Exp : 8 Electrical CircuitEML 211

Title:Experiment 8 : Electrical Circuit

A. Measure Voltage, Current, Power & Resistance In The Direct Current (DC) Circuit

Objective

1. To measure the voltage, current, power and resistance using a volt and ammeter.2. To understand the theory in series and parallel connections in the direct current circuit.

Apparatus1. Multimeter2. DC power (100, 220, 470, 1K, 2.2K, 4.7K, 10K, 68K, 100K)3. Three test resistances (provided)

1. Serial Resistors

Figure (1)

Theory When two resistors, R1 and R2 are connected in the series as shown in the figure (1) above, the equivalent resistor, R12 in this circuit are:

R12 = R1 + R2 (1)

The current, through ammeter can be define as, when voltage, V is read using a volt meter (provided). (2)

Procedure

1.1 Section I

I. R1, R2 and V were chosen with suitable values and the value of R1 + R2 is estimated.II. Resistor R1 and R2 are connect in series as shown in figure(1) above.III. Current, is calculated.IV. Battery is connected to the circuit.V. Reading of both ammeter and voltmeter are recorded.V=4.8 V=0.015AVI. Verified if the current read on ammeter is equivalent to the estimated current using theory, ie, = ammeter reading =

1.2 Section III. Polarity of the battery is reversed.II. Reading on the ammeter and the voltmeter is recorded.III. A review is written.

2. Parallel Resistors

Figure (2)Theory

When two resistors, R1 and R2 are connected in series as shown in the figure (2) above, the equivalent resistor, R12 in this circuit are:

(3)

The current, thru R1 can be define as,

(4)The current, thru R2 can be defined as,

(5)The sum of the current from battery can be defined as:

(6)Procedure

I. V,R1 and R2 are selected with suitable values.II. Resistor R1 and R2 are connected in parallel shown in figure (2) above.III. Current that pass thru resistor R1 is calculated.IV. Current that pass thru resistor R2 is calculated.V. Reading on ammeter is observed and had been recorded.

3. Measurement of Voltage, Current, Power & Load Resistance Using A Voltmeter

Figure (3)

3.1 Measure the voltage across the load using a voltmeter.I. Voltmeter and resistor are connected as shown in figure (3).II. Reading on the voltmeter is taken and recorded.III. Step 1 and step 2 are repeated for three test resistors.

3.2 Measure the current though the load using a voltmeter.

Figure (4)

I. A known resistor (100) is connected in series with the unknown load resistor as shown in figure (4).II. Voltage reading on the voltmeter across the known resistor (100) is taken for three test resistive load.III. Current through the loads are calculated.

3.3 Measuring the power delivered to the load using a voltmeter.

Figure (5)I. A voltmeter is connected across the unknown load as shown in figure (5).II. Voltage reading, V1 is taken and recorded.III. Voltmeter is then connected to the known resistor load (100). IV. Voltage reading, V2 is taken and recorded.V. Power delivered to the load is calculated.

3.4 To measure the load resistance RL, using a voltmeter.I. Circuit shown in figure (5) is referred.II. By using the reading V1 and obtained from section 3.3, we estimated the unknown load resistance, RL.

4. Measurement of Voltage, Current, Power & Resistance Load Using an Ammeter

Figure (6)

4.1 Measuring the current through the load with an ammeter.I. Connection is figure (6) is connected.II. Ammeter reading is taken.III. The ammeter reading indicates the current through the load.

4.2 Measure the voltage across the load with an ammeter.

Figure (7)I. Connection in figure (7) is connected. Use the know resistor, R where the value is 100.II. Ammeter reading is recorded.III. Voltage across the load is estimated by using the formula IV.

(7)

4.3 Measuring the power delivered to the load with an ammeter.

Figure (8)

I. Connection in figure (8) is connected. Use the know resistor, R where the value is 100.II. Current thru the unknown load is called A1 and the value of it is recorded. Use the know resistor, R where the value is 100.III. The connection as shown in figure (9) is connected.

Figure (9)

IV. Current pass thru the know resistor is called A2 and it is been recordedV. Voltage across the load is calculated by using formula (8).

