Lecture 8 Voltage and Current Division Rule

Lecture 8 Voltage and current division rule  Connection of elements!! Resistance in series. Voltage division rule. Resistance in parallel. Current division rule. Lecture 8 2  Series resistance and Voltage division...

Lecture 8 Voltage and current division rule  Connection of elements!! Resistance in series. Voltage division rule. Resistance in parallel. Current division rule. Lecture 8 2  Series resistance and Voltage division Let V1,V2 be the voltages Across R1 and R2 respectively. V1= iR1 and V2= iR2  (eq 1) V1=V*R1/(R1+R2) V-V1-V2=0  (eq2) V-iR1-iR2=0  (eq3) V2=V*R2/(R1+R2) V=i(R1+R2) i=V/(R1+R2) (eq4) Lecture 8 3  Parallel resistance and Current division Let i1,i2 be the current through R1 and R2 respectively. i1= V/R1= i. Req/ R1 =i(R1R2)/(R1+R2) V= i1R1 and V= i2R2 =iR2/(R1+R2) i1=V/R1 and i2=V/R2 (eq1) i= i1+i2 = V/R1 + V/R2 i1=i*R2/(R1+R2) = V ( 1/R1 + 1/R2) = V/Req i2=i*R1/(R1+R2) V= i. Req (eq2) Lecture 8 4  Exercise 1. Find current through the branches. Step 1. Find Req i= V/Req Current division rule 10 Ω || 40 Ω = 8 Ω = 10/12 i1=i*R2/(R1+R2) 12 Ω ||6 Ω = 4 Ω =0.833 A Req= 8+4= 12 Ω i2=i*R1/(R1+R2) Lecture 8 5  Continued… i1= 0.833 * 6/(6+12) = 0.278 A i2= 0.833* 12/ (6+12) =0.553A Similarly i3= 0.833*40/(10+40) =0.66A i4=0.833*10/(10+40) =0.166A Lecture 8 6  Exercise 2. Calculate VAB for the network. Applying Voltage division rule. V1=V*R1/(R1+R2) V21=V*R2/(R1+R2) Apply KVL in loop ABC VAC= R3.V/ (R1+R3) VAB+VBC+VCA=0 = 15*20/(15+10) VAB+VBC-VAC=0 =7.5V VBC= R4.V/(R2+R4) VAB= VAC- VBC =10.20/(40+10) =7.5-4 =4V =3.5V Lecture 8 7 Thank you Amrita Vishwa Lecture Vidyapeetham 8 8  Lecture 9 Mesh Analysis  Mesh Analysis Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it. Lecture 9 2  Mesh Analysis Steps to Determine Mesh Currents: 1. Assign mesh currents i1, i2,.....in to the n meshes. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. 3. Solve the resulting n simultaneous equations to get the mesh currents. Lecture 9 3 Paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The first step requires that mesh currents (assumed) are assigned to meshes Apply KVL. For mesh 1, For mesh 2, V1  R1i1  R3 (i1  i2 )  V2  R2i2  R3 (i2  i1 ) V1  i1 ( R1  R3 )  i2 R3  V2  i1R3  i2 ( R2  R3 ) Lecture 9 4 The third step is to solve for the mesh currents. We are at liberty to use any technique for solving the simultaneous equations.  R1  R3  R3   i1   V1   R       3 R2  R3  i2   V2  It can be solved to obtain the mesh currents i1 and i2 To distinguish between the two types of currents, we use i for a mesh current and I for a branch current. Note: If mesh currents sign is positive, our assumed current direction is correct. Otherwise actual mesh current flowing in opposite direction. Lecture 9 5  For the circuit in figure, find the mesh currents using mesh analysis. We first obtain the mesh currents using KVL. For mesh 1, 15  5i1  15(i1  i2 ) 15  