AI Lect 2 Fall 2024/2025 Electrical Circuit Analysis PDF

Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical...

Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Ch.1 : Electrical Circuit Analysis Lect.2 : Analysis of simple Electrical Circuits 2.1. Some definitions There are certain theorems, which when applied to the solutions of electric networks, wither simplify the network itself or render their analytical solution very easy. These theorems can also be applied to an a.c. system, with the only difference that impedances replace the ohmic resistance of d.c. system. Different electric circuits (according to their properties) are defined below : 1. Circuit. A circuit is a closed conducting path through which an electric current either flows or is intended flow. 2. Parameters. The various elements of an electric circuit are called its parameters like resistance, inductance and capacitance. These parameters may be lumped or distributed. 3. Liner Circuit. A linear circuit is one whose parameters are constant i.e. they do not change with voltage or current. 4. Non-linear Circuit. It is that circuit whose parameters change with voltage or current. 5. Bilateral Circuit. A bilateral circuit is one whose properties or characteristics are the same in either direction. The usual transmission line is bilateral, because it can be made to perform its function equally well in either direction. 6. Unilateral Circuit. It is that circuit whose properties or characteristics change with the direction of its operation. A diode rectifier is a unilateral circuit, because it cannot perform rectification in both directions. 7. Electric Network. A combination of various electric elements, connected in any manner whatsoever, is called an electric network. 8. Passive Network is one which contains no source of e.m.f. in it. 9. Active Network is one which contains one or more than one source of e.m.f. 10. Node is a junction in a circuit where two or more circuit elements are connected together. 11. Branch is that part of a network which lies between two junctions. 12. Loop. It is a close path in a circuit in which no element or node is encountered more than once. 13. Mesh. It is a loop that contains no other loop within it. Fig. 2.1 has four branches, two nodes, six loops and three meshes. Fig.2.1 (Lect.2) Page 1 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul It should be noted that, unless stated otherwise, an electric network would be assumed passive in the following treatment. We will now discuss the various network theorems which are of great help in solving complicated networks. Incidentally, a network is said to be completely Fig.2.2 Symbols 2.2 Ohm’s Law Ohm’s Law applies to electric conduction through good conductors. Ohm’s Law may be stated as The ratio of potential difference (V) between any two points on a conductor to the current (I) flowing between them, is constant, Ohm’s Law , VI = R where R is the resistance of the conductor between the two points considered 2.3 Kirchhoff’s Laws These laws are more comprehensive than Ohm’s law and are used for solving electrical networks which may not be readily solved by the latter. Kirchhoff’s laws, two in number, are particularly useful (a) in determining the equivalent resistance of a complicated network of conductors and (b) for calculating the currents flowing in the various conductors. The two-laws are : (Lect.2) Page 2 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul 2.3.1. Kirchhoff’s Point Law or Current Law (KCL) It states as in any electrical network, the algebraic sum of the currents meeting at a point (or junction) is zero. Put in another way, it simply means that the total current leaving a junction is equal to the total current entering that junction. It is obviously true because there is no accumulation of charge at the junction of the network. Consider the case of a few conductors meeting at a point A as in Fig. 2.3 (a). Some conductors have currents leading to point A, whereas some have currents leading away from point A. Assuming the incoming currents to be positive and the outgoing currents negative, we have I1 + (−I2) + (−I3) + (+ I4) + (−I5) = 0 or I1 + I4 −I2 −I3 −I5 = 0 or I1 + I4 = I2 + I3 + I5 or incoming currents = outgoing currents Similarly, in Fig. 2.3 (b) for node A + I + (−I1) + (−I2) + (−I3) + (−I4) = 0 or I= I1 + I2 + I3 + I4 We can express the above conclusion thus : Σ I = 0....at a junction Fig.2.3 2.3.2. Kirchhoff’s Mesh Law or Voltage Law (KVL) It states as follows : The algebraic sum of the products of currents and resistances in each of the conductors in any closed path (or mesh) in a network plus the algebraic sum of the e.m.fs. in that path is zero. In other words, Σ IR + Σ e.m.f. = 0...round a mesh It should be noted that algebraic sum is the sum which takes into account the polarities of the (Lect.2) Page 3 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul voltage drops. The basis of this law is this : If we start from a particular junction and go round the mesh till we come back to the starting point, then we must be at the same potential with which we started. Hence, it means that all the sources of e.m.f. met on the way must necessarily be equal to the