In this experiment, we will be investigating the concepts of voltage and current division. Voltage and current division is an application of Kirchoff s Laws. Kirchoff s Voltage Law Kirchoff s Voltage Law (KVL) states that the sum of all voltage drops around any loop in any circuit sum to zero. In mathematical form, n v i = i= 1 0 (2.1) Where the v i in Equation (2.1) are the voltages across the individual components in any circuit. As an example of how to use Kirchoff s Voltage Law to solve a circuit, consider the circuit shown in Figure 2.1. V 1 R 1 V s R 2 R 3 V 2 V 3 Figure 2.1: Application of Kirchoff s Voltage Law. If we apply Equation (2.1) to the loop contained in the left half of Figure (2.1), and then the right half of the circuit, while traversing each loop in a clockwise direction, we obtain the two uations 0 = V V V s 1 2 0 = V2 V3 This approach produces two uations for which there are three unknowns. We could use the uation for the outer loop Figure 2.1 to obtain a system of three uations and three unknowns. This would work, but there is an easier way. Let us do away with all the voltages shown in Figure 2.1 except the source and define currents flowing in the two main loops. This is shown in Figure 2.2. Since we are defining the currents, we are free to choose their directions. Call the currents I 1 and I 2. 21
R 1 R 2 R 3 V s I 1 I 2 Figure 2.2: KVL Example. Let us now write uations by applying Kirchoff s Voltage Law for the left and right loops as before, but using our defined currents. Each loop is traversed in the clockwise direction. 0 = Vs R1I1 R2( I1 I2) 0 = R2( I2 I1) R3I2 We now have two uations with two unknowns that can easily be solved. To obtain the voltages as shown in Figure 2.1, just use Ohm s Law. These voltages are given as V1 = R1I1 V2 = R2( I1 I2) V = R I 3 3 2 Solving a circuit by defining your own currents allows you to configure the currents that would provide an easy solution. Once these are obtained, any quantity desired can be solved for. Also, by defining your own currents, you will make fewer mistakes, as it is easier to keep track of polarities of components in your uations. Kirchoff s Current Law Kirchoff s Current Law (KCL) is similar to his voltage law. It states that all currents entering or leaving any circuit node (connection point) sum to zero. Mathematically, this is n i i = i= 1 0 (2.2) To solve a circuit with Kirchoff s Current Law, consider the circuit shown in Figure 2.3 where we have arbitrarily chosen the currents as I 1, I 2, and I 3. 22
R 1 V x V s I 1 I 2 R 2 R 3 I 3 Figure 2.3: KCL Example. Applying Equation (2.2) at the node symbolized by V x, I1 = I2 I3 We can then substitute for the currents using Ohm s Law. This yields, V V R V V = R R s x x x 1 2 3 Thus we now have one uation with one unknown (V x ). This allows us to easily solve the circuit and obtain any circuit voltage or current desired. Equivalent Resistance Consider the circuit shown in Figure 2.4. R R 1 R 2 R 3 I V Figure 2.4: Resistors in Series. Since the resistors have the same current flowing through them, by Ohm s Law 23
V = I R I R I R V = I( R R R ) V = I R 1 2 3 1 2 3 Where R = R R R 1 2 3... Thus, we can conclude the uivalent resistance of any number of resistors in series is the sum of the resistances. For resistors in parallel, consider the circuit shown in Figure 2.5. I R V I 1 R 1 I 2 R 2 I 3 R 3 Figure 2.5 Resistors in Parallel. By Kirchoff s Current Law, I = I1 I2 I3 V V V I = R1 R2 R3 1 1 1 I = V R 1 R 2 R 3 V R = I 1 R = 1 1 1 R R R 1 2 3 The formula for the uivalent resistance of resistors in parallel that is given above can be extended to any number of resistors by adding another term to the denominator. If two resistors are in parallel, this formula can be reduced to a much more convenient form. Simplifying, the uivalent resistance of two resistors in parallel is given by 24
R (2 resistors in parallel) = R RR 1 2 R 1 2 Instructional Objectives 2.1 Take voltage readings at various points in a circuit. 2.2 Take current readings at various points in a circuit. 2.3 Explain the operation and function of a potentiometer. 2.4 Identify and verify voltage and current division configurations in a circuit. 2.5 Verify Kirchoff s Laws. Procedure 1. Adjust the DC power supply to output 10V. Obtain two resistors and one 27kΩ resistor as shown in Figure 2.6. Measure the actual values of these resistors and the power supplies voltage. Record your data in Table 2.1. Be sure to keep track of resistors and not mix them up since you have measured their values. V s 10V V 1 V 3 27kΩ V 2 Figure 2.6: Circuit to Verify Kirchoff s Voltage Law. 2. Measure the voltages as shown in Figure 2.6 and record their values in Table 2.1. 25