20i1  15i2 For mesh 2, Solve two equation, we get  20  10i2  15(i2  i1 ) i1  0.273 A  20  15i1  25i2 i2  0.636 A Lecture 9 6  For the circuit in figure, find the mesh currents and current I using mesh analysis. We first obtain the mesh currents using KVL. For mesh 1, 10  6i1  2(i1  i2 )  3(i1  i3 ) 10  11i1  2i2  3i3 For mesh 2, For mesh 3, 5  2i2  2(i2  i1 )  5  4i3  3(i3  i1 ) 5  2i1  4i2  5  3i1  7i3 Lecture 9 7 Solve three equation, we get i1  1.189 A i2  1.844 A i3  0.205 A I  i3  0.205 A Lecture 1 8  Mesh Analysis with current sources Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. The presence of the current sources reduces the number of equations When a current source exists only in one mesh: That assumed mesh current is equal to that current source value and the sign of that mesh current depends upon the direction of source current. When a current source exists between two meshes: We create a supermesh by excluding the current source and any elements connected in series with it. Lecture 9 9  For the circuit in figure, find the mesh currents and voltage drop across 1Ω resistor using mesh analysis. Current source exists only in one mesh. Direction of assumed current and current source direction are same. For mesh 1, i1  2 A For mesh 2,  10  1(i2  i1 )  2(i2  i3 )  3(i2  i4 ) 10  1i1  6i2  2i3  3i4  8  6i2  2i3  3i4 Lecture 9 10 For mesh 3, 10  3  2(i3  i2 )  4(i3  i4 ) 7  2i2  6i3  4i4 For mesh 4, 0  3(i4  i2 )  4(i4  i3 )  5i4 Solve three equation, we get 0  3i2  4i3  12i4 i2  1.054 A i3  0.823 A i4  0.011A Lecture 9 11 Voltage drop across 1Ω resistor V1  1* (i1  i2 ) V1  1* (2  1.054) V1  3.054V Lecture 9 12 References: Charles K. Alexander and Matthew N. O. Sadiku, “ Fundamentals of Electric circuits” McGraw-Hill Science Engineering ,5th edition,2012 Solving equation in Scientific calculator : https://www.youtube.com/watch?v=YnToAojliO4 Lecture 9 13 Thank You Lecture 9 14  Lecture 11 Nodal Analysis Lecture 11 1  Introduction Nodal analysis is a procedure for analyzing circuits. This procedure utilizes node voltages as circuit variables. Lecture 11 2  Node ???? Node Node Node A Node is the point of connection between two or more branches Lecture 11 3  Circuit with current source The figure shows a circuit with two current sources I1 and I2. The circuit also consists of 3 resistors R1, R2 and R3. Lecture 11 4  Nodes There are 3 nodes in the circuit Node 0, Node 1, Node 2 Lecture Lecture 10 11 5  Reference Node Node 0 is considered as reference node Lecture Lecture 10 11 6  Non reference Nodes Node 1 and Node 2 are considered to be non reference nodes Lecture Lecture 10 11 7  voltage at Non Reference Nodes v1 and v2 are voltages at nodes 1 and 2 respectively Lecture Lecture 10 11 8  Current flow in a resistor i = (vhigher − vlower)/R i1 = (v1 − 0)/R1 i2 = (v1 − v2)/R2 i3 = (v2 − 0)/R3 Lecture Lecture 10 11 9  Application of K. C. L. at Node 1 Applying K. C. L. at node 1: I1= i1+ i2 +I2 --(1) Lecture Lecture 10 11 10  I1= i1+ i2 + I2------(1) i1 = (v1 − 0)/R1 I1- I2= (v1 − 0)/R1+ (v1 − v2)/R2 i2 = (v1 − v2)/R2 i3 = (v2 − 0)/R3 ----------------- (2) Lecture Lecture 1110 11 Application of K. C. L. at Node 2 Applying K. C. L. at node 2: I2 + i2= i3 ------ (3) i1 = (v1 − 0)/R1 i2 = (v1 − v2)/R2 i3 = (v2 − 0)/R3 I2+(v1 − v2)/R2= (v2 − 0)/R3--------------(4) Lecture Lecture 10 11 12  I1- I2= (v1 − 0)/R1+ (v1 − v2)/R2 ------------- (2) I2+(v1 − v2)/R2= (v2 − 0)/R3 --------------(4) On solving (2) and (4) v1 and v2 is calculated. Once v1 and v2 is calculated current flowing through all resistances can be calculated. Lecture Lecture 10 11 13 Calculate the current In R1, R2 and R3 in the circuit shown in figure The value of the resistances is given I1 = 4A I2 = 6A R1=10Ω R2=5 Ω R3=20 Ω Lecture Lecture 10 11 14  I1- I2 = (v1 − 0)/R1+ (v1 − v2)/R2-------- (2) Equation (2)( Node1 equation) I2+(v1 − v2)/R2= (v2 − 0)/R3 --------------(4) Equation (4)(Node2 equation I1 = 4A I2 = 6A R1=10Ω R2=5 Ω R3=20 Ω Lecture Lecture 10 11 15  I1- I2 = (v1 − 0)/R1+ (v1 − v2)/R2 ------------- (2) I2+(v1 − v2)/R2= (v2 − 0)/R3 --------------(4) I1 = 4A I2 = 6A R1=10Ω R2=5 Ω R3=20 Ω 4-6= (v1 − 0)/10+ (v1 − v2)/5 ------------- (2) -2= v1 /10+ (v1 − v2)/5------(5) Lecture Lecture 10 11 16 From previous slide equation(5) is obtained for Node 1 I1 = 4A -2= v1 /10+ (v1 − v2)/5------(5) I2 = 6A Multiply (5) by 10 R1=10Ω -20= v1+ 2*(v1 − v2) R2=5 Ω -20= v1+ 2*v1−2*v2 R3=20 Ω 3*v1−2*v2= -20--------------(6) Lecture 11 17 Lecture 10  3*v1−2*v2= -20--------------(6) At Node 2 from equation (4) I2+(v1 − v2)/R2= (v2 − 0)/R3 ---------(4) 6+(v1 − v2)/5= (v2)/20 ----------------(7) Multiply (7) by 20 120+4*v1 − 4*v2= v2 -4*v1+ 5*v2=120-----------------(8) Solving (6) and (8) v1 = 140/7= 20v Sub v1 in equation (6) 3*(20)−2*v2= -20--------------(6) v =(80/2)=40V Lecture 11 Lecture 10 18 i1 = (v1 − 0)/R1 R1=10Ω i2 = (v1 − v2)/R2 R2=5 Ω i3 = (v2 − 0)/R3 R3=20 Ω v1 = 20v v2=40V i1= (20-0)/10=2A i2=((20-40)/5)=-4A(negative sign indicate current direction is opposite to assumed direction of current. i3=((40-0)/20)=2A Lecture 11 19 Thank You Lecture 11 20  Lecture 12 Nodal Analysis  Introduction Analyse the circuit shown in figure and calculate v0. 2 Lecture 12  Circuit with Voltage source The Node ‘0’ is considered as reference Node. V0 is the voltage at node 1. Application of K. C. L. is required. 3 Lecture 12  Calculation of Nodal voltage I1= (30-v1)/2kΩ) I2= (20-v1)/5kΩ) I3= (v1/4kΩ) I1+I2=I3------------(1) ((30-v1)/2kΩ)+((20-v1)/5kΩ)=(v1/4kΩ) (30/2kΩ)-(v1/2Ω)+(20/5kΩ)-(v1/5kΩ)=(v1/4kΩ) ------------(2) Equation (2)*20KΩ 10*30-10*v1+4*20-4*v1=5*v1 v1=20V=v0 4 Lecture 12 Analyze the circuit and calculate V1 and V2 using Nodal analysis 5 Lecture 12 Currents in each connection to each Node At Node 1 applying K. C. L. (4-v1)/5=(v1-v2)/10+(v1/15) ----------- (1) Multiply equation (1) by 30 24 − 6v1 = 3v1 − 3v2 + 2v1------------ (2) 6 Lecture 12  Current flow in a resistor 11v1 − 3v2 = 24 ---------------(3) At node 2 (v1-v2)/10 =(v2-6)/12+v2/8 --(4) Lecture 1 7 Lecture 12 Application of K. C. L. at Node 2 At node 2 (v1-v2)/10 =(v2-6)/12+v2/8 --(4) Multiply (4) by 120 12v1 − 12v2 = 10v2 − 60 + 15v2 ----- (5) Lecture 1 8 Lecture 12 12v1 − 37v2 = −60 ------ (6) 11v1 − 3v2 = 24 ---------(3) on solving v2 = 2.55 V v1 = 2.88 V Lecture 1 9 Lecture 12 Thank You 10 Lecture 12  Lecture 13 Superposition Theorem  Superposition Theorem (Definition) It states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to EACH independent source acting alone. The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. Lecture 13 2  Steps to Solve Circuits using Superposition Theorem 1. To find the current/ voltage in any branch, take one source at a time and replace rest of the sources by their internal resistances (if given). 2. Calculate the current/voltage by any method(mesh/nodal/KVL/KCL). 3. Now calculate the current/ voltage in the same branch by taking the other source in the circuit and replacing rest of the sources by their internal resistances. 4. Repeat steps 1,2 till all the sources have been considered. 5. Total current/voltage in the given branch= algebraic sum of all the currents/voltages in the branch due to all the current sources. Lecture 13 3 Implementation: Replacing the V or I sources when internal R not given Replace V sources by Short Circuit if their internal R is not given Replace I sources by Open Circuit if their internal R is not given Lecture 13 4 Example: Calculate the current through 1Ω resistor 1V 1V 1Ω 1Ω I1 1Ω I2 I total 2V 2V (a) (b) (c) Step 1:Only 1V source present Replace 2v source by short circuit (SC) I1 = 1A Step 2: Only 2V source present (Replace 1V source by SC) I2 = 2A Step 3: I total = 1+2 = 3A Lecture 13 5 Example: Calculate the current through 1Ω resistor 1A 1A 1Ω 1Ω I1 1Ω I2 I total 2V 2V (a) (b) (c) Step 1: 1A source acting alone (replace 2V source by SC) I1 = 1A Step 2: 2V source acting alone (replace 1A source by OC) I2 = 0A Adding the two currents considering their signs I total = I1+I2=1+0 = 1A 6 Lecture 13 Example 2: Find voltage Vx using superposition theorem 6Ω 4Ω + 42V 3Ω Vx 10V - Lecture 13 7  6Ω I1 4Ω + 42V 3Ω Vx - Lecture 13 8  6Ω 4Ω + 3Ω Vx 10V - I2 Only10Vsourceconnected(42VsourcereplacedbySC) (6 || 3) 2 Vx(10V )   10   10 (6 || 3)  4 24  3.333V Lecture 13 9  6Ω 4Ω + 42V 3Ω Vx 10V - TotalVoltage  Vx  Vx( 42V )  Vx(10V )  9.333  3.333  6V Lecture 13 10 Example 3: Use superposition to find ix Lecture 13 11  Step2: Step 1: Only 2A source connected(3V source is Only 3V source connected(2A SC) source is OC) i x'' = 2x6/(6+9)=0.8 A i x' =3/15=0.2 A Step 3:Totalcurrent=0.2+0.8=1A Lecture 13 i x = 1.0 A 12 13 Step 1: Only 10ma source connected in circuit(3mA source OC) I=10mA Step2: Only 3mA source connected(10mA source OC) I’=3mA Total Current:10mA-3mA=7mA Lecture 13 14 Thank you 15