DC NETWORK THEOREM Current: “Rate of flow of electric charge.” Charge I= Coulombs/Sec or Ampere time Note:1. Direction of current is same as the direction of motion of +Ve charge or opposite to the direction of motion of –Ve charge. Voltage: “Energy required in transferring a charge of one coulomb from one point to another point.” Energy(W) V= Joule/Coulomb or Volts Charge(Q) EMF (Electromotive force): “The EMF of a voltage source is the energy imparted by the source to each coulomb of the charge passing through it.” Energy(W) Joule/Coulomb or Volts E= Charge(Q) Potential Difference: “The pd between two points is the energy required in transferring a charge of coulomb from one point to another point.” Energy(W) pd = Joule/Coulomb or Volts Charge(Q) Voltage drop: “The voltage drop between two points is the decrease in energy required in transferring a charge of coulomb from one point to another point.” Energy(W) Voltage drop = Joule/Coulomb or Volts Charge(Q) Resistance: “Electric resistance is the property of material which offers opposition to the flow of current and dissipates energy.” l Ohm or Ω (Law of resistance) R=ρ a l = Length of the wire Where a = cross − sectional area of the wire ρ = Resistivit y or Specific Resistance of the material Note:1. Resistance also depends on temperature. Ohm’s Law: “The current passing through a conductor is directly proportional to potential drop across its ends provided physical conditions are same.” I ∞V 1 I= V Or R Or V = IR 1 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Conductance (G): It is reciprocal of resistance G= a a 1 = =σ l Resistance (R) ρl Power & Energy in an electric circuit: Power P = VI = I 2R V2 = R Energy E = P × time mho, Ω −1 , Siemens Watt, KW, MW Joule, watt-sec, watt-hours, KWH Series Circuit: I R1 R2 R3 V1 V2 V (a) V3 We know Total voltage V1 = IR1 Req ⇒ I V (b) Fig.1 V2 = IR2 V3 = IR3 V = V1 + V 2 + V 3 ⇒ IReq = IR1 + IR2 + IR3 ⇒ Req = R1 + R2 + R3 Note:1. Same current means resistances are in series. Example 1: Three resistors are connected in series across a 12V battery. The one resistance has a value of 1 ohm, second has a voltage drop of 4 Volts & third has power dissipation of 12 W. Calculate value of each resistance & circuit current. Solution: Hint V = V1 + V2 + V3 12 = I + 4 + 12 / I I = 2 Or 6 Amperes When I=2 Amp R1 = 1 R2 = 2 R3 = 3 When I=6 Amp R2 = 2/3 R3 = 1/3 R1 = 1 2 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Parallel Circuit: I1 R1 I2 R2 I3 R2 Req ⇒ I V (b) I V (a) We know I1 = V R1 Total current Fig.2 V V I3 = R2 R3 I = I1 + I 2 + I 3 V V V V ⇒ = + + Req R1 R2 R3 I2 = ⇒ 1 1 1 1 = + + Req R1 R2 R3 Note:1. 