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2205

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Electrical Engineering

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Apr 3, 2024

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McMaster University 1EL3 Lab 4 Series-Parallel Resistive Circuit Introduction In this experiment, the relationship between voltage and current in each circuit component as well as the series-parallel circuit are both being studied. A voltage source and five resistors arranged in series and parallel. The goal is to confirm Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL).

McMaster University 1EL3 Results Step 1 Colours Expected Resistance Measured Resistance R1 Red Black Brown 200 Ω 199.6 Ω R2 Brown 100 Ω 99.9 Ω

McMaster University 1EL3 Black Brown R3 Brown Red Brown 120 Ω 117.7 Ω R4 Brown Red Brown 120 Ω 118.4 Ω R5 Red Yellow Purple 240 Ω 239.6 Ω Step3,4 Theoretical Resistance Measured Resistance Equivalent Resistance 266.67Ω 267.9 Ω Step 6, 7 and 8 E (supply) V1 V2 V3 V4 V5 Measured Voltage 9.21 v 6.84 2.311 v 1.344 v 0.915 0.924 I T (supply) I1 I2 I3 I4 I5 Measured Current 35.55 mA 33.60 mA 22.45 mA 10.91 mA 7.58 mA 3.44 mA Part 2: Short circuit I T (supply) I3 Measured Current 45.00 mA 0 mA MultiSim Circuit Part I

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McMaster University 1EL3 Sample Calculation Part I Equivalent Resistance: - R T = R1+(R2||(R3+(R4||R5) -1 ) -1 ) -1 = 200 + (100||(200))= 266.667 Ω

McMaster University 1EL3 Analysis Questions 1. For the circuit in Figure 1.1, calculate the total resistance, resistors’ currents, resistors’ voltages, total circuit power, and power dissipated in each resistor. - R T = R1+(R2||(R3+(R4||R5) -1 ) -1 ) -1 = 200 + (100||(200))= 266.667 Ω Current: - I T = 9.00v/ 266.667Ω = 0.0337 A= 33.7 mA I 1 = I T = 33.7 mA - I 2 : (R4||R5) -1 +(R3) = 200 Ω CDR: (200 Ω / R2+ 200 Ω) x (33.7) = 22.467 mA - I 3 : R 2 = 100 Ω CDR: (100 Ω / 200 Ω + 100 Ω) x (33.7) = 11.233 mA - I 4 : CDR but I T = 11.233 mA I 4 = ( 240 Ω / (240 Ω +120 Ω)) x (11.233mA) = 7.487 mA - I 5 = (120 Ω / (240 Ω +120 Ω)) x (11.233mA) = 3.743 mA Voltage: - V 1 : simplify circuit until you have R1 in series with (R2||(R3+(R4||R5)= 66.67 Ω VDR: (200 Ω / (66.67 Ω + 200 Ω)) (9v) = 6.75 v - V 2 = I 2 x R 2 = 22.467 x 100 = 2.25 v - V 3 = I 3 x R 3 = 1.35 v - V 4 = I 4 x R 4 = 0.898 v - V 5 = I 5 x R 5 = 0.898 v Total Power - P = V T x I T = 9v x 33.7 mA = 303.3 mW Power dissipated in each resistor - P1 = V 1 x I 1 = 227.5 mW - P2 = 50.6 mW

McMaster University 1EL3 - P3 = 15.2 mW - P4 = 6.72 mW - P5 = 3.35 mW 2. From your readings of the voltages, is Kirchhoff's voltage law satisfied? Justify your answer. 3. From your readings of the currents, is Kirchhoff's current law satisfied? Justify your answer. 4. Calculate the theoretical values of the supply current (IT) and I3 when R2 is short circuited (Figure 2.1). Compare these values to the ones that you measured in step 10. Conclusion

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