VR=A2R(8)

VI. Power delivered to the load ,PL is estimated by using formula (9)

PL= Voltage x Current= (A2R)(A1)=A1A2R Watt (9)

4.4 Measuring the load resistance, RL using an ammeter.I. Reading obtained from section 4.3 is used in this section.II. Resistance of the unknown load is estimated by using formula (10)(10)

5. Measurement of Voltage, Current, Power & Load Resistance Using a Voltmeter & Ammeter

I. Voltmeter, load and load are connected as shown in figure (10).

Figure (10)

II. Readings obtained from the ammeter as A and voltmeter as V recorded.

Current through the load = A

Voltage across load = V

Power delivered to the load = V x A (Watt)

Resistance of the load, R = Ohms

Result and Calculation 1. Serial Resistor

1.1 Section I

[i]R1 =100R2 =220V =5VR1 + R2 = 320[ii]Current obtained from calculation, = 15.6mA[iii]Voltmeter reading, V=4.8V[iv] Ammeter reading, =14.25mA[v] Theoretical ammeter reading, i, = 15.0mA[vi] Theoretical Voltmeter reading, V, = 5V

1.2 Section II I. Once the batterys polarity in reversed, the needle of multimeter was towards left (backwards) of zero value.II. There is current flow even the batterys polarity is changed. The ammeter and voltmeter needle shows towards (backwards) zero is due to negative value of current and voltage. This proved when we used the digital multimeter which showed a negative value. To solve this, we changed the polarity of ammeter and voltmeter which is similar to power supply. III. Reading of ammeter and voltmeter were taken, there isnt any difference from section (A), where the ammeter and voltmeter reading were 14.25mA and 4.8V respectively.IV. What I could learn from this is, the polarity of power supply (battery or etc) must be similar (positive to positive or negative to negative) to obtain a positive value of current and voltage. By the reversing the polarity, we could obtain a negative value which means the current flow in opposite directions.

2. Parallel Resistor[i]R1=100R2=220V=5V[ii]From calculation, =50mA[iii]From calculation, =22.7mA[iv]From ammeter reading, =44mA=21mA=65mAi = i1+i2= 72.7mA

3.Measurement of voltage, current, power and resistance of load using one voltmeter.

3.1. Measuring the voltage across load resistors using voltmeter.

[i]Readings of voltmeter from the 3 resistance test :

V1= 5.0VV2= 5.0VV3= 5.0V

3.2. Measuring current across load using voltmeter.

[ii]The reading of voltmeter across the 100 resistors for the 3 resistance of load tests:V1=0.9VV2=0.21VV3=0.1V

[iii]Current across the load :

I1= 9.0mA

I2= 2.1mA

I3= 1.0mA

3.3. Measuring power supplied to load using voltmeter.

[i] V1 = 4.9V ( R = 4.7k)

V2 = 0.1V ( R = 100)

[ii]Power supplied to load :

= 4.9mW

3.4. Measuring load resistor ,RL, using voltmeter.

[iii]

4. Measurement of voltage, current, power & resistance load using an ammeter.

4.1Measuring the current through the load with an ammeter.

[i]Ammeter reading

I = 45mA

4.2. Measure the voltage across the load with an ammeter[ii]Ammeter reading

I = 45mA

[iii]Estimate the voltage, V = IR

V = IR

V= 45(10-3)x 100

V=4.5V

4.3 Current thru the unknown load,A1

[i]A1 = 9.6mA

[ii]Current through the known resistor(100), A2

A2 = 5.6mA

[iii]Voltage across the load, VR= A2R

VR = 5.6(10-3) X 100 VR = 0.56V

[iv]Power delivered to the load, PL= Voltage x Current = A1A2R

PL = 9.6(10-3) X 5.6(10-3) X 100 PL = 5.4mW

4.4. Measuring the load resistance, RL using an ammeter.

[i]RL = Ohm RL =

RL = 58.3

5. Measurement of voltage, current, power & load resistance using a voltmeter & ammeter.

[i]Current through the load = 48mA

Voltage across load = 4.8V

Power delivered to the = V x A Watt= 0.23W

Resistance of the load, R = Ohms = = 100

Discussion

1. Experiment 1.1 Section (I)

[i] The ammeter reading obtained from experiment is i = 14.25mA which is near to the theoretical reading i = 15mA.

[ii] The percentage of deviation = = 5% which is relatively small.

2. Experiment 1.1 Section (II)

[i] When the battery polarity is reversed, the voltmeter and ammeter reading is reversed also. To solve this problem, reversed the polarity of ammeter and voltmeter is required so that the needle of both the ammeter and voltmeter can function properly.

[ii] The value of the ammeter and voltmeter reading is almost the same as the value obtained from Section (I). The reading is V= 4.8V and =14.25mA.