2. Same voltage means resistances are in parallel. If two resistances are in parallel and R1=R2=R then Req = R 2 Series Parallel circuit: I1 R1 A I3 R3 I2 R2 B C I4 R4 I5 R5 D I6 R6 A RAB B RBC C RCD D VAB VBC VCD I V (b) I V (a) RAD I V (C) (c) Fig.3 RAB = So R1R2 R1 + R2 RAD = RAB + RBC + RCD And I = VAB = IRAB VBC = IRBC VCD = IRCD 3 1 1 1 1 = + + RCD R4 R5 R6 RBC = R3 V RAD From fig.3b From fig.3b From fig.3b Prepared by: Nafees Ahmed www.eedofdit.webnode.com I1 = VAB R1 I2 = VAB R2 I3 = VBC =I R3 From fig.3a VCD V V I 5 = CD I 6 = CD R4 R5 R6 Power consumed by whole circuit P = I 2 RAD I4 = From fig.3a From fig.3c Or = I R AB + I R BC + I R CD From fig.3b Or = I12 R1 + I 22 R2 + I 32 R3 + I 42 R4 + I 52 R5 + I 62 R6 From fig.3a 2 2 2 Example 2: Two resistances of 20 & 30 ohms respectively are connected in parallel. These two parallel resistances are further connected in series with a third resistance of 15 ohm. If current through 15 ohm resistance is 3 Amperes. Find Current through 20 & 30 ohm resistances (i) (ii) Voltage across whole circuit I1 20 (iii) Total power consumed. 3 A 15 30 Solution: Hint I2 B C Req = (20||30) + 15 A = 12 + 15 =37 I V AB = 12x3 = 36 I 1 = 36 / 20 = 1.8 A I 2 = 36 / 30 = 1.2 A (i) V (ii) V AC = V AB + V BC = 36 + 15 × 3 = 81Volts Fig. 4 (iii) Power = VI = 243 Watts Circuit or Network Elements: R, L, C Network or Circuit: “Path followed by an electric current.” Or “Any interconnection of circuit elements with or without energy sources.” Note:1. 2. A circuit must have a closed energized path. A network may not have a closed path i.e. T-Network 3. So every network may not be circuit (i.e. T-Network) but every circuit is a network. Fig.5-T-Network Loop: “Any closed path in the network.” Mesh: “A closed path which does not have any closed path inside it.” Node: “It is a junction where 2 or more branches are connected together.” 4 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Note:1. 2. A junction where 3 or more branches connected together is known as principal node or essential node. For example B A C Loop but not Mesh H G F D E Loop or Mesh Loop or Mesh Fig.6 Total no of nodes No of principal nodes No of Loops No of Meshes = 6+2 = 8 =2 =3 =2 (A, B, C, D, E, F, G, H) (B & E) (ABEFA, BCDEB, ABCDEFA) (ABEFA, BCDEB) Energy Sources: Basically Two types (A) Independent Energy Sources a. Voltage Source i. AC/DC ii. Ideal/Practical b. Current Source i. AC/DC ii. Ideal/Practical (B) Dependent Energy Sources a. Voltage Source b. Current Source (A) Independent Energy Sources: a. Voltage Source r V V (a) Ideal DC Voltage Source (b) Practical DC Voltage Source (Internal Resistance, r=0) (With Internal Resistance, r) r + V _ (c)Ideal AC Voltage Source (Internal Resistance, r=0) 5 Prepared by: Nafees Ahmed + V_ Fig.7 (d) Ideal AC Voltage Source (Internal Resistance, r) www.eedofdit.webnode.com b. Current Source I R I (a) Ideal Current Source (b) Practical Current Source (Internal Resistance, R=infinity) (Internal Resistance, R) Fig.8 (B) Dependent Energy Sources Consider some branch of a circuit I1 R1 V1 aV1 + _ bI1 + _ Volts Volts (b) Current Dependent Voltage Source (CDVS) (a) Voltage Dependent Voltage Source (VDVS) dI1 cV1 Amperes Amperes (d) Current Dependent Current Source (CDCS) (c) Voltage Dependent Current Source (VDCS) Fig.9 Kirchhoff’s Current Law (KCL): “At any instant of time, the algebraic sum of currents at any node is zero.” ∑I = 0 Or “Total incoming currents = Total outgoing currents.” Kirchhoff’s Voltage Law (KVL): “At any instant of time, the algebraic sum of voltages in a closed path is zero.” 6 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Consider following circuit A I1 R1 B I2 R6 R3 I3 V1 F E R2 C V2 R4 D Fig.10 Apply KCL at node B I1 - I2 - I3 = 0 Or I1 = I2 + I3 Apply KVL in loop ABEFA I 1 R 1 + I 3 R3 − V1 + I 1 R6 = 0 Apply KVL in loop BCDEB I 2 R 2 − V 2 + I 2 R 4 − I 3 R3 = 0 Apply KVL in loop ABCDEFA I 1 R1 + I 2 R2 − V2 + I 2 R4 − V1 + I 2 R6 = 0 Example 3: In the following circuit calculate current in each branch by KCL & KVL (Fig.11a). Solution: I1+I2 B A C I1 I2 2Ω 2Ω 4Ω 4Ω 6Ω 6Ω 40V 40V 44V 44V F D E (a) (b) Fig.11 Apply KVL in loop ABEFA − 40 + 2 I 1 − 4 I 2 + 44 = 0 − − − − − − − −(1) − 44 + 4 I 2 + 6( I 1 + I 2 ) = 0 − − − − − − − (2) Apply KVL in loop BCDEB 34 28 62 Solving (1) & (2) I1 = A I2 = A I1 + I 2 = A 11 11 11 Mesh Analysis: Steps are 1. Identify the total meshes 2. Assume some mesh current in each mesh (clockwise or anticlockwise) 3. Apply KVL in each mesh. 4. Solve the above equations. 7 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Example 4: Solve Example 3 by Mesh Analysis. Solution: Let the two meshes are having clockwise currents as shown in following figure Apply KVL in loop 1 − 40 + 2 I 1 − 4( I 1 − I 2 ) + 44 = 0 − − − −(1) 2Ω Apply KVL in loop 2 − 44 + 4( I 2 − I 1 ) + 6 I 2 = 0 − − − −(2) 40V Solving (1) & (2) 34 28 62 I1 = A I 2 − I1 = A I2 = A 11 11 11 4Ω I2 I1 6Ω 44V Fig.12 Example 5: Find the current through R3=4 ohm resistance by mesh analysis (Fig.13a) Solution: 2Ω 2 2Ω R3 10V 5A 10V (a) 2 I1 R3 I2 I3 5A (b) Fig.13 Let the three meshes are having clockwise currents as shown in following figure Apply KVL in mesh 1 − 10 + 2 I1 + 2( I1 − I 2 ) = 0 − − − − − −(1) Apply KVL in mesh 2 + 2( I 2 − I1 ) + 4( I 2 − I 3 ) = 0 − − − − − −(2) Clearly From mesh 3 I 3 = −5 A − − − − − −(3) Solving (1), (2) & (3) Current in R3 = I2-I3 = (-3)-(-5) = 2 A I 1 = −1 A I 2 = −3 Node Analysis: Steps are 1. Identify the total Principal Nodes 2. Assume one node as reference node (Voltage of this node = 0 Volts) 3. Assume some node voltages for other remaining nodes w.r.to reference node. (V1, V2, V3 etc). 4. Assume some branch currents in different branches. 5. Apply KCL at different nodes and make the equations in terms of node voltages and circuit elements. 6. Solve the above equations. 