Parallel Circuits. Objectives: Summary of Theory:

Transcription

1 Parallel Circuits Objectives: 1. Demonstrate that the total resistance in a parallel circuit decreases as resistors are added. 2. Compute and measure resistance and currents in parallel circuits. 3. Explain how to troubleshoot parallel circuits. Summary of Theory: a parallel circuit is one in which there is more than one path for current to flow, it can be thought of as two parallel lines, representing conductors, with a voltage source and components connected between the lines. This idea is illustrated in figure 9-1.The source voltage appears across each component. Each path for current is called a branch. The current in any branch is dependent only on the resistance of that branch and the source voltage. As more branches are added to a parallel circuit, the total resistance decreases. If the total current in a circuit increases, with no change in source voltage, the total resistance must decrease according to

2 ohm s law. When connected in parallel the voltage is the same across all branches but the current is divided at branches. E = V1 = V2 =V3 1/RT = 1/R1 + 1/R2 + 1/R3 1/RT = R2R3/R1R2R3 + R1R3/R1R2R3 + R1R2/R1R2R3 RT = R1R2R3/ ((R2+R3) + (R1R3) + (R1R2)) If R1 = R2 = R3. RT = R 3 /3R 2 = R/3 In Parallel: RT< Rmin Kirchhoff s current law: The sum of the currents entering a circuit junction is equal to the sum of the currents leaving the junction. The current leaving the source must be equal to the sum of the individual branch currents. While Kirchhoff s voltage law is developed in the study of series circuits and the current law is developed in the study of parallel circuits, both laws are applicable to any circuit. In parallel circuit, it is the current that is divided between the resistances. Keep in mind that the larger the resistance, the smaller the current. The general current divider rule.

3 In parallel: IT = I1 + I2 RT = R1R2/ (R1+R2) V1 = V2 = E I1 = IT I2 I1 = V1/R1 = E/R1 = ITRT/R1 = IT (R1R2/R1+R2)/R1 = ITR2/R1+R2 I2 = ITR1/R1+R2 So Ii = ITRj/ Ri+Rj (This is the equation of current divider rule). Materials Needed: Resistors: one 2.2kΩ, one 2.7kΩ, one 1kΩ, one 0.2kΩ. One dc ammeter, 0-10 ma. Procedure: 1. Obtain the resistors listed in Table 9-1. Measure and record the value of each resistor.

4 Table 9-1 Component Listed Value Measured Value R1 = 2.2kΩ 2.19kΩ R2 = 2.7kΩ 2.684kΩ R3 = 1kΩ 1.002kΩ R4 = 0.2kΩ 0.219kΩ 2. In Table 9-2 you will tabulate the total resistance as resistors are added in parallel. Enter the measured value of R1 in the table. Then connect R2 in parallel with R1 and measure the total resistance as shown in figure 9-3. Enter the measured resistance of R1 in parallel with R2 in Table 9-2. Table 9-2 R1 R1 R2 R1 R2 R3 R1 R2 R3 R4 RT(measured) 2.19kΩ 1.205kΩ 0.547kΩ 0.156kΩ IT(measured) 5.45mA 9.91mA 21.88mA 77.1mA 3. Add R3 in parallel with R1 and R2. Measure the parallel resistance of all three resistors. Then add R4 in parallel with the other three resistors and repeat the measurement. Record your results in Table Complete the parallel circuits by adding the voltage source and the ammeter as shown in figure 9-4. Measure the total current in the circuit and record it in Table 9-2.

5 5. Measure the voltage across each resistor. How does the voltage across each resistor compare to the source voltage? 6. Use ohm s law to compute the branch current in each resistor. Use the source voltage and the measured resistances. Tabulate the computed currents in Table 9-3. Table 9-3 I1 = Vs/R1 I2 = Vs/R2 I3 = Vs/R3 I4 = Vs/R4 I (computed) ma ma ma Use the general current divider rule to compute the current in each branch. Use the total current and total resistance that you recorded in Table 9-2. Compare the calculation using the current divider rule with the results using ohm s law. Show your results in Table 9-4. Table 9-4 I1 = RTIT/R1 I2 = RTIT/R2 I3 = RTIT/R3 I4 = RTIT/R4 I (computed) 5.45 ma ma ma ma The same result. 8. Simulate a burned-out resistor by removing R4 from the circuit. What is the new total current? IT = ma Conclusion: When connect the circuit in parallel the voltage is the same across all branches but the current is divided at branches.

6 Total resistance less than the smallest resistance in the circuit.

7 10 Series- Parallel Combination Circuits Objectives: 1. Use the concept of equivalent circuits to simplify series-parallel circuit analysis. 2. Compute the currents and voltages in a series-parallel combination circuit and verify your computation with circuit measurements. Summary of theory: Many circuits can be analyzed by applying the ideas developed for series and parallel circuits to them. In this experiment, the circuit elements are connected in composite circuits containing both series and parallel combinations. The key to solving these circuits is to form equivalent circuits from the series or parallel elements. The components that are in series or parallel may be replaced with an equivalent component. For example, in figure 10-1 (a) we see that the identical current must flow through both R2 and R3. We conclude that these resistors are in series and could be replaced by an equivalent resistor equal to their sum. Figure 10-1(b) illustrates this idea.

8 Materials Needed: Resistors: one 2.2kΩ, one 2.7kΩ, one 1kΩ, one 0.2kΩ. Procedure: 1. Measure and record the actual values of the four resistors listed in Table Table 10-1 Component Listed Value Measured Value R1 = 2.2kΩ 2.19kΩ R2 = 2.7kΩ 2.684kΩ R3 = 1kΩ 1.002kΩ R4 = 0.2kΩ 0.219kΩ 2. Connect the circuit shown in figures Then answer the following questions.

9 (a) Are there any resistors for which the identical current will flow through the resistors? Answer yes or no for each resistors: R1 Yes, R2 No, R3 No, R4 Yes. (b) Does any resistor have both ends directly to both ends of another resistor? Answer yes or no for each resistors: R1 Yes, R2 Yes, R3 Yes, R4 Yes. 3. Compute the total resistance of this equivalent circuit and enter it in the first two columns of Table Then disconnect the power supply and measure the total resistance to confirm your calculation. Table 10-2 Computed Voltage Divider Ohm s Law Measured RT 3.137KΩ 3.137KΩ 3.137kΩ IT 3.825mA V V V V V2,3 3.8 V V V V V I2 I3 VT 12.0 V 12.0 V V