Basic Electrical Engineering GEC 210 PDF

Faculty of Engineering Baze University Abuja BASIC ELECTRICAL ENGINEERING GEC 210 Asia’u T...

Faculty of Engineering Baze University Abuja BASIC ELECTRICAL ENGINEERING GEC 210 Asia’u Talatu Belgore Email: asia’[email protected] Office: D25 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja There are two general approaches to circuit/ network analysis : (i) Direct Method- Here, the network is left in its original form while determining its different voltages and currents. Such methods are usually restricted to fairly simple circuits and include Kirchhoff's laws, loop analysis, Nodal analysis, superposition theorem, Compensation theorem and Reciprocity theorem etc. (ii) Network Reduction Method- Here, the original network is converted into a much simpler equivalent circuit for rapid calculation of different quantities. It can be applied to simple as well as complicated networks. Examples are: Delta/Star and Star/Delta conversions, Thevenin's theorem and Norton's Theorem etc. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Source Transformation An equivalent circuit is one whose v-I characteristics are identical with the original circuit. There are 2 types of sources; voltage and current sources. They can either be dependent(value of V/I depend on the amount of V/I elsewhere in the circuit) or independent(voltage/current is fixed) Source transformation is the process of replacing a voltage source VS in series with a resistor R by a current source IS in parallel with a resistor R, or vice versa. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Source Transformation Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Example: Replace the given network with a single current source and a resistor. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Mesh Analysis Used for analyzing circuits using mesh currents as circuit variables Steps for determining mesh currents 1. Obtain the number of meshes m 2. Assign mesh currents to the meshes i.e., i1, i2, …, im Choose current flow direction to be clockwise for convenience (could be anti- clockwise). 3. Define voltage drop polarities based on the direction in each mesh. 4. Apply KVL to each of the m meshes and use Ohm’s Law to express the voltages in terms of mesh currents. 5. Solve the simultaneous equations to obtain unknown mesh currents Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Mesh Analysis Note A circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. You can use one method to check your results of the other method. Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja 1. find the current in A.B branch, using mesh analysis Applying KVL in mesh; 30 − 10𝐼1 − 10 𝐼1 − 𝐼2 = 0 30 − 10𝐼1 − 10𝐼1 + 10𝐼2 = 0 20𝐼1 − 10𝐼2 = 30 … … … … … … … ….. 1 Electrical Engineering Department Lecture 2 - EEE405 9 By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Applying KVL in loop 2: −5𝐼2 − 5 𝐼2 − 𝐼3 − 10 𝐼2 − 𝐼3 = 0 10𝐼1 − 20𝐼2 + 5𝐼3 = 0 … … … … ….. 2 In loop 3; 𝐼3 = −12𝐴 Substituting 𝐼3 = −12𝐴 in equation 2; 10𝐼1 − 20𝐼2 = 60 … … … … … … … … …. (2) In matrix form; 20 −10 𝐼1 30 = 10 −20 𝐼2 60 Find 𝐼1 &𝐼2 There fore, current in A-B branch = 𝐼2 = −3𝐴 Electrical Engineering Department Lecture 2 - EEE405 10 By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Mesh Analysis Use mesh analysis to find ix Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Mesh Analysis Identify all meshes Assign currents to all the meshes Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Mesh Analysis Apply KVL around mesh 1 3𝐼1 − 2𝐼2 = 1 … … … … … … (1) Apply KVL around mesh 2 2𝐼1 − 9𝐼2 = 3………..…………..(2) Solve (1) and (2) for unknowns 3 −2 𝐼1 1 = 2 −9 2 𝐼 3 Find 𝐼1 & 𝐼2 𝐼𝑥 = 𝐼1 − 𝐼2 𝐴 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja QUIZ!!! Use mesh analysis to find i1 and i2 Take note of current source 3Ω 6Ω 5A Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Mesh Analysis Set i2 = -5A since there is a current source and write mesh equation for the other mesh For Mesh 1 Apply KVL around mesh 2 Solve for i1 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Nodal Analysis Used for analyzing circuits using node voltages as circuit variables Steps for determining nodal voltages 1. Obtain the number of nodes n 2. Select a node as reference node (usually the ground node) and assign voltages to the remaining nodes i.e., v1, v2, …, vn-1 Ground node is assumed to have 0 potential. 