8 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Example 6: Solve Example 3 by Node Analysis. Solution: Apply KCL at node V1 V1 − 40 − 0 V1 − 44 − 0 V1 − 0 + + =0 2 4 6 372 V1 = Volts 11 2 34 I1 = − A 11 40V 28 I2 = − A 11 62 I3 = A 11 So I1 V1 I2 4 6 44V Fig.14 Reference or Datum Node, V=0 I4 I1 V1 2Ω I2 Example 7: Solve Example 5 by Node Analysis. Solution: Apply KCL at node V1 10V I3 I3 R3 2Ω 5A V1 − 10 − 0 V1 − 0 V1 − 0 + + =5 2 2 4 V1 = 8 Volts V=0 Fig.15 V − 0 I3 = 1 = 2 Amps 4 Superposition theorem: “In a linear circuit, containing more then one independent energy sources, the overall response (Voltage or current) in any branch of the circuit is equal to sum of the response due to each independent source acting one at a time while making other source in-operative.” So I R2 R1 V1 I’ R2 R1 I1 ⇒ V1 (a) (b) + R1 I” R2 (c) Fig.16 So according to superposition theorem 9 Prepared by: Nafees Ahmed I=I’+I” www.eedofdit.webnode.com I1 Example 8: Solve example 5 by superposition theorem. Solution: I’ R3 2 2 I” R3 2 2 10V 5A (a) (b) Fig.17 Consider 10 V Voltage source only (Fig.17a) 10 2 = 1 Amps I'= 2 + 4 2 + (2 || 4) Consider 5A Current Source only (Fig.17b) 1 I" = 5 = 1 Amps 1 + 4 So I=I’+I”=1+1=2 Amps Example 9: Find I in the following circuit by superposition theorem (Fig.18a). Solution: 5 5 I1 Ω I I’ 20A 20A 12 12 4 8 4 8 40 V 10V (a) Consider 20A current source only (Fig.18b) By current division rule 8 20 = 10 I 1 = 8 + 5 + (4 || 12) Again bye current division rule 5 4 I'= 10 = Amps 2 4 + 12 (b) 5 10V I2 8 Ω 4 (c) 5 V1 Consider 10V Voltage source only (Fig.18c) 10 5 = − I 2 = − 8 8 + 5 + (4 || 12) By current division rule 5 4 5 I" = Amps − = − 32 4 + 12 8 10 Prepared by: Nafees Ahmed I” 12 4 8 I” ’12 40V (d) Fig.18 www.eedofdit.webnode.com Consider 40V Voltage source only (Fig.18d) Apply KCL at node V1 V1 − 0 V1 − 40 − 0 V1 − 0 + = 4 8+5 12 6240 V1 = V 256 V − 0 65 I '''= 1 = Amps 12 32 5 5 65 35 I = I'+ I' '+ I' ' ' = + − + = Amps 2 8 32 8 So Thevenin’s theorem: “Any linear two terminal circuits can be replaced by an equivalent network consisting of a voltage source (VTh) in series with a resistance (RTh).” Any Linear Two terminals Network A A IL RL ⇒ IL RL RTh VTh B (a) Fig.19 B (b) Thevenin’s Equivalent Circuit Where VTh = Open circuit voltage at load terminals RTh = Equivalent resistance of the network at load terminals when sources are made in-operative. VTh IL = RTh + RL And Example 10: Solve example 5 by Thevenin’s Theorem. Solution: VTh 2 2 2 2 Ω RTh 10V Ω 5 A V=0 (a) Fig.20 11 VTh the Prepared by: Nafees Ahmed (b) www.eedofdit.webnode.com Find RTh (Fig.20a) RTh = (2 || 2) = 1Ω Find VTh (Fig.20b) Apply KCL at node VTh A So I L = IL RL RTh VTh − 10 − 0 VTh − 0 + =5 2 2 VTh = 10 Volts VTh B (c) Thevenin’s Equivalent Circuit VTh 10 = = 2 Amps RTh + RL 1 + 4 Example 11: Find Thevenin’s equivalent circuit across AB of following circuit (Fig.21a). Solution: 4A A A 3Ω 5Ω 20 Ω 3Ω 5Ω 20 Ω 30V 2Ω 2Ω B (a) 4A A Find VTh(Fig.21c) 30 VTh = −3 × 4 + 5 20 + 5 VTh = −6 Volts B (b) Find RTh (Fig.21b) RTh = (20 || 5) + 3 + 