Experiment No: 2 Date: 23.10.2023 Name-Surname: MARWA ABOUARRA BME2301 — Circuit Theory Laboratory 1) Introduction Ohm’s Law Was discovered by Georg Simon Ohm, a German physicist. It explains how current, resistance, and voltage all work together in an electrical circuit. ππππ‘πππ (ππππ‘π , π ) = π ππ ππ π‘ππππ (πβπ, Ω) × πΆπ’πππππ‘(π΄πππππ, π΄) π½=πΉ×π° Figure 1.Ohm's Law Graph [1]. Figure 2.Ohm's Law Triangle [1]. Kirchhoff’s Voltage Law or KVL (AKA loops/mesh analysis) The idea of a loop is important for KVL. Any closed path through the circuit that encounters no node more than once is referred to as a loop. Basically, to make a loop, start at any node in the circuit and follow that path until you return to the starting node[2]. Figure 3. Loop identification example [2]. π1 − π4 − π6 − π3 = 0 ππππ 1 ππ π1 πππ π‘βπ πβππ ππ ππππππ‘πππ ππ ππππ’ππ 3. −π2 + π5 + π4 = 0 ππππ 2 ππ π2 πππ π‘βπ πβππ ππ ππππππ‘πππ ππππ’ππ 3. π1 − π2 + π5 − π6 − π3 = 0 ππππ 3 ππ π3 πππ π‘βπ πβππ ππ ππππππ‘πππ ππππ’ππ 3. Kirchhoff’s Current Law or KCL (Nodal Analysis) States that the total sum of currents entering a node is equal to the total sum of currents leaving the node ∑ ππππ‘πππππ π‘βπ ππππ = ∑ πππππ£πππ π‘βπ ππππ . Figure 4. Three wires connected at a node with different currents travelling down each wire [3]. I1 + I2 = I3 ππ π‘βπ πΎπΆπΏ ππ πΉπππ’ππ 4. Page 1 of 7 BME2301 — Circuit Theory Laboratory 2) Theoretical Analyses a. 1st Circuit Theoretical analyses of 1st circuit Figure 5. i. Calculate all node voltages in Fig. 5 using node voltage method and write your results in Table-1. Using nodal analysis π½ππ = π½π = ππ½ (ππ2 − ππ ) ππ1 ππ2 − ππ3 + + =0 π 1 π 2 π 3 π 2 π 3 (ππ2 − ππ ) + π 1 π 3 ππ2 + π 1 π 2 (ππ2 − ππ3 ) = 0 7.557ππ2 − 0.726ππ3 = 35.64 π ππ2 − 0.096ππ3 = 4.716 π (1) (ππ3 − ππ2 ) ππ3 ππ3 − ππ4 ππ4 + + + =0 π 3 π 4 π 5 π 6 π 4 π 5 π 6 (ππ3 − ππ2 ) + π 3 π 5 π 6 ππ3 + π 3 π 4 π 6 (ππ3 − ππ4 ) + π 3 π 4 π 5 ππ4 = 0 3093.048ππ3 − 1224ππ2 = 4025.5566 π 2.527ππ3 − ππ2 = 3.288 π (2) Adding 1 to 2 to find ππ3 2.431ππ3 = 8.004 π π½ππ = π. πππ π½ Substituting ππ3 in 1 or 2 to find ππ2 . Therefore, π½ππ = π. πππ π½ (ππ4 − ππ3 ) ππ4 + =0 π 5 π 6 π 6 ((ππ4 − ππ3 ) + π 5 ππ4 = 0 8.6ππ4 = 22.3856 π π½ππ = π. ππππ π½ Page 2 of 7 BME2301 — Circuit Theory Laboratory ii. iii. iv. v. vi. Determine voltages across each component in Fig. 5 using calculated node voltages in the previous question and write them into related fields on Table-2. Show that Kirchhoff’s Voltage Law is satisfied for all meshes in Fig. 5 on related fields in Table-3. Calculate all mesh currents and write them in the related fields on Table-4. Determine