3. Experiment 2.

[i] Experimental result is meanwhile theoretical result is

[ii] The percentage deviation is = 10.59%

The error could be the(i) Poor calibration caused the inaccuracy of the reading. Correct scale should be used to measure ammeter and voltage reading, not too big or not too small. (ii)Parallax error caused by the observer using incorrect eye level to observe the reading of the instrument. The position of observers eye should be perpendicular to the reading scale.(ii) The overheating of the circuit and resistor which increase the resistance for current flow.

4. Precautionary steps to overcome the errors faced. The observers eye must be perpendicular to reading scale. The reflection of the needle in ammeter or voltmeter on the mirror and needle should overlap to obtain accurate reading. Run the experiment and obtain the results for a few times and get the average reading to obtain accurate reading. Ensure the power supply is switched when its not in use. This is to reduce the chances of getting overheat which would spoil the reading. Avoid direct sunlight contact or heating element which can increase the surrounding temperature, maintain room temperature at 25 degree Celcius. Adjust the range of ammeter and voltmeter to the suitable value to avoid multimeter to malfunction. Start from high range then reduce to suitable range one by one.

Conclusion

1.In series arrangement, [i]The equivalent resistance of R1 and R2 is R1+R2[ii]Current pass through all resistors are the same.[iii]The value of voltage drop across the resistor is proportional to the resistance and follow the voltage division rule.

2.In parallel arrangement, [i]The equivalent resistance of R1 and R2 is .[ii]The current pass through the resistors follow the current division rule.[iii]The voltage drop across the all resistors are the same.

References1. EML 211/2 Engineering Laboratory Manual , Dr.Elmi Abu Bakar, Assoc.ProfDr.Roslan Ahmad2. Electric circuit analysis, by David E Johnson, 1999 year3. Electric circuit theory, by R. Yorke, 1986 year

II Superposition Theorem, Thevenin and Norton

Theories :

1. Superposition theoremThe superposition theorem states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to each independent source acting alone. This theorem is a way to analyse a linear circuit with more than one independent source by calculating the contribution of each source separately. In order to negate all but one power source for analysis, replace all independent voltage sources with short circuit and replace all independent current sources with open circuit.2. Thevenin theoremThevenin theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTH in series with a resistor RTH , where VTH is the open-circuit voltage source at the terminals and RTH is the input or equivalent resistance at the terminals when the independent sources are turned off. When applying the Thevenin theorem, all independent voltage sources will be short circuited and all independent current sources will be open circuited. By replacing the whole circuit to a simple circuit with one voltage source and one resistance, calculation on the load resistor will be easier.

3. Norton theoremNorton theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit with an independent circuit source, IN in parallel with a resistance, RN. IN is the current pass through the two terminals and RN is the equivalent resistance at the terminal when the independent source is turned off. The method of applying Norton theorem is almost the same but only Thevenin theorem is used to calculate voltage but Norton theorem is used to calculate current.

Objectives

1.Introduce fundamental theorems commonly used to simplified a circuit for the purpose of evaluating or designing2.Understand the methods or usage of this basic theorems in practice in a circuit or network of direct current

Apparatus

1.DC power supply (2)2.Multimeter (3)3.Resistor (100, 470, 1k, 1.5k, 4.7k)

Introduction

To understand if the way of a circuit is r the purpose of designing or analyzing, and commonly used technique to facilitate the circuit for an univalent circuit. The three fundamental theorems usual used for the simplification of a circuit are:[i]Superposition Theorem[ii]Norton Theorem and[iii]Thevenin Theorem

1.0 Superposition Theorem

Objective

To learn the concept of superposition theorem in the network circuit fixed torrent. Theory Consider the circuit network shown in Figure(1).

Figure(1)

Remove the sources V1 and the network circuit above will be as shown in Figure(2).

Figure(2)Since R1 and R2 are parallelThus , R1||R2 = R12 = Therefore, the network circuit can be summarised as shown in Figure(3).

Figure(3)Thus, Referring to Figure(2), , And ,Referring back to Figure(1), now removed only V3. The circuit to be is as shown in Figure(4).

Figure(4)Now, R2 and R3wil be parallel resistors. Thus, R2||R3 = R23 =

Therefore, the network circuit can be summarised as shown in figure(5).

Figure(5)Therefore,

Also,

AndThus, when both V1 and V2 areconnected as shown in Figure(5) above, t current through the resistor R1 are defined asThe current though resistor R2 are define as The current though resistor R3 are defined as If , the current flow direction will follow the s flow direction.

2.0Thevenin & Norton Theorem

Objective 1. To learn the concept of Thevenin and Norton theorem in sample network.2.Learn the concept of an ideal and actual current source.