Component/ Nominal R (k:) R (k:) Voltage (V) Calculated Voltage (PreLab) (V) R 1, V 1 R 2, V 2 R 3, V 3 Table 2.1: Data for Figure 2.6. 3. Using your measured values for R 1, R 2, R 3, and V s, calculate the voltage drops for V 1, V 2, and V 3. Place these in the Actual Calculated Volts column in Table 2.2. Calculate the % error of your measured voltages to the prelab voltages (data in Table 2.1) using the prelab values as the accepted values. Also calculate the % error of your actual calculated values to the measured values (from Table 2.1). Use the measured values as the accepted values in your calculations. Place these results in Table 2.2. Component/ Actual Calculated Volts (V) % Error Calculated to % Error Actual Calculated to R 1, V 1 R 2, V 2 R 3, V 3 Table 2.2: Calculated Data and Errors for Figure 2.6. 4. The errors you obtained in Table 2.2 should be less than 5%. If they are not, try to find the reason why. 5. Construct the circuit shown in Figure 2.7. Measure the actual value of the supply voltage and the values of the resistors you use. V s 26
10V I 1 I 2 27kΩ I 3 Figure 2.7: Circuit to Verify Kirchoff s Current Law. 6. Repeat steps 2 and 3 for the currents I 1, I 2, and I 3. Record your data in Tables 2.3 and 2.4. Always turn off power when you are making changes in the circuit, such as moving the ammeter to measure a different current. Component/ Nominal R (k:) R (k:) Current (ma) Calculated Current (PreLab) (ma) R 1, I 1 R 2, I 2 R 3, I 3 Table 2.3: Data for Figure 2.7. Component/ Actual Calculated Amps (ma) % Error Calculated to % Error Actual Calculated to R 1, I 1 R 2, I 2 R 3, I 3 Table 2.4: Calculated Data and Errors for Figure 2.7. 27
7. In this part of the experiment, we will be investigating the properties of the potentiometer. The potentiometer is a variable resistor. A functional representation of a potentiometer is shown in Figure 2.8. Measure the resistance between the upper and lower terminals of the potentiometer. Record this value. Also record what happens to the measured resistance as you rotate the shaft of the device. Record what happens to the resistance between the wiper and the upper terminal and the wiper and lower terminal as the shaft of the pot is rotated. upper terminal center or "wiper" terminal lower terminal Figure 2.8: Potentiometer. 8. Construct the circuit shown in Figure 2.9. Adjust the potentiometer until V out is 1V. Record the value of V s and V out. V s V out V s =6V 9. Shut off the power and disconnect the power supply from the circuit. Carefully, without moving the arm of the potentiometer, measure the resistance between the upper and center terminal and then between the lower and center terminal. Record these values. V out Figure 2.9: Potentiometer Circuit. R uppercenter R lowercenter 28
10. This is a design problem. Your task is to design a circuit that will deliver 1.5V ± 5% across a load resistor of. Figure 2.10 demonstrates the problem. The resistors available for use are listed in Figure 2.10. As with most engineering design problems, there are constraints. In this case, your design must cost less than 20 cents. Each resistor costs 6 cents. Assume your time is free. Design and document a circuit that will meet the design criteria. Record your proposed solutions and brainstorm until you have found a solution. Construct your proposed circuit. Verify that it meets the design criteria by measuring V in and V out. When your circuit is working, demonstrate the design to the lab instructor. You may use more than one resistor of a particular value if you wish. Your Circuit Design Available Resistors V in = 5V 1kΩ 6.2kΩ 15kΩ 27kΩ R load V out = 1.5V / 5% 5% Figure 2.10: Circuit Design Schematic for Step 10. V in V out Post Lab Questions 2.1. Draw a schematic diagram similar to Figure 2.6 except label the elements with the measured resistance and computed voltage. Compare the voltages you calculated to the voltages you measured for the circuit in Figure 2.6. Explain why they may not be exactly ual. Verify that Kirchoff s Voltage Law applies to this circuit using your measured values. 2.2 Compare the currents you calculated to the currents you measured for Figure 2.7. Explain why they may not be exactly ual. Verify that Kirchoff s Current Law applies to this circuit. 2.3 For the potentiometer circuit of step 8 and 9, draw the uivalent voltage divider circuit and label the resistances with their actual measured values. Use the measured values and the voltage divider relationship to show that the potentiometer functions as a voltage divider. 29
2.4 Draw the circuit you designed in step 10. Explain the reasoning you used to get to your final solution and discuss how you verified that the circuit met the design criteria. 2.5 Assume that the supply voltage was exactly 5V and the resistors were ideal in step 10. For what range of R load will your circuit deliver 1.5V ± 5%? In other words, what is the maximum and minimum resistance of R load for which the circuit will still operate? 210
Name: Section: PreLab #3: Linearity, Proportionality, and Superposition 1. For the circuit shown in Figure 3.0a, calculate the proportionality constant that relates the output voltage to the input voltage, k=v out /V in. V in V out Figure 3.0a: Circuit for Problem 1. 2. Calculate the voltages V 5V and V 15V as shown in Figures 3.0b and 3.0c respectively. 3.9kΩ 4.7kΩ 3.9kΩ 4.7kΩ 5V V 5 V 6.2k Ω V 1 5V 6.2k Ω 15V Figure 3.0b: Circuit 1 for Problem 2. Figure 3.0c: Circuit 2 for Problem 2. 211