Measure the current in the ckt .(It).
9. Calculate the internal resistance of the battery using the formula
R= Er
It
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Kirchhoff’s Loop Rule
Objective
To use kirchhoff’s voltage law to analyze a series parallel circuit.
Equipment Required
1. Variable Power supply
2. Multimeter
3. Patching unit
Accessory Parts
a. Resistor R1, R3 2200 Ohm ¼ Watt.b. Resistor R2, 1500 Ohm ¼ Wattc. Resistor R4, 3300 Ohm ¼ Watt.d. Resistor R5, R6 1000 Ohm ½ Watt.
Introduction
We can consider e.m.f to be energy per unit charge transferred in to electrical energy and p.d to be energy transferred from electrical energy. We know that energy is always conserved. In a circuit, the electrical energy supplied by the battery is used in the circuit no surplus energy arrives back at the battery. Kirchoff’s loop rule recongiser this and is stated as In any closed loop in a circuit the sum of the e.m.f is equal to the sum of the P.ds.
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Consider the circuit shown above
We assume that the battery (Supply) has negligible resistance if we apply Kirchhoff’s loop rule to the complete loop from the source to R1 and back again to the source as shown in the diagram, then the p.d across the resistor equals the e.m.f of the source .This is true for each resistor, so if we now apply kirchoff’s function rule we get
I= I1+I2+I3= VR1 +VR2+VR3
Since I= V/R Where is the total resistance we get
V/R= V/R1 + V/R2+ V/R3
Now divide through out by V and we get
1/R= 1/R1+1/R2+1/R3
Procedure
1. Measure and record the resistance of each resistor
R1 R2 R3 R4 R5 R6R Measure
2. Connect the circuit shown below.
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3. Compute, then measure and record the effective circuit resistance in fig1 above.
4. Set the supply voltage as given by your instructor.
5. Measure and record the voltage drop across each resistor in fig 3.
R1 R2 R3 R4 R5 R6 ItE(v)T(MA)
6. Measure and record the current through each branch and the total current.
7. Turn off the power supply and reverse the power supply connections to the circuit.
8. Repeat steps 5 and 6 .Record the data in the table above.
Report Instructions
a. Compare the measured voltage drops for the two different source polarities. Explain any differences and similarities.
b. Compare the sum of the voltage drops in the branches Explain any differences.
c. Compare the voltage at the function of R1 and R2 with the sum of the voltage drops across R3, R4 and R5.3Explain any differences and similarities.
d. Calculate the voltage drops in the circuit and compare them with measured voltage drops. Explain any differences.
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Kirchhoff’s Current Law
Objective
To verify Kirchhoff’s current law
Equipment Required
1. Variable power supply from IDL 800 Digital Lab.
2. Circuit Patching Unit
3. Multimeter
Accessory Parts
a. Resistor R1, R3 2200 ohm ¼ wattb. Resistor R2,1500 ohm ¼ watt c. Resistor R4, 3300 ohm ¼ wattd. Resistor R5, R6, 1000 ohm ½ watte. Interconnecting Wires
Introduction
When an electric current arrives at a function, the current divides in to two or more parts, with some electrons going in one direction and the rest going along the other paths. This is true no matter how complicated the function or the circuit may be electrons can not appear or disappear so charged is said to be conserved.
A battery does not produce electric charge it simply pumper the charge around the circuit .A large number of electrons enter the battery at the positive terminal every second, and the same number leave the battery at the negative terminal every second. Similarly, the rate at which electrons arrive at one end of a wire is exactly the same as the rate at which they leave the other. This is all summarized in kirchhoff’s function rule which states that:
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“The total current flowing in to a point is equal to the total current flowing out of that point”.
This can be written as I1+I2 – I3 =0
Notice that I3 Has a negative sign .By convention, currents going in to a function are positive but currents leaving a function are negative .The sum of the currents at any function is zero.
I1
I1+I2 – I3= 0
I3
Procedure
1. Measure and record the resistance of each resistor.
R1 R2 R3 R4 R5 R6R Measure
2. Connect the circuit shown below
R1 R2 R3 R4 R5 R6 ItE(V)I(ma)
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3. The source voltage must be given.