voltage drops in the resistances, every voltage being given its proper sign, plus or minus. Fig.2.4 2.4. Determination of Voltage Sign In applying Kirchhoff’s laws to specific problems, particular attention should be paid to the algebraic signs of voltage drops and e.m.fs., otherwise results will come out to be wrong. Following sign conventions is suggested : (a) Sign of Battery E.M.F. A rise in voltage should be given a + ve sign and a fall in voltage a −ve sign. Keeping this in mind, it is clear that as we go from the −ve terminal of a battery to its +ve terminal (Fig. 2.5), there is a rise in potential, hence this voltage should be given a + ve sign. If, on the other hand, we go from +ve terminal to −ve terminal, then there is a fall in potential, hence this voltage should be preceded by a −ve sign. It is important to note that the sign of the battery e.m.f. is independent of the direction of the current through that branch. Fig.2.5 (Lect.2) Page 4 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul (b) Sign of IR Drop Now, take the case of a resistor (Fig. 2.6). If we go through a resistor in the same direction as the current, then there is a fall in potential because current flows from a higher to a lower potential. Hence, this voltage fall should be taken −ve. However, if we go in a direction opposite to that of the current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign. It is clear that the sign of voltage drop across a resistor depends on the direction of current through that resistor but is independent of the polarity of any other source of e.m.f. in the circuit under consideration. Fig.2.6 Consider the closed path ABCDA in Fig. 2.7. As we travel around the mesh in the clockwise direction, different voltage drops will have the following signs : I1R2 is − ve (fall in potential) I2R2 is − ve (fall in potential) I3R3 is + ve (rise in potential) I4R4 is − ve (fall in potential) E2 is − ve (fall in potential) E1 is + ve (rise in potential) Using Kirchhoff’s voltage law, we get −I1R1 −I2R2 +I3R3 −I4 R4 −E2 + E1 = 0 or Fig.2.7 I1R1 + I2 R2 - I3R3 + I4R4 = E1 −E2 (Lect.2) Page 5 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Examples Solved Example 2.1 By applying Kirchhoff’s laws Find the current through the resistance R L=20Ω when connected as shown in Fig. 2c.1. Fig.2c.1 Solution Mesh ABEFA 5I+10I1 =30 (1) Mesh BCDEB 3(I - I1)+20 (I - I1) -10 I1=0 23I – 33I1 =0 (2) 115 I + 230 I1 = 690 (3) 115 I- 165 I1 =0 (4) 395I1=690 I1= 690/395 = 138 / 79 From (1) 5I= 30-10*138/79 I = 6- 276/79 = Therefore IRL = I - I1 = 6-276/79 - 138/79 = (474 -276-138) /79 = 60/79 A (Lect.2) Page 6 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Example 2.2 By applying Maxwell’s theorem Find the current through the resistance R and the voltage across it when connected as shown in Fig. 2c.2. Fig.2c.2 Solution Loop ABEFA 15 I1 -10 I2 = 30 (1) Loop BCDEB -10 I1 +33 I2 = 0 (2) Therefore, 150 I1 -100 I2 =300 (3) -150 I1 +33*15 I2 =0 (4) So, 395 I2 = 300 then I2= 300/395 = 60/79 A (Lect.2) Page 7 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Example 2.3 By applying Thevenin’s theorem Find the current through the resistance R L=20Ω when connected as shown in Fig. 2c.3. Fig.2c.3 Solution I= 30 / (5+10) = 2 A Vth= (3*0 + 10 *I) = 20 V Rth = 3 + 10 // 5 = 3+ (10*5) /(10+5) = 3 + 50/15 = 3+ 10/3 = 19 / 3 Ω Therefore, IRL = Vth /( Rth + R L) = 20 /(19/3 + 20) = 20 /(79/3) = 60/79 A (Lect.2) Page 8 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Example 2.4 Use Thevenin’s theorem to find current in the branch AB of the network shown in Fig.2c.5. Fig.2c.5 Solution To find Vth I= 10 / (4+3+2+4) = 10/13 A Vth= 10-I*4 = 10-4*10/13 = (130-40)/13 = 90/13 V To find Rth Rth= 4//(3+2+4) = 4//9 = 4*9/(4+9) = 36/13 Ω So, Vth= 90/13 V , Rth= 36/13 Ω Therefore, the current through 1 Ω resistance I R=1Ω = Vth /(Rth +1) = (90/13 ) / {(36/13) +1} = (90/13) / (49/13) = 90/49 = 1.84 A (Lect.2) Page 9 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Example 2.5 In the circuit shown in Fig. 2c.6 Find the current that would flow if a 5-Ω resistor (RL) was connected between points A and B by using Thevenin’s theorem (The two batteries have negligible resistance). Fig. 2c.6 Solution I = 4 / (3+1+12) = 4/16=1/4 Ω Vth= 4 -3*I= 4-3*1/4 = 13/4 V Rth= (12+1) //3 = 13 // 3 = 13*3 /(13+3) =39/16 Ω Therefore , Vth= 13/4 V , Rth= 39/16 V The current through load 5 Ω = IR=5Ω = Vth / (Rth +5) = (13/4) / {(39/16) +1} = (13/4) / (55/16) = 52/55 A (Lect.2) Page 10 of 11 Delta University for Science and Technology Basic Electricity and Electronics College of Artificial Intelligence BAS116 Basic Science and Computer Dep. Lect.2 Analysis of simple Electrical Fall semester 2024/2025 circuits Prepared by ---------- Assoc. Prof. Adel Zaghloul Solved Example 2.6 Find the current that would flow if a 5/2 -Ω resistor was connected between points A and B by using Thevenin’s theorem. (The two batteries have negligible resistance). Fig.2c.7 Solution I=(4-2)/ (3+1+2) = 2/6 =1/3 A Vth= 4 - 3*I = 4 -3*1/3 = 4-1 = 3 V Rth= (1+2) // 3 =3 // 3 = 3*3 /(3+3) = 3/2 Ω Therefore the current through the load (R L=5/2 Ω) between points A and B IRL = Vth / (Rth+RL) = 3/ (3/2+ 5/2) =3/4 A (Lect.2) Page 11 of 11