2.1Thevenin Theorem

Procedure1.Thevenin equivalent circuit and its equivalent values for the circuit network given in Figure(6) is connected. The equivalent circuit is drawing the space provided.

Figure (6)

2.Power supply is connected to the circuit above but do not connect RL. Voltage across the terminal XY is measured and recorded. This is the Thevenin voltage, ETH.EOC=ETH3.Power supply is removed and the resistance across x-y is measured. Value of resistance is recorded. This is the Thevenin equivalent resistance RTH.RSC=RNO =RTH4. Voltage drop across the RL load is calculated after it is connected, the following formula is used.

5.Power supply s connected with the RL load in the network. The voltage drop across the load resistor is measured and recorded.6.Step 1 to Step 5 are repeated for the circuit in Figure(7) by using Thevenin s theorem.

Figure(7)

2.2Norton Theorem

1.Norton equivalent circuit is drawn for the circuit in Figure(6).

2.Power supply is connected to the circuit. Load resistor is replaced with an ammeter(0-10)mA. Ensure the meter is connected to the correct polarization. Current is recorded as ISC. This current is the Norton current.ISC=IN=

Note: Resistance range is equal to RN that has been measured before.

3.Voltage drop across the RL from the Norton equivalent circuit is calculated byusingthe following formulas

4.RL load is connected and voltage drop is measured.5.Step 1 to Step 5 are repeated for the circuit in Figure(7) by using Nortons s theorem.

3.0ExerciseFor the circuit shown below,

Figure(8)1.Get a Thevenin equivalent Circuit.

2. Calculate the voltage AB, E1 by opened the current source.

3. Calculate the voltage AB, E2 by closed the circuit.Therefore, the equivalent Thevenin voltage is , ETH.ETH = E1+E2

4.Now, by removing the power supply and closing circuit at AB, calculate the equivalent resistor, RTH.

Result and Calculation

A. Superposition Theorem Experimental value: () () ()

Figure 11.602.500.90

Figure 2-1.000.601.60

Figure 42.501.900.60

Table 1

Theoretical value (referring to figure 2):

Theoretical value (referring to figure 4):

Overall theoretical value:

Thevenin Theorem

For Figure(6)

1.Thevenin voltage, EOC =ETH =4V2.Thevenin resistance, RSC = RNO = RTH = 6003.Voltage drop across the RL by calculation,

4.Measured voltage drop across load RL,

For Figure(7)

5.Thevenin voltage, EOC = ETH = 7.8V6.Thevenin resistance, RSC = RNO = RTH = 4007.Voltage drop across the RL by calculation,

8.Measured voltage drop across load RL,

2.2Norton Theorem

For Figure (6)

9.Norton current, ISC = IN = 6.7mA10.Voltage drop across the RL

11.Measured voltage drop across load RL,

For Figure (7)

12.Norton current, ISC = IN = 10.5mA13.Voltage drop across the RL

14.Measured voltage drop across load RL,

Discussion

1. Based on the experimental, the experimental results obtained were almost the same compared to theoretical values with a difference in range of 0.1 to 0.5.

For Thevenin Theorem experiment, the value of measured is 1.8 which is the same as the value calculated of the voltage drop across the load when it is connected.

For Norton theorem experiment, the value of measured is 1.8 which is higher compared to the calculated value ( 1.5 ) of the voltage drop across the from the Norton equivalent circuit.

.The measured value is different from calculation value because there are some error:

The resistance value of the resistor is not exactly the same as shown by the body colour of the resistor, there are some deviation. So, the resistance we using in the calculation is possible to be higher or lower than actual.

The apparatus used (ammeter and voltmeter) may consists calibration error.

The parallax error caused by the observer.

Some precautions are needed in this experiment:

Do not connect the circuit with the power supply for a long time because this will cause the overheating of wire.

When take the reading from the ammeter or voltmeter, must ensure that the eye level is correctly with the scale.

Connect the resistors properly to the circuit.

Conclusion

Throughout the experiment, we learnt usage and fundamentals therorems of Superposition,Thevenin and Norton are applied in a circuit. There is a small difference in experimental and theoretical results due to some errors occurred but since the percentage of error is small then it is satisfying.

References

1. EML 211/2 Engineering Laboratory Manual , Dr.Elmi Abu Bakar, Assoc.Prof Dr.Roslan Ahmad2. http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Source/Source2.html3. http://www.accesscomms.com.au/reference/polarity.htm

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