4. Measure the total current and the current through each resistor .Record the data in the table above.
Report Instruction
a. Compare the total measured current with the sum of the branch currents. Explain any discrepancies.
b. Compute the current through the branches compare it to the measured values.
c. Compare the computed effective resistance.
Measuring Instruments
Theoretical Background
An ammeter has a very low resistance and is placed in series in a circuit.
The basic galvanometer described can be converted in to an ammeter by adding a low resistance “shunt” which is usually fitted inside the casing of the instrument
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A
and consists of a short length of quite thick wire .Most of the current takes this low resistance shunt route, and only a tiny proportion thickles through the coil to rotate the pointer.
A variety of different range settings can be achieved by varying the shunt resistance so that the amount of current that goes through the shunt varies. The table overleaf shows readings for an instrument that reads to full-scale deflection 1 MA, 10MA, and 1000 MA.The coil in such a meter will always read to a maximum of 1.
Range (MA) I coil up to I shunt up to Fraction in coil0-1 1MA 0 10-10 1MA 9MA 1/90-100 1MA 99MA 1/990-1000 1MA 999MA 1/999
A Voltmeter has a very high resistance and is placed in parallel with the component .you can convert a galvanometer to be a voltmeter by adding a large resistance in series with the meter as shown below.
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Calculating Shunt and multiplier values for use with a meter to give different current and voltage ranges
1.Eg a galvanometer of full scale deflection 5 MA is to be converted in to a 0-10 A ammeter .If its coil has a resistance of 50 Ω. What value of shunt must be fitted?
Draw the circuit with a current of 10A flowing under these circumstances we want the pointer of the meter just to reach the end of its scale, and this will mean 5MA (0.005A) flowing through it.
Use kirchhoff’s function rule to work out the current that must go through the shunt.
10-0.005 = 9.995 A
9.995A 50Ω
10A
X y
The p.d between X and (V) must be sufficient to drive 0.005A through the 50 Ω of the coil. Thus Vxy= IR= 0.005X50= 0.25V
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R
Now this p.d is also across the shunt, so to work out R We must ask what size is needed inorder that, with a p.d of 0.25 v across it a current of 9.995 A will flow through it.
This gives R= V/I =0.25 =0.025Ω
9.995
2.A galvanometer of resistance 50Ω and full-scale deflection 5MA is to be made into a 0-10 v voltmeter .How can this be done?
Imagine that the voltmeter is to be used to measure a 10V battery. If the pointer of the galvanometer is just to reach the end of its scale when connected to the battery a current of 0.005 A must be drawn from the 10v battery, the total resistance of the whole circuit must be given by:
R= V/I = 10/0.005= 2000Ω
The galvanometer already provides 50Ω of this so x= 2000-50= 1950Ω
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MultimetersObjectiveTo measure resistance, current and voltage with multimeter.Equipment Required1. Variable D-C Power supply2. Circuit Patching unit 3. Multimeter 4. Accessory parts
a. Resistor R1, 22,000 ohm ¼ watt.b. Resistor R2, 15,000 ohm ¼ watt.c. Resistor R3, 47,000 ohm ¼ watt.d. Resistor R4, 3300 ohm ¼ watt.e. Interconnecting leads.
Introduction A Multimeter is an instrument that combines the function of a multi- Range voltmeter ,ammeter , and ohmmeter in one package the separate functions and ranges can be selected through a switching arrangement .The multimeter is delicate and expensive.Therefore, all precautions pertaining to the individual instruments must be observed or the meter may be damaged.
Notea. To measure current the Ammeter is connected in series with circuit.b. To measure voltage the voltmeter is connected in parallel with the component.c. To measure ohms the power of the circuit must be turned off and the resistor
must be disconnected from the circuit and the ohmmeter should be placed across (in parallel) with the component.
Procedure
1. Set up the circuit shown below 31
Warning: Make certain power supply is off when making connections, or a shock could result.