2 = 9 Ω VTh = −3 × 4 + 5 I RTh 3Ω 5Ω 20 Ω 30V I 2Ω B (c) B RTh So Its Thevenin’s equivalent circuit is shown in fig.21d, It is to be noted here that terminal B is +Ve and A is –Ve. 12 Prepared by: Nafees Ahmed VTh A (d) Thevenin’s Equivalent Circuit www.eedofdit.webnode.com Norton’s theorem: Fig.21 “Any linear two terminal circuits can be replaced by an equivalent network consisting of a current source (IN) in parallel with a resistance (RN).” A A Any Linear Two terminals Network IL RL ⇒ IN IL RL RN B (b) Norton’s Equivalent Circuit B (a) Fig.22 Where IN = Isc = Short circuit current at load terminals RN = Equivalent resistance of the network at load terminals when sources are made in-operative (=RTh). RN IL = IN R N + RL And Example 12: Solve example 5 by Norton’s Theorem. Solution: Find RN (Fig.22a) Same as RTh RTh = (2 || 2) = 1Ω Find IN (Fig.22b) 10V 10 I N = 5A + A 2 I N = 10 Amps RN IL = IN R N + RL 1 IL = 10 = 2 Amps 1+ 4 the 2 2 IN = Isc 5A (a) A IN RN IL RL B (b) Norton’s Equivalent Circuit Fig.23 Example 13: Find current and voltage across 5 ohm resistance by Norton’s theorem (Fig.24a). Solution: 4Ω 3Ω 2Ω 15V 6A (a) 13 3Ω 4Ω 5Ω Prepared by: Nafees Ahmed 2Ω RN (b) www.eedofdit.webnode.com Find RN (Fig.24b) V1 13 R N = (4 || 2) + 3 = Ω 3 Find IN (Fig.24c) 15V Apply KCL at node V1 V1 − 15 − 0 V1 − 0 V − 0 + −6+ 1 =0 4 2 3 117 V1 = Volts 13 V − 0 IN = 1 = 3Amps 3 (13 / 3) 13 3 = Amps IL = (13 / 3) + 5 9 13 VL = I L RL = Volts 3 4Ω 3Ω 6A 2Ω (c) IN A IN IL RL RN B (d) Norton’s Equivalent Circuit Fig.24 Example 14: Find Norton’s equivalent circuit at A-B of example 11. 12 V Solution: V1 Find RN (Fig.21b) 3Ω 20 Ω RN = RTh = (20 || 5) + 3 + 2 = 9 Ω 5Ω IN Find IN (Fig.25a) 30V Apply Node analysis at node V1 after applying 2Ω source transformation V=0 V1 − 30 − 0 V1 − 0 V 1 − 12 − 0 (a) + + = 0 A 20 3 5 76 V1 = V IN RN 3 12 0 2 V − − IN = 1 = A 20 3 B (b) Norton’s Equivalent Fig.25 Circuit Example 15: Use Norton’s theorem to find out current in 6 ohm resistance and verify it with Thevenin’s theorem (Fig.26a). Solution: 4 4 4 I3 A 12 V 2 4 2A 4 6 B (a) 14 Prepared by: Nafees Ahmed 12V 4 2A 2 (b) 2 I1 I2 (c) www.eedofdit.webnode.com By Norton’s Theorem: Find RN (fig.26b) RN=2 Ohms Find IN (fig 26c) Clearly from mesh 1 I1 = 2 A Apply KVL in mesh 2 2(I2+I1)-12=0 So I2 = IN = 4 Amp 2 IL = 4 = 1 Amps 2+6 By Thevenin’s Theorem: Find RTh (fig.26b) RTh = RN=2 Ohms Find VTh (fig 26d) Clearly from mesh 1 I1 = 2 A So VTh = 12-2I1= 8 Volts 8 IL = = 1 Amps 2+6 4 12V 4 2A A 2 I1 B (d) Fig.26 Hence verified Maximum Power Transfer Theorem: “Maximum power is transferred by a circuit to a load resistance (RL), when RL is equal to Thevenin’s equivalent resistance (RTh) of that network.” A So for maximum power IL RL = RTh RTh RL And maximum power will be VTh 2 V Pmax = Th 4 RL B Fig.27 Thevenin’s Equivalent Circuit Proof: Load current will be IL = VTh − − − − − (1) RTh + R L Power P = I L2 RL 2 VTh RL = RTh + RL RL = VTh2 − − − − − ( 2) ( RTh + RL ) 2 Differentiating equation equa (2) w. r. t. RL and put dP/dRL = 0 15 Prepared by: Nafees Ahmed www.eedofdit.webnode.com ( R + RL ) 2 × 1 − RL × ( RTh + RL ) dP = VTh2 Th − − − − − (3) dRL ( RTh + RL ) 4 RL-RTh = 0 RL = RTh Put this in equation (2) 2 V Pmax = Th 4 RL Or Example 16: Find out the value of R for maximum power transfer to this load and find out the value of maximum power (fig.28a). Solution: 6V 6 3 15 2A 15 R 6 3 RTh 8V (a) (b) 6V Find the RTh (Fig.28b) RTh = (15 + 6) || 3 = 21 Ω 8 2A I1 15 6 I2 + VTh 3 Find VTh (Fig.28c) 8V - (c) Clearly I1 = 2 A By KVL in mesh 2 + 6 I 2 + 6 + 3I 2 + 15( I 2 − I1 ) = 0 Fig.28 So I2 = 1 A VTh = 3I 2 + 8 = 11Volts Hence for maximum power transfer to R 21 V2 RL = RTh = Ω Pmax = Th = 11.524Watt and 8 4 RL Star-Delta and Delta-Star transformation: 1 1 R1 R3 R31 ⇔ R2 R12 3 2 R23 (b) Delta ( ∆ ) Connection 2 3 (a) Star or WYE (Y) Connection Fig.29 16 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Star to Delta R R + R2 R3 + R3 R1 R12 = 1 2 R3 Delta to Star R12 R31 R1 = R12 + R23 + R31 R23 = R1 R2 + R2 R3 + R3 R1 R1 R2 = R12 R23 R12 + R23 + R31 R31 = R1 R2 + R2 R3 + R3 R1 R2 R3 = R23 R31 R12 + R23 + R31 Proof: (Equivalent resistance at 1 - 2)Y = (Equivalent resistance at 1 - 2 )∆ R1 + R2 = (R23 + R31 ) || R12 R1 + R2 = R12 (R23 + R31 ) − − − − − (1) R12 + R23 + R31 Similarly (Equivalent resistance at 2 - 3)Y = (Equivalent resistance at 2 - 3)∆ R2 + R3 = (R12 + R31 ) || R23 R (R + R31 ) R2 + R3 = 23 12 − − − − − ( 2) R12 + R23 + R31 (Equivalent resistance at 3 - 1)Y = (Equivalent resistance at 3 - 1)∆ R3 + R1 = (R12 + R23 ) || R31 R (R + R23 ) R3 + R1 = 31 12 − − − − − (3) R12 + R23 + R31 Delta to star Equation (1) + (2) + (3) 2(R12 R23 + R23 R31 + R31 R12 ) R12 + R23 + R31 R R + R23 R31 + R31 R12 R1 + R2 + R3 = 12 23 − − − − − ( 4) R12 + R23 + R31 R12 R31 R1 = − − − − − (5) (4) - (2) R12 + R23 + R31 R12 R23 R2 = (4) - (3) − − − − − ( 6) R12 + R23 + R31 R23 R31 R3 = (4) – (1) − − − − − (7 ) R12 + R23 + R31 2(R1 + R2 + R3 ) = 17 Prepared by: Nafees Ahmed www.eedofdit.webnode.com Star to Delta From Equation (5), (6) & (7) R12 R23 R31 (R12 + R23 + R31 ) (R12 + R23 + R31 )2 R12 R23 R31 R1 R2 + R2 R3 + R3 R1 = − − − − − (8) (R12 + R23 + R31 ) R R + R2 R3 + R3 R1 R12 = 1 2 (8) / (7) R3 R R + R2 R3 + R3 R1 R23 = 1 2 (8) / (5) R1 R R + R2 R3 + R3 R1 R31 = 1 2 (8) / (6) R2 Example 17: Find the equivalent resistance between the terminals a-b of the bridge circuit of the fig.30a Solution: a R1=2 a 6 4 2 R3=2/3 R =1 R1 R2 + R2 R3 + R3 R1 = ⇒ b 10 14 (a) 2 b 10 14 (b) Fig. 30 Apply delta to star transformation R12 R31 6× 4 R1 = = = 2Ω R12 + R23 + R31 12 R12 R23 6× 2 R2 = = = 1Ω R12 + R23 + R31 12 R23 R31 4× 2 2 R3 = = = Ω R12 + R23 + R31 12 3 2 Rab = 2 + (1 + 14 ) || + 10 3 Rab = 8.234Ω 18 Prepared by: Nafees Ahmed www.eedofdit.webnode.com