ELECTRICAL ELECTRONIC PRINCIPLES EXPERIMENT #2 SERIES-PARALLEL DC CIRCUITS INTRODUCTION Using Problem Based Learning Most circuits are combinations of series and parallel circuits. This laboratory experiment will introduce students to problem

solving of these combination circuits. Since the concepts of series- parallel circuits have not been discussed this experiment is one using problem-based learning: learning concepts through experimentation. EQUIPMENT Resistors: 1-680Ω, 1-1ΚΩ, 1-1.8ΚΩ, 2-2.2ΚΩ, 1-3.3ΚΩ Digital Multi meter DC Power Supply PROCEDURE 3.3K 2.2K 1. Check the color codes of the resistors to insure that you are selecting the proper ones. BEFORE YOU PLACE THEM IN THE CIRCUIT measure the individual resistances and place the values in Table 1. 2. Calculate values in Table 1; begin with calculating the total resistance and total current. Use the nominal values to calculate circuit parameters. Should R6 be used in the calculation if it is an open circuit with no current flows through it? Why? Why not? (Note: Follow the current as it flows out of the source through the first resistor, causing a voltage drop across the resistor, R1) 3. Using current divider and/or voltage divider rule and/or Ohm's Law, calculate the voltage at point A, point B and the voltage between the two points, VAB = (VA - VB) 4. Calculate the voltage at point C. What type of circuit is point C? Does current flow through it? If no current flows through it is there a voltage across R6? 5. Draw the circuit using MultiSim. Use the simulation to verify the current, voltage and resistance values specified in the Data Table 1. 6. Construct the circuit in figure 1, without power. Do not wire DC power! Use the ohmmeter to measure the total resistance of the circuit before you wire the 10 volt supply to the circuit. 7. Set the DC power supply for 10 volts. Wire the DC supply to the circuit and measure the voltage across the source and across each resistor. From your measurements is Kirchoff's voltage law satisfied around the path? Which path? Explain. 8. Measure the total current in the circuit and the current through the resistors in the Table. (Remember: the Ammeter must be placed in series with the resistors, not in parallel). Do the currents in the branches add up to the total current? Is Kirchoff Current Law satisfied? Remember: As you prepare your report, think about the theories and laws this circuit proves. The Ojective used should be specific to this circuit. Make sure you include all of the parts of the report (no graph required.) The questions above are discussion points for your lab. TABLE 1 PARAMETER R₁ 680 R₂ 1K R3 1.8K R4 2.2K R5 3.3K R6 2.2K CALCULATED VALUES MEASURED And Color Code 65922 994.722 1.79 k 2.187 KS2 3.2731-2 2.2 25K RT 2.535k 2.503 V₁ 2.6181 2.68V V2 2.61V 2.612V V3 4.70 V 4.702V VA 2.926V 2.936 V V5 4.389 V 4.383 V V6 7.32V 7.3131 ΣV Around a loop: Pick a loop, then pick the second loop ESOURCE lov VA 4.71V VB 4.394V 9.937V 4.702 V 4.383V VAB 0.316 Vor 316 MV 0.319 V or 319MV Vc 9.32V I₁ 3.94 MA 7.313 V 3.97 MA 12 2.61 MA 2.63MA 4 1.33 MA 1.32 MA ITotal** 3.97 MA 3.94MA *Note: Think about Kirchoff's Voltage Law **Note: Think about Kirchoff's Current Law Use the nominal value of resistors for the "calculated" column. Single subscripted variables refer to the resistor number, e.g. the current through R4 is 14. What is the name for point C? V₂ = (123) (R2) = (2.61 MA) (1K√2) = 2.61V V3= (12,3) (R3) = (2.61MA) (1.8K-2) = 4.70 V Series-Parallel Circuits V4 (14,s) (Ru) = (1.33MA) (2.2K√2) = 2.926V Vs = (I4,5) (Rs) = (1.33MA) (3.3k-2) = 4.389V I6=0A R6 W-VC 2.2ΚΩ VC = ++ 7.32V R1 VI2,3 = 2.61MA √145=1.33MA w 680 Ω R2 R4 He 1.21-2 2.2ΚΩ Esource 10 V IKR VAB= 4.71V A 4.394V B R₁ 6802 R3 1.8kQ R5 3.3 ΚΩ IT3.94MA R₁ To V 680 5.5k 2 R2-5 1.855kQ ↓I2,3 R2, 3 7812 2.8k 2 Iчs R4,5 H+ R2-5 R2-5 = E2,3 + 1855 R2, 3 R4,5 + 2.8k 5.5k 7.32-2.61MA 2.8k 2 12,3 = EU39= V₁ = I + (R) V₁ = (3.94mA) (680-2) V₁ = 2.68 V 3.94mA = I₁ = IT" 14,5 = E4,5 7.32 V-1.33mA R4,5 = 5.5k V 2,3 = 4,5 = VT - V₁ = 7.32 V RT = R1-5 = 2.535 KR R2-5=1.855K-R =10V RT 2.535k Er lov IT R = 2.535K-R IF I₁ = 3.94MA