2. Switch the multimeter to the ohmmeter function .set the range scale at RX1 and zero the meter.
Note: Be sure the zero- ohms adjustment is made before switching to other ranges.
3. Measure and record the resistance of the resistors.
Note: If meter indications are at the congested area of the meter, more accurate measurements may be obtained by switching to the next range.
Resistor R1 R2 R3 R4 EtR(Ω)I(ma)E(Volts)
4. Switch the Multimeter to the d-c ammeter function connect the meter into the circuit to measure the total current.
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Caution: Make certain that the proper precautions with respect to connections, polarity, and power are observed.
5. Set the power supply as instructed by your instructor
6. Measure and record the total current and the current through each resistor. Record the data in step 3 table.
Turnoff the power and disconnect the multimeter.
Caution: Make certain the range selector in set at the highest current measuring range when measuring an unknown current .If the indication is a low value switch to the next low range. The meter could be damaged if the procedure is neglected.
7. Switch to the d.c voltmeter function.
Caution: Make sure the range selector is set at its highest range when measuring an known voltage if the meter indication is low, switch to the next range observe proper polarity.
Report Instruction
a. Describe the basic elements of the multimeter.
b. Explain the precautions that should be observed when using the multimeter.
c. Explain the use of the function switch.
d. Describe the method of using the multimeter to measure current, voltage, and resistance.
e. Describe the scales on the multimeter.
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Theoretical Back Ground
The Wheat Stone Bridge
The basic bridge circuit is shown below .The fundamental concept of the wheat stone bridge is that two voltage or potential, dividends in the same circuit are both supplied by the same input as shown below .The circuit output is taken from both voltage divider outputs.
In its classic form, a galvanometer is connected between the output terminals and is used to monitor the current flowing from one voltage divider to the other. If the two voltage dividers have exactly the same ratio (R1/R2=R3/R4) then the bridge is said to be balanced. And no current flows in either direction through the galvanometer. If one of the resistors changes even a little bit in value, the bridge will become unbalanced and current will flow the galvanometer. Thus, the galvanometer becomes a very sensitive indicator of the balance condition.
In its basic application a,d,c voltage (E) is applied to the wheatstone bridge and a galvanometer(G) is used to monitor the balance condition. The value of R1and R3 are precisely known , but do not have to be indentical R2 is calibrated variable resistance, the current value of which may be read from a dial or scale.
An known resistor,Rx is connected as the fourth side of the circuit as shown and power is applied R2 is adjusted until the galvanometer, G reads zero current at this point.
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Rx= R2 X R3
R1
The potentiometer
A potentiometer has a sliding contact and acts as an adjustable potential divider. The basic circuit is shown below.
You know that the current through the two series resistor R1 and R2 will be the same. By adjusting the position of the sliding contact you adjust the values of R1 and R2 and hence the value of the potential differences V1 and V2.
35
If we now put a lowel resistance R1 in parallel with R2 as shown below.
Then the potential difference VL across RL can be calculated using the equation.
VL= R2 RL Vs
R1RL+R2RL+R1R2
How ever if RL is large in comparison with R1 and R2 (as it would be in practical application such as the input to an operational amplifier) then this equation can be simplified to
VL= R2 Vs
R1+R2
Similarly if RL is in parallel with R1, the potential difference VL is given by
VL= R1
You can see that the supply voltage (e.m.f) is divided by this circuit in proportion to the values of R1 and R2.
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The balanced Bridge
Objective
1. To determine the voltage distribution and resistance ratio in the balanced bridge.
2. The use of balanced bridge circuit in making resistance
Equipment Required
Student’s trainer board (IDL-800 Digital Lab) voltmeter or multimeter, Galvanometer, test probe.
Accessory parts:R1= 22 KΩ, R2 47 KΩ, R3= 4.7 K
R4=0.50K Potentiometer, connecting leads.
