ECOR1043_ Circuits Lab 3 - Group 14 (1)

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Lab #3: Kirchhoff’s Laws & Loop/Nodal Analysis ECOR1043: Circuits Lab Group No: 1043 5.2 Hardware Measurements R 1 = 119.2 Ω R 2 = 267.0 Ω R 3 = 325.1Ω R 4 =98.8 Ω R 5 = 99.5Ω R 6 = 148.8 Ω

5.2.1 6. V R1 ≈ 0.578 V I R1a ≈ 4.849 mA I R1b ≈ -4.849 mA Current calculated across R1a I R1a = V / R I R1a = (0.578 V) / (119.2 Ω) I R1a = 4.849 mA Current calculated across R1b I R1b = V / R I R1b = (0.578 V) / (-119.2 Ω) I R1b = -4.849 mA 5.2.2 5. I R1 ≈ 3.649 mA I R2 ≈ 1.985 mA I R3 ≈ 1.630 mA V R1 (V ab ) ≈ 0.435 V V R2 (V cb ) ≈ 0.530V V R3 (V bc ) ≈ 0.530V 5.2.3 5. I R1 ≈ 2.299 mA I R2 ≈ 1.345 mA I R3 ≈ 0.954 mA V R1 ≈ 0.274 V V R2 ≈ 0.359 V V R3 ≈ 0.310V I R4 ≈ 0.476 mA I R5 ≈ 0.472 mA I R6 ≈ 2.319 mA V R4 ≈ 0.047V V R5 ≈ 0.047V V R6 ≈ 0.345V 5.3.1 8.

I R1a = 8.333 mA I R1b = -8.333 mA 14. V ab = 1 V V ba = -1 V 5.3.2 4. V ab = 446.93 mV V bc = 553.07 mV V ca = 553.07 mV 6. 8. V a = 1 V V b = 553.07 mV V c = 552.07 mV 9.

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11. I R1 = 3.7244 mA I R2 =2.0484 mA I R3 = 1.6760 mA 13. 5.3.3 3.

I R1 = 2.3373 mA I R2 = 1.3664 mA I R3 = 970.87 μA V ab = 280.47 mV V bc = 368.93 mV V cd = 320.39 mV 4. I R4 = 485.44 μA I R5 =485.44 μA I R6 = 2.3373 mA V db = 48.544 mV V de = 48.544 mV V ea = 350.59 mV 5. 5.3.4 4. V 1 = 80 V V 2 = 65V V 3 = 20V V 0 = 0V

5. 6. V R5 Ω = 60 V V R26 Ω = 65 V V R90 Ω = 45 V V R8 Ω = 20 V V R30 Ω = 60 V V R80 v = 80 V 10. V 1 = 60 V V 2 = 45 V V 3 = -20 V V 0 = 0V 11.

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12. V R5 Ω = 15V V R26 Ω = 65V V R90 Ω = 45V V R8 Ω = 20V V R30 Ω = 60V V R80 v = 80V Discussion Questions: 1. You took actual hardware measurements and took simulated measurements for the same three circuits. What differences were observed between simulations and measurements? List the reasons for these discrepancies between the measurements taken with actual instruments and measurements using simulations. In the simulation, you have the freedom to observe any part of the circuit and analyze the entire model comprehensively. In contrast, when conducting measurements, your accuracy relies solely on your understanding and familiarity with the measuring equipment. The measured values may not be as precise as those obtained in the simulation due to various factors, such as voltage fluctuations or the quality of the components, such as resistors. 2. In section 5.3.4, look at the values of the node voltages in step 4 and 10 and volt- ages across the elements in step 6 and 12 (before and after the reference point was changed). What do

you deduce from this about the voltage at a node and the voltage difference between two elements (or points)? As the ground reference was altered, it influenced the current path and resulted in the emergence of new voltage values, without introducing or eliminating any new components. Consequently, the voltage levels across all elements remained consistent. The only observable alteration pertained to the reference point, while the overall value remained unchanged. 3. For the circuit of Fig. 10, verify KVL for loops 1, 2 and 4 using simulated values. I R1 = 2.3373 mA I R2 = 1.3664 mA I R3 = 0.97087 mA I R4 = 0.48544 mA Loop 1: -1V + (Ir1 + Ir4) * 120 + Ir1 * 270 = 0 -1V + (2.3373 + 0.48544) * 120 + 2.3373 * 270 = 0 -1V + 969.7998 =0 -1V/-1 = -969.7998 /-1 V = 969.7998 Loop 2: (Ir2 + Ir4) * 330 + (Ir2-Ir3) * 100 + Ir2 * 270 = 0 (1.3664 + 0.48544) * 330 + (1.3664 - 0.97087) * 100 + 1.3664 * 270 = 0 Loop 4: -1V + (Ir1 + Ir4) * 120 + 120 + (Ir1+Ir2) * 330 + (Ir1 + Ir3) * 100 = 0 -1V + (2.3373 + 0.48544) * 120 + 120 + (2.3373 + 1.3664) * 330 + (2.3373 + 0.97087) * 100 = 0 -1V + 2011.76679 = 0 -1V/-1 = -2011.76679/-1 V = 2011.76679 4. Using the measured voltages across elements show (using your calculations) that the total power supplied to the circuit is equal to total power absorbed by the elements in the circuit

of Fig. 12. (Hint: To calculate the power associated with the voltage source, the current can be found by using KCL at node 1.) Current across R5Ω I R5 = V / R I R5 = (15 V) / (5 Ω) I R5 = 3 A Current across R90Ω I R90 = V / R I R90 = (45 V) / (90 Ω) I R90 = 0.5 A Current across R8Ω I R8 = V / R I R8 = (20 V) / (8 Ω) I R8 = 2.5 A Current across R30Ω I R30 = V / R I R30 = (60 V) / (30 Ω) I R30 = 2 A Current across V 80 0 = - I V80 + I R5 + I R30 I V80 = 3A + 2 A I V80 = 5 A Power supplied = Power absorbed V I = I 2 R 5 + I 2 R 26 + I 2 R 90 + I 2 R 8 + I 2 R 30 (80)(5) = ((3 A) 2 (5 Ω)) + ((2.5 A) 2 (26 Ω)) + ((0.5 A) 2 (90 Ω)) +((2.5 A) 2 (8 Ω)) + ((2 A) 2 (30 Ω)) 400 W = 400 W Therefore, power supplied to the circuit equals power absorbed.

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