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 with more than one independent source. • It is used to determine the value of some circuit variable ( 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 in response to applied current (input)

v11 i R (for applied current i1) and

viR22 (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 / 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 : Replace by its internal resistance (for non-ideal source) or (for ideal source). • To turn off a : 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) 723 R() ||2 RRRR  eqaccbabbg 88 3 IA1.043 23 8 => current through a-b is given by 7 24 IA a to  b  0.913 () 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: VVVVcgbgcb (2 1.043)(40.13)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 IA atob 3() 1(3)dueto 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 IA atob  () 3(1)dueto V votagesource 7

Step 4: Consider only the 3A current source on the right 26 IA atob 3()   4(3)dueto 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

IIIIab1(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 RRThL

Power transferred to the load is

2 2 2 2 VVRTh Th L 2 VTh p i RLL  R  iRL  RRRRRR22 2  2 th L Th L Th L RTh 2RRTh L RL

31 MAXIMUM POWER TRANSFER

2 DERIVATION dp 2 ()2RRRThLL () RR ThL  V Th 4 dRRRLThL ()

2 (2RRRThLL ) V Th 3 0 ()RRThL

0 (2RRRRRThLLThL ) ()

RRLTh 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 sourcepIRRIRmax LLThLL()2 • Efficiency under maximum power transfer condition is given by 2 IRLL pmax  2 10050% 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); I4 ;4(15)60 thabth 151515 10 15

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:

iii000'''(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: -3i123 + o 6i -1i-5i '=0(3)

Loop 3: -5i12 -1i+10i3o +5i '=0(4)

Node 0: iiii3100 '4'(5) Substituting (2) and (5) into (3) and (4) gives two simultaneous equations

32ii20 '8(6)

ii205 '20 (7) 52 => iA' (8) 0 17 39 NUMERICAL

To obtain i0″ we turn off the 4-A current source.

KVL in upper loop: 65''0(9)iii450

KVL in lower loop: iii450 10205''0(10) But i5 = − i0″. Substituting this in Eqs. (9) and (10) gives

64ii40 ''0(11)

ii405  ''20(12) 60 => iA''(13) 0 17

i0 i 0' i 0 ''= (52 – 60)/17 (1) = -8/17 = -0.47 A

40 NUMERICAL Q2. Use superposition to find current flowing through the 3-Ω resistor.

41 NUMERICAL

Solution: Since there are three sources, iiii123 iT

+ 20 V 1 Ω 4 Ω - First consider 20V source only 4 (3 1) 2 Ω i1 R  2  4  3 Ω eq 4 3 1 20 5 4 i 5 A ; i   2.5 A ( currentdivision ) T 41 1 3 4 42 NUMERICAL

Then consider 2 A source only

- 2 A + 2 V Source 1 Ω 4 Ω Transformation 4 Ω 1 Ω

2 Ω i2 2 Ω i2 3 Ω 3 Ω

42 R  1  3  5.33  eq 42 2 i 0.375 A ; i  i  0.375 A TT5.33 2 43 NUMERICAL

1 Ω 4 Ω

16 V

Finally consider 16V source only 2 Ω - i3 + 3 Ω 42 R  1  3  5.33  eq 42 iT 16 3 2 i 3 A ;  i   1 A ; i   1 A T 5.3333 2 1 3

44 NUMERICAL => Current flowing through 3-Ω resistor is

iiiiA123 2.50.37511.875

45 NUMERICAL

Q3. For the circuit shown in fig, the value of Vs1 and Is are fixed. When Vs2 = 0, the current I = 4A . Find the value of I when Vs2 = 32 V.

46 NUMERICAL Solution: The current flowing through 6Ω resistor due to the voltage and current sources are given by (assuming circuit linearity):

IIdue'()''()'''() to VIduesss to12 VIdue to I

IVVIsss12 (1) where, parameters α, β and η represent constants

When Vs2=0, I = 4 A => II due'()'''() to VI 4 ss due1 to I

=> IVIss1   4 (2)

47 NUMERICAL

Vs2 32 When Vs2 = 32 V is acting alone: IAT  4 R(8eq || 8)4  I 8 48 IIcurrent''()2 divisionAT vs2 8816

IVA''2(3) s2

Current flow through 6Ω resistor when Vs2 = 32 V is

I Vsss12   VIA   4 2 6

48 NUMERICAL

Q4. Find the value of RL for maximum power transfer and compute the maximum power delivered

49 NUMERICAL Solution: Step 1: Calculation of Thevenin Voltage and Thevenin Resistance across terminals a-b

6 12 R 2  3  6 ||12   59 -12 + 18i1 - 12i 2 = 0,iA 2  2 Th 18 => i1 = −2∕3.