Procedure: Connect the circuit below Have the instructor check the circuit before applying the power. Be sure and measure each resistor.
a. Apply power and adjust R4 until the galvanometer shows no deflection. The bridge is then balanced.
b. Disconnect one end of R4 (at point B) with an ohmmeter measure and record the value of R4.
c. Using the formula R1/R2 =R3/R4 Calculate the value of R4 __________Ohms.
Are the measured and calculated values approximately equal?
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d. Reconnect R4 with a 12k resistor, adjust R4 until the bridge is balanced.
e. Disconnect R4 measure and record its values _____________ohms.
f. Using the formula of step 1c calculate the value of R4 ____________Ohms.
How do the measured and calculated values compare?
g. Reconnect R4 and measure the voltage drops across R1,R2,R3 and R4 and record below.
1. Voltage across R1_____________Volts.
2. Voltage across R2_____________Volts.
3. Voltage across R3_____________Volts.
4. Voltage across R4_____________Volts.
h. Write in the space below an equation for the voltage relationship in a balanced bridge circuit interms of ER1, ER2, ER3 and ER4.
Summery
The principle of the Wheatstone bridge circuit is that an unknown resistance can be determined by placing it in the circuit along with the three known resistances and then finding the point at which there is no current flowing through a galvanometer because the two potential dividers have exactly the same ratio.
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Rheostatus
Objective:
1. To study the difference between rheostats and potentiometers.
2. To learn the use of each
Equipment: IDL-800 Digital lab, patching board multimeter.
Access parts: Rheostats, resistors, connecting wires.
Introduction
A Rheostat is generally used as a current controlling device while a potentiometer is used to control voltage. There are two ways of using a variable resistor. It may be used as a rheostat for controlling the current in a circuit, when only one end connection and the sliding contact are required .It can also act as a potential divides for controlling the p.d applied to a device all three connections then being used.
Procedure
1. Determine the action of a series circuit as follows
a. Connect the circuit shown
b. have the instructor check the circuit
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c. With the switch still open, turn the knob of the rheostatfully counter
clockwise.
d. with power switch closed, set the power supply to 12 V using voltmeter.
e. Record the current reading _________________________ma
f. Calculate and record the total resistance____________ohms.
g. Turn the knob if the rheostat fully clockwise and record the current reading
_____________ma.
h. Calculate and record the total resistance now in the circuit ___________ohms.
i. Explain why there was a difference in the amount of current in the circuit
with the rheostat fully counter clock wise and with the rheostat fully
clockwise.
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Transformer Turns Ratio
Objective
To determine the voltage, current and turns ration of a transformer.
Equipment Required
1. Variable power supply2. Oscilloscope 3. Multimeter4. Circuiting patching board5. Accessory parts
A. Initial Value1. Transformer 2. Resistor RL 1000 ohms ½ watt3. Interconnecting leads
B. Additional values
1. Resistor 1500 ohms ½ watt
2. Resistor 10k Ω ½ watt
3. Resistor 100kΩ ½ watt
Introduction
The ratio between the number of primary and secondary coil turns determines whether a transformer is of the step-up or step down type the turns ratio of an idea transformer determines the voltage ratio and the current ratio .However, in a practical transformer, because of losses in the circuit, the turns ratio is never precisely the same as the voltage ratio is never precisely the same as the voltage ratio or the current ratio. The current ratio is the inverse of the voltage ratio. A step up in voltage in the secondary occurs in a transformer which has a step-down in the current of the secondary circuit, and vice versa. The secondary of the transformer does not generate power, but it takes power from the primary circuit.
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Important Equations
N=NP NE= EP NI= IL
NS EL IP
Where: N = turns ratio
NP=Number of turns in the primary coil
NS=Number of turns in the secondary coil
NE= Voltage turns ratio
EP=Primary voltage in volts
EL= Secondary (load) voltage in volts
NI= Current turns ratio
IL= load (Secondary) current in amperes
IP= Primary current in amperes.
Procedure
1. Set up the circuit shown below
2. Measure and record the input voltage. Measure and record the actual load resistance and the voltage across the load resistor of 1000 ohms, 10 kilohms, 100 kilohms.
