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Superposition Theorem Application ELL 100 - Introduction to Electrical Engineering LECTURE 8: NETWORK THEOREMS SUPERPOSITION AND MAXIMUM POWER TRANSFER SUPERPOSITION THEOREM APPLICATION Circuit with more than one energy/power supply units 2 SUPERPOSITION THEOREM APPLICATION System with more than one energy sources 3 SUPERPOSITION PRINCIPLE • Helps us to analyze a linear circuit with more than one independent source. • It is used to determine the value of some circuit variable (voltage across or current through a particular impedence) • It is applied by calculating the contribution of each independent source separately. • The output of a circuit is determined by summing the individual responses of each independent source. • The idea of superposition rests on the linearity property (specifically, additivity) 4 SUPERPOSITION LINEARITY – ADDITIVE PROPERTY • The response to a sum of inputs is the sum of the responses to each input applied separately. 5 SUPERPOSITION LINEARITY – ADDITIVE PROPERTY • Example: Ohm’s Law Voltage (output) developed across a resistor in response to applied current (input) v11 i R (for applied current i1) and v22 i R (for applied current i2) Then applying current (i1 + i2) gives v() i1 i 2 R i 1 R i 2 R v 1 v 2 6 SUPERPOSITION STATEMENT • In any linear bilateral network containing two or more independent sources (voltage and/or current sources), the resultant current / voltage in any branch is the algebraic sum of currents / voltages caused by each independent source (with all other independent sources turned off). 7 SUPERPOSITION Superposition theorem for two independent sources (either voltage or current). 8 SUPERPOSITION THINGS TO KEEP IN MIND WHILE APPLYING SUPERPOSITION: • To turn off a voltage source: Replace by its internal resistance (for non-ideal source) or short circuit (for ideal source). • To turn off a current source: Replace by its internal resistance (for non-ideal source) or open circuit (for ideal source). • The dependent sources should not be zeroed. They remain the same for every particular solution with each independent source. 9 SUPERPOSITION THINGS TO KEEP IN MIND WHILE APPLYING SUPERPOSITION: • To turn off a voltage source, • To turn off a current source, 10 SUPERPOSITION STEPS TO APPLY • Step-1: Retain one source at a time in the circuit and replace all other sources with their internal resistances. • Step-2: Determine the output (current or voltage) due to the single source acting alone using any circuit analysis techniques (mesh, node, transformations etc.). • Step-3: Repeat steps 1 and 2 for every independent source. • Step-4: Find the total contribution by adding algebraically all the contributions due to all the independent sources. 11 SUPERPOSITION Example 1: Consider the network shown in figure. Calculate Iab and Vcg using superposition Theorem. 12 SUPERPOSITION Step 1: Voltage source only (turn off current source) 7 23 R (RRRR ) || 2 eq ac cb ab bg 88 3 IA1.043 23 8 => current through a-b is given by 7 24 I 0.913 A ( a to b ) ab 8 23 IAacb 1.043 0.913 0.13 13 SUPERPOSITION Step 1: Voltage source only (turn off current source) Voltage across c-g terminal: VVVVcg bg cb (2 1.043) (4 0.13) 2.61 Note : ‘c’ is higher potential than ‘g’. 14 SUPERPOSITION Step 2: Current source only (turn off voltage source) 15 SUPERPOSITION Step 2: Current in the following branches: (14 / 3) 2 3 resistor= 1.217A ; 4 resistor =2-1.217=0.783A (14 / 3) 3 2 1 resistor = 0.783=0.522A (b toa) 3 Voltage across 3 resistor (c & g terminals)Vcg =1.217 3=3.651V 16 SUPERPOSITION Step 3: Add contributions from both sources Current flowing from ‘a’ to ‘b’ due to voltage source only Iab= 0.913 A Current flowing from ‘a’ to ‘b’ due to current source only Iab= -0.522 A The total current flowing through 1Ω resistor (due to the both sources) from ‘a’ to ‘b’ = 0.913 - 0.522 A = 0.391 A Voltage across the current source due to voltage source Vcg= 2.61 V. Voltage across the current source due to current source Vcg= 3.651 V Total voltage across the current source Vcg= 2.61 + 3.65 V = 6.26 V 17 SUPERPOSITION Example 2: Calculate the current Iab flowing through the 3Ω resistor using the superposition theorem. 