KVL around the outer loop: -12 + 6i1 +3i 2 +2(0)+V Th =0 VVTh 22

50 NUMERICAL For maximum power transfer,

R LTh= R = 9 

The maximum power is,

2 2 VTh 22 pWmax  13.44 44RL 9 

51 NUMERICAL

Q5. Find the value of RL that absorbs maximum power from the circuit and the corresponding power under this condition.

52 NUMERICAL

Solution: Obtain VTh and RTh for the circuit below

We can use superposition theorem to find VTh

53 NUMERICAL Considering the 20V source only

The current through ‘b-c’ branch 20/20 = 1A (from ‘b’ to ‘c’)

=> voltage across the ‘b-a’ terminal Vba = 1 × 10 = 10 V (’b’ is higher potential than ‘a’) ∴ Vab = −10 V

54 NUMERICAL Considering the 10V source only Note: No current is flowing through the ‘c-b’ branch

∴ Vab = 5 V (‘a’ is higher potential than ‘b’)

Considering only 2 A source only Note that the current flowing through the ‘c-a’ branch is zero

∴ Vab =10 V 55 NUMERICAL The voltage across the ‘a’ and ‘b’ terminals due to the all sources

VTh = Vab (due to 20 V) + Vab (due to 10 V) + Vab (due to 2A source) = -10 + 5 + 10 = 5 V

RRTh  ab   5  5 || 10   10 || 10  5  5  10 

56 NUMERICAL Thevenin equivalent circuit is drawn below:

Maximum power transfer occurs for RL = Rth = 10 Ω 2 2 2 2 VVTh Th 5 Power delivered to the load is Pmax  i RLL  R    0.625 W RRRth L4 th 4 10 57 NUMERICAL Q6. Applying Norton’s theorem, calculate the value of R that results in maximum power transfer to the 6.2Ω resistor. Find the maximum power dissipated by the resistor 6.2Ω under that situation.

58 MAXIMUM POWER TRANSFER Solution: Step-1: Short the terminals ‘a’ and ‘b’ after disconnecting the 6.2Ω resistor. The Norton’s current for the circuit shown is computed by using ‘mesh analysis’.

Loop 1: 36 24 20R 123()I11 RI 0 2 I => II;   1215 8RR 15 8 Loop 2: 10  5(I2  I 3 )  3(I 2  I 1 )  0, note I 3   2 A 59 MAXIMUM POWER TRANSFER The Norton resistance is obtained between the terminals ‘a’ and ‘b’

3R RR(|| 3)55 N 3 R

60 MAXIMUM POWER TRANSFER Norton equivalent circuit is drawn below:

The maximum power will dissipate in load resistance when load resistance = Norton’s resistance i.e. RRNL 6.2 3R R 6.2   5; R  2  N 3 R 2 12 1 24 20R Pmax  INL RL    R  6.61 watts 4 4 15 8R  61 SUPERPOSITION

Q1. Using the superposition theorem, find va in the circuit of Fig.

(Ans. 14 Volts)

62 SUPERPOSITION Q2. Using the superposition theorem, calculate the value of source current

Ix that yields I = 0 if VA and VC are kept fixed at 7 V and 28 V

(Ans. IX = -5.833A) 63 SUPERPOSITION Q3. For the circuit shown below, it follows from linearly that we can write

Vab=αIx+βVA+ηVB, where α,β and η are constants. Find the values of (1) α,(2)β, and (3)η.

(Ans. α = -1,β=0.063, and η=-0.063

64 SUPERPOSITION Q4. Using superposition theorem, find the current i through 5Ωresistor as shown in fig.

Ans. -0.538A

65 MAXIMUM POWER TRANSFER Q5. Consider the circuit of fig.

(a)Find the linear relationship between Vout and input sources Vs and Is (b)If Vs=10V and Is = 1, find Vout (c)What is the effect of doubling all resistances values on the coefficients of linear relationship found part (a)?

(Ans. (a) Vout=0.333Vs+6.66Is; (b) ) Vout=9.999V (c) Vout=0.333Vs+6.66Is;

66 MAXIMUM POWER TRANSFER

Q1.Determine the value of RL that will draw the maximum power from the rest of the circuit in Fig. Calculate the maximum power.

Answer: 126.67 Ω, 96.71 mW.

67 MAXIMUM POWER TRANSFER Q2.Find the maximum power that can be delivered to the resistor R in the circuit of Fig..

Answer: 1.6Ω, 5.625W 68 MAXIMUM POWER TRANSFER Q3. The variable resistor R in Fig. is adjusted until it absorbs the maximum power from the circuit. (a) Calculate the value of R for maximum power. (b) Determine the maximum power absorbed by R

Answer: 25 kΩ, 49 mW 69 MAXIMUM POWER TRANSFER Q4. For the circuit in Fig, what resistor connected across terminals a-b will absorb maximum power from the circuit? What is that power?

Answer: 8 kΩ, 152 W 70 MAXIMUM POWER TRANSFER Q5. Find the maximum power transferred to resistor R in the circuit of Fig.

Answer: 20.77 W 71