E in = _____________volts, ac
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RL nom (kΩ) 1 10 100RL meas (kΩ) 1k 10k 100kEL (VAC)
3. Calculate the voltage ratio for load resistance of 1k Ω,10kΩ,100kΩ and record in table above.
4. Calculate and record in the table above the load current for load resistance of 1kΩ, 10kΩ,100kΩ.
5. Insert in the primary a 1500 ohm resistor Rp.Measure and record the actual value of Rp with a 10 kΩ load resistor in the circuit, measure and record the voltage across Rp and RL.
Rp=____________Ohms ERP= Volts ac ERL _______________volts ac
6. Calculate the primary current the load current and the current ratio for a load resistance of 10kΩ.
Report Instruction
1. Is the transformer used in this exercise a step up or a step down type? What must be done to make it the opposite type of transformer?
2. In an ideal transformer EP/ES =IS/IP =N what would be the ratio of output power to input power?
What would be the input impedance as seen by the generator interms of N and RL?
3. How does the voltage ratio vary with the load current?
4. How would you expert the current ratio to vary with load current.
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Resistive and alternating currents (AC)
Objective
To analyse the phase relation of voltage and current across a resistor in an AC Circuit.
Equipment Required
Signal Generator and amplifies Oscilloscope Connecting Probes Patching unit (Bread board) Multimeter Accessory: Connecting wires, Resistor.
Introduction
When we consider a.c circuits we need to know whether the current and voltage are in phase with each other or out of phase. In a resistor, the voltage and current increase and decrease directly with each other as you would expect from ohm’s law .we say they are in phase .If we use the r.m.s values for the current and voltage through the resistor, then for ordinary currents and frequencies, the behavior of a resistor in an a.c circuit in the same as the behavior of a resistor in a.d.c circuit.
Key Words
Root mean square (r.m.s) value: A value for the current or voltage that would be equivalent to the effective steady value.
Irms = I peak = I peak x 0.707 where I rms is the r.m.s current and I Peak is the maximum value of the current in a cycle (the peak current).
Vrms= Vpeak= Vpeak xd 0.707 where vrms is the r.m.s
Potential difference and Vpeak is the maximum value of the potential differences in a cycle (the peak potential difference) .Alternating current has a sinusoidal waveform this means that its magnitude is varying continuously but its average magnitude is zero the half cycle average current is given by the relation.
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I avg= 0.637x Ip
Where Ip is he peak value of the current
Peak current: The maximum value of the current in a cycle peak potential difference the maximum value of the voltage in a cycle.
r.m.s Current : the value of the current that would be equivalent to the effective steady value.
r.m.s Potential Difference :the value of the potential difference that would be equivalent to the effective steady value.
Procedure
1. Connect the circuit shown below
I=IP
2 I V
V=VP
2
2. Apply signal from the signal generator3. Connect an ammeter in series with the ckt.(A.C Range).4. Connect an oscilloscope across the resistor to observe.5. Vary the amplitude of the signal generator and observe wave form on the
oscilloscope and the current on multimeter. 6. Is the phase of the voltage and current the same
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Capacitive Circuits and alternating Current
Objective: To determine the phase angle of voltage and current across the capacitor.
Equipment Required
Signal Generator and amplifier Oscilloscope Multimeter Patching Unit (Bread board)
Accessory Parts: Connecting leads, capacitor
Introduction
The capacitor will draw current to oppose any change of voltage across itself. Inductors store energy in their magnetic fields while capacitors store energy in their electric fields. The strength of this field is determined by the amount of charge on each body and the bodies physical size.
The capacity for storing energy in an electric is commonly called capacitance. The amount of capacitance of a capacitor depends up on the area of the plates, the distance between the plates and the type of dielectric material that is used.
Xc= 1/I+C= 1/WC
Procedure
1. Set up a circuit shown below.
2. Use a signal generator as the a.c supply.
3. Connect an oscilloscope so that it shows you the current and the voltage in the circuit. You should see a trace on the screen.