3. Apply KCL to each of the non-reference nodes and use Ohm’s Law to obtain branch currents in terms of node voltages. 4. Solve the simultaneous equations to obtain unknown node voltages. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Nodal Analysis Note Current flows from a higher potential to a lower potential in a resistor. If a voltage source is connected between the reference node and a non-reference node, the voltage at the non- reference node is the same as that of the voltage source. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja steps Identify the principle nodes or junctions present in the circuit Assign a junction potential on each junction with respect to the assigned reference junction having V0 = 0v Assuming all the currents in out going direction from each junction for the KCL Solve the equations to calculate the value of junction potential. Using individual junction potential, find the values of the required electrical quantity Electrical Engineering Department Lecture 2 - EEE405 18 By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Nodal Analysis Find the node voltages using nodal analysis Find the value of i0 using nodal analysis Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Nodal Analysis Using nodal analysis, find the current in the 4Ω branch; Assume all currents at nodes are out-going At node A; Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Electrical Engineering Department Lecture 2 - EEE405 21 By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Solving equations 1 & 2; Electrical Engineering Department Lecture 2 - EEE405 22 By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Example: find the current in the 100Ω resistor. Applying KCL at all nodes; At node 1, 𝐼1 + 𝐼2 = 𝐼3 ∴ 𝐼1 + 𝐼2 − 𝐼3 = 0 Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja 60 − 𝑉1 𝑉1 − 𝑉2 +1− =0 20 30 At node 2; 𝐼3 + 𝐼4 = 𝐼5 ∴ 𝐼3 + 𝐼4 − 𝐼5 = 0 𝑉1 − 𝑉2 40 − 𝑉2 𝑉2 − 0 + − =0 30 50 100 Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Current through the 100Ω = 𝐼5 𝑉2 −0 𝑉2 = 100 100 Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Star and Delta Connections One experiences great difficulty due to a large number of simultaneous equations that have to be solved. However, such complicated networks can be simplified by successively replacing delta meshes by equivalent star system and vice versa. Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja A Star (or Wye) connection has one end of each resistor connected together and the other end of the resistor left open for external connections. This connection looks like the letter Y. A Delta connection has each resistor end connected end-to-end to form a closed loop. The connection looks like the Greek letter Delta (Δ). Star Connection Delta Connection Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Star-Delta Conversion Rba = 𝑅𝑐 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Rac = 𝑅𝑏 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Rcb = 𝑅𝑎 Note that the neutral node is eliminated by the conversion The equivalent delta resistance between any two terminals is given by the sum of star resistances between those terminals plus the product of these two star resistances divide by the third star resistances. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Star-Delta Conversion In case that the resistances are identical, the delta connection can further be simplified as R = (RY2 + RY2 + RY2)/RY = 3RY Please note that the neutral node is eliminated by the conversion Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Delta-Star Conversion 𝑅𝑏𝑎 𝑅𝑎𝑐 Ra = 𝑅𝑎𝑏 +𝑅𝑏𝑐 +𝑅𝑎𝑐 𝑅𝑐𝑏 𝑅𝑏𝑎 Rb = 𝑅𝑎𝑏 +𝑅𝑏𝑐 +𝑅𝑎𝑐 𝑅𝑎𝑐 𝑅𝑏𝑐 Rc = 𝑅𝑎𝑏 +𝑅𝑏𝑐 +𝑅𝑎𝑐 The resistance of each arm of the star is given by the product of the resistances of the two delta sides that meet at its end divided by the sum of the three delta resistances. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Delta-Star Conversion In the event that the resistances are identical, the equivalent star connection can further be simplified as RY = R 2 /(R + R + R ) = R /3 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Star-Delta Conversion Example: Find the Delta equivalent of the Star connection below with Ra = 5 Ω, Rb = 10 Ω and Rc = 10 Ω. 