Sirindhorn International Institute of Technology Thammasat University at Rangsit


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1 Sirindhorn International Institute of Technology Thammasat University at Rangsit School of Information, Computer and Communication Technology COURSE : ECS 304 Basic Electrical Engineering Lab INSTRUCTOR : Dr. Prapun Suksompong WEB SITE : EXPERIMENT : 01 DC Measurements I. OBJECTIVES 1. To measure dc voltage and current in electrical circuits.. To verify Ohm s law. 3. To study voltagedivider and currentdivider circuits. 4. To verify Kirchhoff s voltage and current laws. II. BASIC INFORMATION 1. The amount of current in a circuit depends on the amount of voltage applied to the circuit and the nature of the conductive path. Ammeters and voltmeters are used to measure current and voltage, respectively. Recall from Experiment 00 that, to measure a current, the ammeter terminals must be connected in series with the current to be measured. In measuring a voltage, the voltmeter terminals must be connected in parallel with the voltage to be measured. Connecting either an ammeter or a voltmeter into a circuit may disturb the circuit in which the measurement is being made. It is therefore important to be aware of these possible meterloading effects.. The voltage across many types of conducting materials is directly proportional to the current flowing through the material. The constant of the proportionality is called resistance. This fact was discovered by George Simon Ohm, a German physicist, and it is thus referred to as Ohm s law. Resistors are one of the most commonly used elements in electric and electronic circuits. Materials often used in fabricating resistors are metallic alloys and carbon compounds. 3. Practical resistors are specified by their nominal resistance value and tolerance value. The most common type of resistors is the carbon composition or carbon film resistor. Resistors are limited in their ability to dissipate heat. Their power ratings commercially available are 1/8, 1/4, 1/, 1, and W. Since heatdissipation capability is related to the surface area, resistors with larger power ratings are physically larger than those with smaller ratings. Ohmmeters are used to measure resistance. It should be noted the resistor
2 must not be connected to any circuit at the time its resistance is being measured by an ohmmeter. Otherwise, the resistance value read by the ohmmeter will be incorrect. 4. By definition, an electric circuit is an interconnection of electrical elements. The interconnection of the elements, however, imposes constraints on the relationships between the terminal voltages and currents. The two laws that state the constraints in mathematical form are known as Kirchhoff s current law (KCL) and Kirchhoff s voltage law (KVL). These two laws are named after Gustav Kirchhoff, a German physicist, and can be stated as follows. KCL: For any electric circuits, the algebraic sum of all branch currents at any node is zero. KVL: For any electric circuits, the algebraic sum of all voltages around any loop is zero. 5. Ohm s law and Kirchhoff s laws find immediate applications in designing voltagedivider and currentdivider circuits. A simple voltagedivider circuit consists of two resistors connected in series with a voltage source. The voltage divides between the two resistors in series such that the voltage across either resistor equals the source voltage multiplied by the resistance of that resistor and divided by the sum of the resistances. A simple currentdivider circuit, on the other hand, consists of two resistors connected in parallel across a current source. The current divides between two resistors in parallel such that the current in either resistor equals the source current multiplied by the resistance of the other branch and divided by the sum of the resistances. The following gives detail information of the concepts mentioned above. 1. Ohm s law i v R Figure 11: The representation of voltage and current of a resistor. From the representation of voltage and current of a resistor in Figure 11, we have v = i R, where v is the voltage across the resistor in volts, i is the current flowing through the resistor in amperes, and R is the resistance in ohms.