18 SUPERPOSITION Solution: Step 1: Consider only the 3A current source on the left 26 I3 A ( atob ) 1(dueto 3 Acurrent source ) 77 Step 2: Consider only the 2V voltage source on the left I2(dueto 2 V votagesource ) 0 A ( atob ) 19 SUPERPOSITION Step 3: Consider only the 1V voltage source 1 I A() atob 3(dueto 1 V votagesource ) 7 Step 4: Consider only the 3A current source on the right 26 I3 A ( atob ) 4(dueto 3 Acurrent source ) 77 20 SUPERPOSITION Step 5: Consider only the 2V voltage source on the right 2 I A() atob 5(dueto 1 V votagesource ) 7 Step 6: Resultant current Iab flowing through 3Ω resistor due to the combination of all sources IIIIab1(3 dueto Avotagesource )2(2 dueto V votagesource )3(1 dueto V votagesource ) II4(dueto 3 Avotagesource ) 5( dueto 2 V votagesource ) 6 1 6 2 9 0 1.285A ( a to b ) 7 7 7 7 7 21 SUPERPOSITION LIMITATIONS 1. It doesn’t work for calculation of power. • Power calculation is not linear operation. Example: 2 • When current i1 flows through resistor R, the power is P1 = Ri1 , 2 When current i2 flows through R, the power is P2 = Ri2 • If current i1 + i2 flows through R, the power absorbed is 2 2 2 P3 = R (i1 + i2) = Ri1 + Ri2 + 2Ri1i2 ≠ P1 + P2. 22 SUPERPOSITION LIMITATIONS EXAMPLE : Using superposition power consumed by the individual sources are PW1( dueto 12 V sourceleft )12 watts ; P W 2( dueto 12 V sourceleft ) 12 watts ; Total power consumed 24watts But current flowing through the resistor is zero, so total power consumed is actually zero! 23 SUPERPOSITION LIMITATIONS 2. Superposition theorem cannot be applied for circuits with nonlinear elements (eg. Diodes and Transistors). 3. In order to calculate load current or voltage for several choices of load resistance, one needs to solve for every voltage and current source in the network several times. With a simple circuit this is fairly easy, but in a large circuit with many sources this method becomes a painful experience! Thevenin/Norton equivalent with Mesh/Node analysis is a better choice in that case. 24 MAXIMUM POWER TRANSFER THEOREM 25 MAXIMUM POWER TRANSFER APPLICATION Solar panel 26 MAXIMUM POWER TRANSFER APPLICATION Communication Systems 27 MAXIMUM POWER TRANSFER APPLICATION Stereo Amplifier Design Speakers 28 MAXIMUM POWER TRANSFER APPLICATION Audio Amplifiers 29 MAXIMUM POWER TRANSFER STATEMENT • Maximum power is transferred to the load from a network when the load resistance equals the Thevenin resistance as seen from the load (RL = RTh). 30 MAXIMUM POWER TRANSFER DERIVATION If the value of the load resistance is RL , the current V flowing through the circuit is i Th RRTh L Power transferred to the load is 2 2 2 2 VVRTh Th L 2 VTh p i RLL R iRL RRRRRR22 2 2 th L Th L Th L RTh 2RRTh L RL 31 MAXIMUM POWER TRANSFER 2 DERIVATION dp 2 (RRRRRTh L ) 2 L ( Th L ) V Th 4 dRL() R Th R L 2 (RRRTh L 2 L ) V Th 3 0 ()RRTh L 0 (RRRRRTh L 2 L ) ( Th L ) RRL Th The power transferred from the source to the load is maximum when the resistance of the load is equal to the internal resistance of the source. This condition is referred to as resistance/impedence matching. 32 MAXIMUM POWER TRANSFER 2 VTh • The maximum power transferred is obtained by pmax 4RTh 22 • The total power delivered by the sourcepmax IL( R L R Th ) 2 I L R L • Efficiency under maximum power transfer condition is given by 2 IRLL pmax 2 100 50% 2 IRLL 33 MAXIMUM POWER TRANSFER Example 1. Find the value of the adjustable resistance R that dissipates the maximum power across terminals a-b. What is the maximum power that can be delivered to this load? 34 MAXIMUM POWER TRANSFER Solution: Obtain Thevenin equivalent of source network 100 VVVIAVV (15); I 4 ; 4(15) 60 th ab15 15 15 10 15 th 10 15 R 5 11 th 10 15 35 MAXIMUM POWER TRANSFER Thevenin equivalent is obtained, with Vth=60V and Rth = 11Ω Rth =11Ω + Vth = 60V - RL Maximum power transfer occurs for RL = Rth = 11Ω Power delivered to the load is 2 2 2 2 VVTh Th 60 Pmax i RLL R 81.82 W RRRth L4 th 4 11 36 NUMERICAL Q1. Find io in the circuit of Fig using superposition. 37 NUMERICAL Solution: i0 i 0' i 0 '' (1) ’ where i0 and i0” are due to the 4-A current source and 20-V voltage source respectively. 38 NUMERICAL ’ To obtain i0 we turn off the 20-V source and apply mesh analysis Loop 1: iA1 4 (2) Loop 2: -3i1 + 6i 2 -1i 3 -5io '=0 (3) Loop 3: -5i1 -1i 2 +10i3 +5i o '=0 (4) Node 0: i3 i 1 i 0' 4 i 0 ' (5) Substituting (2) and (5) into (3) and (4) gives two simultaneous equations 3ii20 2 ' 8 (6) ii205 ' 20 (7) 52 => iA' (8) 0 17 39 NUMERICAL To obtain i0″ we turn off the 4-A current source. KVL in upper loop: 6i4 i 5 5 i '' 0 0 (9) KVL in lower loop: i4 10 i 5 20 5 i 0 '' 0 (10) But i5 = − i0″. Substituting this in Eqs. (9) and (10) gives i0 i 0' i 0 '' (1) 6ii40 4 '' 0 (11) ii405 '' 20 (12) 60 => iA'' (13) 0 17 = (52 – 60)/17 = -8/17 = -0.47 A 40 NUMERICAL Q2.
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