4. Connect an a.c ammeter in series with the capacitor.
5.Vary the frequency using the signal generator turning knobe you see the wave shape on the scope and the current magnitude on the Ammeter They vary in
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opposite direction .eg increasing frequency will decrease the amplitude of the wave but it will increase the current.
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Inductive circuits and alternating currents
Objective
To determine the phase angle of voltage and current across an inductor.
Equipment Required
Signal generator and amplifier Oscilloscope Multimeter Patching unit (Bread board) Accessory parts: Connecting leads, Inductor
Introduction
In alternating current circuits the opposition to the flow of current is called impedance in an AC circuit may also be due to a component called a reactor there are two distinct types of reactors---- Inductors and capacitors. These components are called reactors because they do not dissipate electric energy in the form of heat, as do resistors, but alternately store energy and then deliver this energy back to the circuit.
Lenze’s law when a voltage is induced in a coil as a result of any variation of the magnetic field with respect to the coil, the induced voltage is in such a direction as to oppose the current change that caused the magnetic variation.
Inductance: is defined as that property of a circuit which opposes any change in the rate of current flow.
L= d2N2 Ω
L
µ= permeability of core
d= diameter of the coil (inches)
N= number of turns48
L= Length
Procedure
1. Set up a circuit as shown below
2. Use a signal generator as the a.c supply
3. Connect an oscilloscope so that it shows you the current and the voltage in the circuit.
4. Connect an AC ammeter in series with the inductor
5. Vary the frequency using the signal generator knob and observe the sine wave on the scope and the current in an ammeter.
6. What do you conculed from your observation?
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Combining an Inductor and a capacitor in a circuit
Objective
To analyze the resonant frequency of an LC circuit
Equipment Required
Oscilloscope Signal Generator Multimeter Bread board (Patching Board)
Accessory Parts
a. Different value of capacitorsb. _________//__________ Inductorsc. Connecting wires
Introduction
In an Lc circuit when the power supply is connected there will be oscillations between the inductor and the capacitor. We know that the phase or diagrams for an inductor in a circuit and a capacitor in a circuit are as shown.
Ic VL
Vc IL
In the circuit shown below the current through the capacitor Ic is the same as the current through the inductor. IL we can combine the phase or diagram as shown.
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VL= ILXC= ICXL VC=ICXC=ICXC
Phasor
Since Vc and Vc are in opposite directions, the impedance of this circuit is given by
Z= (XL)2-(XC)2
There will therefore by a frequency,f , at which XL=XC and the impedance will be zero this will happen when zx+L = 1/2+C This value of f is called the resonant frequency of the circuit and the circuit will conduct extremely well at this frequency .In practice , resistors are often added to such ckts in order to damp the oscillations.
Procedure
1. Set up the circuit above.
2. Investigate the behavior of the circuit with different values for L and C and use a signal generator as the a-c supply so that you can see the effect of changed in frequency.
3. Can you find the resonant frequency for a circuit? Does you find a free with your theoretical value given above?
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Combining resistor, capacitors and Inductors series RCL resonant.
Objective
To analyze the effect of changing circuit constants in a series resonant circuit.
Equipment Required
Oscilloscope Signal generator Multimeter Patching unit (Bread board)
Accessory unit
a. Inductor b. Capacitor C. Resistor D. Connecting wires
Introduction
A series resonant circuit is used in frequency discrimination applications .the characteristics of a series resonant circuit are dependent on the values of the circuit elements and the operating frequency .Conditions at resonance may be analyzed entirely by mathematics .However, the results usually do not agree exactly with an experimental analysis because of component tolerances and other unknown factors. Series resonant circuits are frequently used in radar, radio, television and any other circuitry operating in the rf range.
Consider the circuit shown.
We can draw a phase or diagram for this circuit as shown below
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IPwL
IPR
IPWC
The peak potential differences for a circuit such as the one shown in fig 1 are given by.