3 . Kirchhoff s current law (KCL) For any electric circuits, the algebraic sum of all branch currents at any node is zero. Let s consider the circuit in Figure 1. The total current I T flows from the source into node A. At this node, the current splits into three branches as indicated in the figure. Each of the three branch currents, I 1, I, and I 3, flows out of node A. KCL states that the total current into node A is equal to the total current out node A, that is, I T = I 1 I I 3 I 1 I T I I T A I 3 B V ps Figure 1: Total current into node A is equal to the sum of currents out of node A. The sum of currents into node B equals the total current out of node B. Now, following the currents through the three branches, we can see that they come back together at node B, where the current I T flows out. KCL applied at this node, therefore, gives the same equation as that at node A. 3. Kirchhoff s voltage law (KVL) For any electric circuits, the algebraic sum of the branch voltages around any loop is zero. For example, in the circuit shown in Figure 13, there are three voltage drops (denoted by V 1, V, and V 3 ) and one voltage source (Vs). By KVL, we can algebraically sum all of the voltages around the circuit to obtain V 1 V V 3 Vs = 0 V 1  V s  V   V 3 3
4 Figure 13: The sum of the voltage drops equals the source voltage, Vs. Alternatively, the previous equation can be expressed by moving the voltage drop terms to the right side of the equation as follows. Vs = V 1 V V 3 In other words, the source voltage is equal to the sum of all voltage drops. 4. The voltagedivider circuit Let s see the circuit in Figure 14. i V s R 1 R V 1 V Figure 14: The voltagedivider circuit. Applying KVL around the closed loop yields Vs =i(r 1R ) or Vs i = (R R ) Next, we use Ohm s law to calculate V 1. 1 Vs R V = ir = R = Vs (R R ) (R R ) From the equation, we can see that V 1 is a fraction of Vs, where the fraction is equal to the ratio of R 1 to R 1 R. Similarly, we can have Vs R V = ir = R = Vs (R R ) (R R ) 1 1 4
5 5. The currentdivider circuit The currentdivider circuit, as shown in Figure 15, consists of two resistors connected in parallel across a current source. i s R 1 i 1 R i v  Figure 15: The currentdivider circuit. The currentdivider circuit divides the source current i s between R 1 and R. We find the relationship between i s and the current in each resistor (i 1 and i ) by directly applying Ohm s law and KCL. First, the voltage across the parallel resistors is By KCL, i s =i 1 i. Therefore v= i1 R 1=iR which gives v v 1 1 i s = v, R1 R R1 R v is 1 1 R R 1 Thus, we have 1 v R1 R i 1= = i s = is R (R 1R ) R R 1 1 v R R1 i = = i s = is R 1 1 (R 1R ) R R 1 5
6 III. MATERIALS REQUIRED  DC power supply  Multimeter  Resistors (1/W): 330, 470, 80, 1k, 1.k, k,.k, 3.3k, 4.7k, and 10k. IV. PROCEDURE Part A: Verification of Ohm's law 1. Turn on the dc power supply.. Use a DMM to measure the output voltage (Vps) of the power supply, and adjust the voltage until the meter reads 10 V. 3. Measure the resistance of a resistor R = k, and record the result in Table Connect the circuit as in Figure 16. Measure the current I flowing through the resistor, and record the result in Table 11. ammeter A  Vps  V I voltmeter R = k Figure 16: The circuit for verifying Ohm's law. 5. Vary Vps to 15 V, 0 V, and 5V, and record the currents corresponding to each case in Table Calculate the value of I in each of the above cases, and record the results in Table
7 Part B: Voltagedivider circuit 1. Let R1 = 10 k and R = k. Measure the resistance of R 1 and R, and record the results in Table 1.. Turn on the power supply. 3. Connect a multimeter across the power supply, and adjust the output voltage Vps of the supply so that Vps = 1V. 4. Connect the circuit in Figure 17. A V ps = 1 V R 1 V 1 R Figure 17: A voltagedivider circuit. 5. Measure the voltage across R1; this is the voltage V1 in the figure. Record the result in measured value of V1 in Table Similarly, measure V in the figure, and record the value in measured value of V in Table Use the measured values of V ps, R 1, and R to calculate V 1 and V by voltagedivider rules: C V V1=V ps R1/RT V =V ps R/RT where RT is the total resistance R1R. Record the values in "calculated value" in Table 1. 7