all resistor currents using calculated mesh currents in previous question and write them into related fields in Table-5. Show that Kirchhoff’s Current Law is satisfied for all nodes in Fig. 5 on related fields in Table-6 KVL π1 : −ππ + ππ 1 + ππ 2 = 0 π2 : −ππ 2 + ππ 3 + ππ 4 = 0 π3 : −ππ 4 + ππ 5 + ππ 6 = 0 π4 : −ππ + ππ 1 + ππ 3 + ππ 5 + ππ 6 = 0 OHM’S Law ππ 1 = π 1 πΌ1 = 0.33πΌ1 ππ 2 = π 2 πΌ2 = 2.2πΌ2 ππ 3 = π 3 πΌ3 = 2.7πΌ3 ππ 4 = π 4 πΌ4 = 100πΌ4 ππ 5 = π 5 πΌ5 = 1.6πΌ5 ππ 6 = π 6 πΌ6 = 6.8πΌ6 KCL π1 : −πΌπ + πΌπ 1 = 0 π2 : πΌπ 1 = πΌπ 2 + πΌπ 3 π3 : πΌπ 3 = πΌπ 4 + πΌπ 5 π4 : −πΌπ 5 + πΌπ 6 = 0 By using series parallel reduction, we find πΌπ πΌπ = ππ π πππ‘ππ = 6 2.15 π°πΊ = π. πππ ππ¨ = π°πΉπ From ohm’s law π½πΉπ = π. ππ × π. πππ = π. ππππ π½ We substitute ππ 1 in the first mesh π1 Page 3 of 7 BME2301 — Circuit Theory Laboratory −6 + 0.9207 + ππ 2 = 0 π½πΉπ = π. ππππ π½ Substitute in ohm’s law πΌπ 2 = ππ 2 5.0793 = 2.2 2.2 π°πΉπ = π. ππππ ππ¨ We substitute πΌπ 1 πππ πΌπ 2 in π2 πΌπ 3 = πΌπ 1 − πΌπ 2 π°πΉπ = π. ππππ ππ¨ Substitute in Ohm’s law ππ 3 = 2.7 × 0.4813 π½πΉπ = π. ππππ π½ We substitute ππ 3 in the second mesh π2 −5.0793 + 1.2995 + ππ 4 = 0 π½πΉπ = π. ππππ π½ Substitute in ohm’s law πΌπ 4 = ππ 4 3.7998 = = 0.0379 ππ΄ 100 100 π°πΉπ = π. ππππ ππ¨ We substitute πΌπ 3 πππ πΌπ 4 in π3 πΌπ 3 = πΌπ 4 + πΌπ 5 πΌπ 5 = 0.4813 − 0.0379 π°πΉπ = π. πππ ππ¨ Substitute in Ohm’s law ππ 5 = 1.8 × 0.443 π½πΉπ = π. ππππ π½ We substitute in the third mesh π3 −ππ 4 + ππ 5 + ππ 6 = 0 −3.7798 + 0.7979 + ππ 6 = 0 π½πΉπ = π. ππππ π½ Substitute in ohm’s law Page 4 of 7 BME2301 — Circuit Theory Laboratory πΌ6 = ππ 6 6.8 π°π = π. ππππ ππ¨ 3) Simulations a. 1st Circuit Figure 6. Simulation Settings. Figure 7. Schematic of the 1st circuit Figure 8. Simulation result for the 1st circuit Page 5 of 7 Calculated [V] Simulated [V] Voltages across Calculated [V] Simulated [V] π½ππ 6.000 6.000 π½πΉπ 0.9207 0.9200 π½ππ 5.032 5.080 π½πΉπ 5.0793 5.0798 π½ππ 3.292 3.788 π½πΉπ 1.2995 1.2916 π½ππ 2..603 2.996 π½πΉπ 3.7798 3.7880 π½πΉπ 0.7979 0.7929 π½πΉπ 2.9819 2.9954 Calculated Simulated Table-3 M1 M2 M3 M4 M1 M2 M3 Table-4 M4 Curr ent at M1 M2 M3 Table-2 Voltages at −π½π + π½πΉπ + π½πΉπ = π −ππ½ + π. πππππ + π. πππππ = π −π½πΉπ + π½πΉπ + π½πΉπ = π −π. πππππ + π. πππππ + π. πππππ = π −π½πΉπ + π½πΉπ + π½πΉπ = π −π. πππππ + π. πππππ + π. πππππ = π −π½π + π½πΉπ + π½πΉπ + π½πΉπ + π½πΉπ = π −ππ½ + π. πππππ + π. πππππ + π. πππππ + π. πππππ = π −π½π + π½πΉπ + π½πΉπ = π −ππ½ + π. πππππ + π. πππππ = π −π½πΉπ + π½πΉπ + π½πΉπ = π −π. πππππ + π. πππππ + π. πππππ = π −π½πΉπ + π½πΉπ + π½πΉπ = π −π. πππππ + π. πππππ + π. πππππ = π −π½π + π½πΉπ + π½πΉπ + π½πΉπ + π½πΉπ = π −ππ½ + π. πππππ½ + π. πππππ + π. πππππ + π. πππππ = π Simulated[mA] Current across Calculated [mA] Simulated [mA] π°π − π°π = π. ππππ π°π − π°π = π. ππππ π°πΉ π 2.790 2.788 π. ππ°π − πππ. ππ°π − ππππ°π = π π°π = π. ππππ°π π. ππ°π − πππ. ππ°π − ππππ°π = π π°π = π. ππππ°π π°πΉ π 2.3087 2.309 π°πΉ π 0.4813 0.478 π°πΉ π 0.037 0.037 π°πΉ π 0.443 0.440 π°πΉ π 0.438 0.440 Calculated [mA] Table-5 Table-1 BME2301 — Circuit Theory Laboratory Page 6 of 7 BME2301 — Circuit Theory Laboratory Calculated n2 Simulated Table-6 n1 n1 n3 n4 n2 n3 n4 4) Comments/Discussion a. 1st Circuit The simulated circuit in figure 7 is functioning properly. From the results we got in the simulated and the theoretically calculated parts in table 1, 2, 3, 4, 5 and 6, we can see that they are fairly close results to each other – thus the results in lined with our expectations- , and that is due to several reasons why most common is that theoretical calculations often assume ideal components, whereas real-world components have tolerances and non-ideal characteristics. For example, resistors may have tolerances that lead to small variations in their resistance values, which can affect voltage calculations. In the lab, we expect to get close values to the calculated ones we got in this preliminary work and the difference will be calculated using Absolute Error equation to check the Error percentage if necessary. REFRENCES [1] “What Is Ohm’s Law: Definition, Formula, Graph & Limitations,” GeeksforGeeks, May 18, 2021. [2] “Kirchhoff’s Voltage Law - Digilent Reference,” digilent.com. [3] Isaac Physics, “Kirchhoff’s Laws,” Isaac Physics. TABLE OF FIGURES FIGURE 1.OHM'S LAW GRAPH [1]. FIGURE 2.OHM'S LAW TRIANGLE [1]. ................................. 1 FIGURE 3. LOOP IDENTIFICATION EXAMPLE [2]. .......................................................................... 1 FIGURE 4. THREE WIRES CONNECTED AT A NODE WITH DIFFERENT CURRENTS TRAVELLING DOWN EACH WIRE [3]........................................................................................................... 1 FIGURE 5. .................................................................................................................................... 2 FIGURE 6. SIMULATION SETTINGS. .............................................................................................. 5 FIGURE 7. SCHEMATIC OF 1ST CIRCUIT ....................................................................................... 5 FIGURE 8. SIMULATIONS OF 1ST CIRCUIT .................................................................................... 5 Page 7 of 7