VR= I peak R
VL= I peak XL
VC= I peak Xc
The sum of the potential differences across the circuit may be written as
VT= √(VR)+(VL−VC )2
= √ ( I peak R )2+(I peak XL−Ipeak xc)2
Therefore the impedance of the circuit may be written as
Z=√R2+(XL−XC)2
The phase angle between the current and the p.d for the circuit is given by
tan XL-XC
R
When XL-XC (this is generally at high frequencies) the phase angle is positive so the current lags behind the applied p.d when XL < Xc the phase angle is negative and the current leads the applied p.d When XL=Xc the phase angle is zero and the reactance in the circuit matches the resistance.
This is when the resonant frequency in reached. The resonant frequency is given by
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f=
As we found for circuits with an inductor and a capacitor. However the presence of the resistor will dampen the resonant oscillations.
Procedure
1. Set up the circuit shown in fig1.
2. Measure and record the d.c resistance of R1 and L1
3. Determine the resonant frequency of the circuit by setting the generator for maximum output and by adjusting the frequency of the generator.
4. Investigate the behaviour of the circuit with different values of R,L and C.
5. Use signal generator as the a.c supply so that you can see the effect of changes in frequency.
Report Instruction
a. Can you find the resonant frequency for a circuit?b. Does the frequency you find agree with the theoretical value given above?c. How does changing the value of R affect the damping of the oscillations.
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Reactance in a series R-L series
Objective
To analyze the effects of variations in frequency on reactance in a series R-L circuit.
Equipment Required
Audio frequency generator Oscilloscope Multimeter Circuit patching unit (Bread board)
Accessory Parts
a. Transformer T1b. Resistor R1 1000 ohms ½ watt.c. Mutual inductance demonstration assemblyd. Interconnecting leads.
Introduction
It is often necessary to have practically all of the voltage drop in a series circuit across an inductor with little of the applied voltage developed across a resistor. The inductor in this case is referred to as a choke coil typical chokes ( inductor in series with a resistor) are used in a.f and r.f circuits .There are many uses of a high voltage generated by opening an inductive circuit. One use is the ignition system of an automobile where a high voltage spark, needed for each cylinder, is produced.
In such circuits, the total opposition to current (impedance is dependent up on the resolution of inductive reactance and resistance. Since inductive reactance is directly proportional to the frequency and inductor value the impedance is dependent also on frequency. You know that the p.d in an inductive circuit leads the current. When there is another component in the circuit, such as a resistor the phase difference between the p.d an the current is not, but an angle .The total opposition to the flow of current is a combination of the resistance from the resistor and reactance from the inductor, we can not use either them in this case. We use
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the term impedance to describe the total opposition to the flow of currents which combine resistors and inductors. Impedance is generally given the symbol Z.
WhereZL= inductive impedance in ohms
El= Voltage across the inductor in volts.
IL= Current through the inductor in amperes
XL= inductive reactance in ohms
F= Frequency in Hertz
L= inductance in hevrys
Q= Quality factor.
Procedure
1. Set up the circuit shown above.
2. Investigate how the p.d across the inductor varies with time by using an oscilloscope to display the p.d
3. How does varying the value of R alter the p.d across the inductor.
4. Vary the frequency of the signal generator and observe the voltage across the inductor.
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Report Instruction
a. How does voltage across the inductor vary with frequency.b. Explain the physical characteristics that determines the inductance and
resistance of the coil.c. Calculate the Q of the coil at a frequency of 500 HZ and 1000 HZ. On what
does the quality factor Q depends?
Reactance in a series R-C circuit
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Objective
To analyze the effect of variations in frequency on reactance in a series R.C circuit.
Equipment Required
1. Audio frequency Generator (Signal Generator)
2. Circuit patching unit (Bread board)
3. Multimeter
4. Oscilloscope
Accessory Parts
a. Capacitor C1, 0.1 µ Fb. Resistor R1, 1000 ohmsc. Connecting wiresd. Oscillscope probes
Introduction
The opposition which a capacitor offers to alternating current flow when a sine-wave voltage is applied is known as capacitive reactance. Like inductive reactance capacitive reactance is measured in reactance is opposite to that on inductive reactance.