8 Part C: Currentdivider circuit 1. Let R 1 =. k, R = 3.3 k, and R 3 = 4.7 k. Measure the resistance of each resistor, and record the values in Table 13. Turn on the power supply, and adjust the supply so that the output voltage Vps = 15 V. 3. Connect the circuit as in Figure 18. I T A R 1 R R 3 V ps V I 1 I I 3 A A A Figure 18: A currentdivider circuit. 4. Measure IT, I1, I, and I3 using a multimeter. Record the values in Table Calculate I 1, I, and I 3 by currentdivider rules: I1 = IT RT /R1 I = IT RT /R I3 = IT RT /R3 IT = Vps/RT where RT is the total resistance R1 // R // R3. Record the results in calculated value in Table
9 Part D: Verification of Kirchhoff s laws 1. Turn on the dc power supply, and adjust the supply so that the output voltage Vps = 15 V. Record this value in Vps (measured) in Table Connect the circuit shown in Figure 19. V 1 V V 3 V 4 V 5 R R ohms I 1 R ohms R 4 R 6 00 ohms R ohms Vps V 330 ohms R 3 80 ohms 1000 ohms R ohms I 8 Figure 19: The circuit for verifying Kirchhoff s laws. 3. Measure the voltages V1, V, V3, V4, and V5, and record the values in Table Calculate Vps (= V1 V V3 V4 V5), and record the result in Vps (calculated) in Table Measure I1, which is the current flowing through R1, and record the value in Table Measure the currents I, I3, I4, I5, I6, I7, and I8, which are currents flowing through R R3, R4, R5, R6, R7, and R8, respectively. Record the values in Table Turn off the power supply. 8. Calculate the sum of I and I3, and record the result in the table. Check your result by comparing the value with I4. 9. Calculate the sum of I5, I6, and I7, and record the values in Table 14. Check your result by comparing it with I Measure the resistance of each resistor, and calculate V1, V, V3, V4, and V5 by using Ohm s law (e.g., V 1 =I 1 R 1 ), and record your results in Table
10 TABLE 11: Verification of Ohm s law R = V (volts) I (amps) Calculated I (amps) TABLE 1: Voltagedivider circuit TA Signature: R 1 = R = Vps V1 V Measured value Calculated value N/A TABLE 13: Currentdivider circuit TA Signature: R 1 = R = R 3 = Measured value Vps I1 I I3 IT Calculated value N/A TABLE 14: Verification of Kirchhoff s laws TA Signature: V PS = (measured). Measured value V PS = (calculated). Calculated value V1 V V3 V4 V5 V1 V V3 V4 V5 Measured value Calculated value I1 I I3 I4 I5 I6 I7 I8 I I3 I5I6I7 10 TA Signature:
11 QUESTIONS: Choose a correct answer for each question. 1. When an additional resistor is connected across an existing parallel connection of resistors, the total resistance. (a) decreases. (b) increases. (c) remains the same. (d) increases by the value of the added resistor.. In a series connection of resistors, each resistor has. (a) the same current. (b) the same voltage. (c) the same power. (d) all of the above. Fill in the blanks with a correct answer. 1. Loop equations are written by applying to each loop in a circuit.. When a 1. k resistor and a 100 resistor are connected in parallel, the total resistance is. When they are connected in series, the total resistance is. 3. A good ammeter should have (value) internal resistance. 4. To calculate the current flowing through a resistor, a voltmeter is connected in with that resistor, and the relationship between the measured voltage and can be used to determine the current value. 5. A good voltmeter should have (value) internal resistance. True or False. 1. A constant voltage is provided by a constant current source across its terminals under all load conditions.. The total resistance of a parallel connection of R1 and R is greater than R1 R. 11
12 Analyze the circuit. I R Vps R 1 I 1 A load 1 0. A 5 V load 0.4 A 0 V Figure 110: The circuit used in the analysis. A voltagedivider circuit in Figure 110, with V ps = 0 V, can supply two loads: load 1 is 0. A at 5 V, and load is 0.4 A at 0 V. Assume that the current I 1 is 0.1 A. The value of R 1 and R must be The resistance of each load is: R 1 =, R =. load 1 s resistance =, load s resistance =. The current I = A. Show the detail of your analysis in the report. 1
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