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Rba = = _______________= ------ Ω 𝑅𝑐 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Rac = = ________________= ____ Ω 𝑅𝑏 𝑅𝑎 𝑅𝑏 +𝑅𝑏 𝑅𝑐 +𝑅𝑐 𝑅𝑎 Rcb = = ________________= _____ Ω 𝑅𝑎 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja CLASSWORK!!! By using mesh resistance matrix, calculate the current in each branch of the circuit below Electrical Engineering Department Lecture 2 - EEE405 33 By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Find the nodal currents using node analysis Electrical Engineering Department Lecture 2 - EEE405 34 By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Network theorems Electrical Engineering Department Lecture 2 - EEE405 35 By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja The mode of analysis of each circuit is based on some rules and theorems; Superposition theorem; aids in finding current and voltage in a circuit which has multiple sources; the effects produced by each of the sources can be summed Thevenin’s theorem; aids in circuit simplification; multiple sources and resistances can be represented by an equivalent circuit with a single voltage source and a single resistor. Norton’s theorem; aids in simplification; multiple sources and resistances can be represented by an equivalent circuit with a single current source and a single resistor. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Superposition Theorem. “In a linear circuit with more than one source, the voltage across (or current through) an element is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone”. ‘’ the total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Superposition Theorem Steps for applying Superposition Theorem 1. Turn off (set to zero) all independent sources except one source. 2. Find the output (voltage or current) due to that source by using any method of analysis 3. Repeat steps 1 and 2 for all other sources 4. Find the total contribution by adding the contributions of the independent sources Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Superposition Theorem Example: use superposition theorem to find voltage vR2 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Superposition Theorem Turn off one source. The order does not matter, we kill the current by open circuit. From voltage divider rule, Back to the original circuit and short circuit the voltage source Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Superposition Theorem Back to the original circuit and short circuit the voltage source From current divider rule, The answer is the sum (superposition) of the partial answers Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja 2. Find the current through resistor R2, using superposition theorem Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Thevenin Theorem: “When viewed from the load, any network composed of ideal voltage and current sources, and of linear resistors, may be represented by an equivalent circuit consisting of an ideal voltage source vequiv in series with an equivalent resistance Requiv.” Load Load Electrical Engineering Department By Abdullahi Yusuf Sada  Faculty of Engineering Baze University Abuja Thevenin’s Theorem Steps for obtaining Thevenin Equivalent of a circuit Open circuit the terminals A–B (remove the load resistance) of the given circuit Calculate the open circuit voltage VAB seen at the terminals A and B Remove (kill) all the sources in the circuit. Calculate the equivalent resistance RAB = Requiv seen at the terminals A and B Replace the entire network by a single Thevenin source, whose voltage is Vth and whose internal resistance is Rth. Connect RL back to its terminals from where it was previously removed. Finally, calculate the current flowing through RL by using the equation. I = Vth/(Rth + RL) Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Example: find the Thevenin equivalent of the circuit below with R2 as the load Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Temporarily remove the load resistance and obtain VTh and RTh as seen from the load The circuit is now in series and voltage supplied is 28 V – 7 V = 21 V, while resistance is R1 + R3 = 5 Ohm. From Ohm’s law, V = IR and I = V/R = 4.2 A. Hence, voltage drop at R1 = 4.2 × 4 = 16.8 V. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Since sum of voltage drops = 0, voltage between the points can be computed from one source and load resistance = 28 - 16.8 - VTh = 0. Hence, VTh = 11.2 V For RTh, R1 and R3 are in parallel = R1||R3 = 0.8 Ohm The Thevenin equivalent of the circuit is given as (b) + - + - + - (a) (b) Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Example: find the Thevenin equivalent of the circuit below with RL as the load, as well as the load current and voltage. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Temporarily remove the load resistance and obtain VTh and RTh as seen from the load 12 Ohm and 4 Ohm resistors are in series because current will not flow across the 8 Ohm resistor since it is open. From Ohm’s law, V = IR and I = V/R = 48/(12+4) = 3 A. Hence, voltage drop at 4 Ohm resistor = 3 X 4 = 12 V. Although no current through 8 Ohm resistor, 12 V flows across it because it is in parallel with 4 Ohm resistor. Therefore, VTH = VAB = 12 V Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem Temporarily remove short voltage source 8 Ohm resistor is in series with a parallel combination of 12 Ohm and 4 Ohm resistors, i.e., 8 + (12 || 4) = 11 Ohm. Hence, RTH = 11 Ohm. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja Thevenin’s Theorem The Thevenin equivalent is given as From Ohm’s law, current across load resistor = IL = VTH/(RTH + RL) = 12/16 = 0.75 A VL = IL X RL = 0.75 X 5 = 3.75 V Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Whereas Thevenin’s theorem reduces a two-terminal active network of linear resistances and emf’s to an equivalent constant-voltage source and series resistance, Norton's theorem replaces the network by an equivalent constant- current source and a parallel resistance. Load Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Norton’s theorem “Any 2-terminal active network containing voltage sources and resistance when viewed from its output terminals is equivalent to a constant-current source and a parallel resistance”. Iequiv Requiv Load Load Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Steps for obtaining Norton’s Equivalent of a circuit Remove the resistance (if any) across the two given terminals and put a short-circuit across them. Compute the short-circuit current Isc. Remove all voltage sources but retain their internal resistances, if any. Similarly, remove all current sources and replace them by open-circuits. find the resistance RN of the network as looked into from the given terminals. The current source (Isc) joined in parallel across RN between the two terminals gives Norton's equivalent circuit. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja simplified sTeps in solving norTon’s Theorem Identify the load resistor RL Replace RL with a short circuit branch The current flowing through this short circuit branch will be the Norton’s current IN Remove RL and replace all the active sources with their internal resistances The equivalent resistance across the open circuit wil be the Norton’s resistance RN Draw the Norton’s equivalent circuit 𝐼𝑁 𝑥𝑅𝑁 Calculate 𝐼𝐿 using 𝐼𝐿 = 𝑅𝑁 +𝑅𝐿 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Example: find the Norton equivalent of the circuit below with R2 as the load Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Temporarily short the load (opposite of Thevenin) and obtain Ishort 𝐼𝑆𝐶 = 𝐼𝑅1 + 𝐼𝑅2 28 7 From Ohm’s law, 𝐼𝑅1 = = 7𝐴 𝑎𝑛𝑑 𝐼𝑅2 = = 7𝐴 4 1 Hence, 𝐼𝑆𝐶 = 7 + 7 = 14𝐴 To calculate Rnorton, we do the same as in Thevenin’s theorem, and 𝑅𝑁𝑜𝑟𝑡𝑜𝑛 = 0.8Ω Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem The Norton equivalent of the circuit is given as INorton Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Example: find the Norton equivalent of the circuit below with RL as the load. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Temporarily short the load and obtain Ishort The 8 Ω and 4 Ω resistors are in parallel and in series with 12 Ω resistor Total resistance, RT = (8||4) + 12 = 14.67Ω. 𝑉 48 Current in the circuit is, 𝐼𝑇 = = = 3.27𝐴 𝑅𝑇 14.67 Current flowing through 8 Ω resistor is the same as Ishort or INorton and it can be found by 4 current divider rule, i.e., 𝐼𝑁𝑜𝑟𝑡𝑜𝑛 = 𝐼𝑇 𝑥 = 1.09𝐴 4+8 Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem Short voltage source and open load resistor Obtain RN by looking at the circuit from the open circuit point It can be seen that the 8 Ω resistor is in series with a parallel combination of the 4 Ω and 12 Ω resistors Hence, RN = 8 + (12||4) = 11 Ω. Electrical Engineering Department By Abdullahi Yusuf Sada Faculty of Engineering Baze University Abuja norTon’s Theorem The Norton equivalent of the circuit is given as IN 5Ω 1.09 A 11 Ω We can find the current through the 5 Ω resistor through current divider rule, i.e., IL = IN X RN/(RN + RL) = 1.09 X 11/(11+5) = 0.75 A We can find the load voltage from Ohm’s law, VL = IL X RL = 0.75 X 5 = 3.75 V Electrical Engineering Department By Abdullahi Yusuf Sada