Capactive reactance is inversely proportional to frequency, where as inductive reactance is proportional to frequency. As the frequency increases, the capacitive reactance decreases. Since the opposition to current flow decreases, the amount of current in the circuit increases.
The capacitor has an extremely high impedance at zero frequency and for this reason capacitors are often used to block d,c. but allow some ac to pass. The resistance expressed in the formula for the Q of a capacitor is the opposition of the capacitor to dc, which is known as leakage resistance this resistance usually has a high value is effectively in shunt with the capacitance.
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Important Equations
Xc=
and the phase angle = tan -1 xc/R
Where : Xc= Capacitive Reactance in ohms
F= frequency in hertz
C= Capacitive in farads
Q= Quality factor
Rc=Capacitive Reactance in ohms.
PROCEDURE
1. Set up the circuit shown below.
2. Measure and record the d-c resistance of the resistor and the capacitor.
R1= _____________Ohms Rc1= __________________ohms.
3. Set the frequency Generator for 500 Hz , 1000 Hz ,1500 H z one at a time.
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4. Measure and record the voltage across the resistor and capacitor at each of the frequencies listed.
5. Measure the current
6. Calculate Xc at each frequency.
Report Instruction
a. How does the voltage drop across the capacitor vary with frequency.b. Compare the calculated and experimentally determined values of the
capacitive reactance.
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Reactance in a parallel R-L circuit
Objective
To analyze the effects of variations in frequency on reactive (Reactance) in a parallel R-L circuit
Equipment Required
1. Audio Frequency Generator
2. Oscilloscope
3. Multimeter
4. Bread Board
5. Accessory Parts
a. Connecting wires
b. Inductor
c. Resistor
Introduction
Inductive reactance increases when either frequency or inductance increases. In contrast with the series circuit, the voltage across the resistor and inductor in the parallel R.L circuit is the same for a given frequency , but the currents are out of phase and generally not equal .In the parallel R-L circuit the value of R equals the maximum value of impedance in the series circuit, however, the value of R is equivalent to the minimum value of impedance .These factors help to determine whether a series or a parallel circuit is chosen as a filter .Because of the resistance offered by the winding of an inductor a purely inductive branch is never obtained in a parallel R-L circuit.
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Procedure
1. Set up the circuit shown below
1K 500 mh
2. Measure and record the d-c resistance of R1 and L1
R1= _____________Ohms L1= _____________Ohms.
3. Vary the Generator frequency and measure the current through the inductor.
4. How does current through the inductor vary with frequency?
5. Observe the wave shape across the inductor using oscilloscope. What happens to the wave shape when you change frequency.
6. Compare the calculated and experimentally determined values of current in the parallel R-L circuit.
XL= 2 Fl
I= EL
XL
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Reactance in a parallel R-C circuit
Objective
To analyze the effects of variations in frequency on reactance in a parallel R-C circuit.
Equipment Required
1. Oscilloscope
2. Signal generator and amplifier
3. Multimeter
4. Patching unit (Bread Board)
5. Accessory Parts
a. connecting wire
b. capacitor 0.1 µ F
c. Resistor 10 k Ω ½ watt
Introduction
A frequency change in a parallel R.C circuit produces a different effect from that created by a similar frequency change in a series circuit. As frequency increases from zero to infinity in a series R-c circuit, the total impedance goes from infinity to an ohmic value equivalent to that of the resistive branch. In the parallel circuit the total impedance ranges from value equivalent to that of the resistive branch to zero as frequency increases zero to infinity.
A common application of the parallel R-C circuit is as a by pass for alternating current. This circuit offers a low impedance to ac without affecting the required d-c levels.
Xc= 1/2+c
It= √ IR2+ Ic2
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Procedure
1. Set up the circuit shown below.4.7k Ω
R1
15k C1 0.1 µ F
R2
2. Measure and record the d.c resistance of R1 and R2.
3. Measure the current through the parallel branches.
4. Using the oscilloscope observe the wave shape across R1 on channel 1, and across c1 on channel Z.
5. Explain how current through the capacitor varies with the frequency.
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