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Advanced electronics.  EMF  Electromotive "force" is not considered a force measured in newtons, but a potential, or energy per unit of charge, measured.

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Presentation on theme: "Advanced electronics.  EMF  Electromotive "force" is not considered a force measured in newtons, but a potential, or energy per unit of charge, measured."— Presentation transcript:

1 Advanced electronics

2  EMF  Electromotive "force" is not considered a force measured in newtons, but a potential, or energy per unit of charge, measured in volts  PD  Potential difference measured between two points (eg across a component) if a measure of the energy of electric charge between the two points also measured in volts Definitions

3  Current The flow of electric charge  Resistance The resistance to current  Capacitor Store charge in circuit Definitions

4  A circuit with a number of ‘elements’ or ‘branches’ is called a NETWORK A network which has one or more sources of EMF is said to be an ACTIVE circuit A network with no source of EMF is said to be PASSIVE Definitions

5  Active: Those devices or components which produce energy in the form of Voltage or Current are called as Active Components Definitions

6  Passive: Those devices or components which do not produce energy are known as Passive. Some components, which may may store or maintain Energy in the form of Voltage or Current are known as Passive Components Definitions

7 Double subscript notation A D C B E DA I AB V BC

8 Kirchoff’s first law

9 The total current is shared by the components in a parallel circuit A1 = A4 = A2 + A3

10 Kirchoff’ second law I =2A 12 V 2Ω2Ω4Ω4Ω IxR = 8V IxR = 4V The sum of all the pd’s around the circuit are equal to the EMF of the supply In this example we are ignoring the internal resistance of the battery

11 Kirchoff’s second law I =1.5A 12 V 2Ω2Ω4Ω4Ω I x R = 6V I x R = 3V This time we are taking the internal resistance of the battery into consideration The sum of the pd’s across the two resistors does not equal the EMF of the cell. The current has dropped to 1.5 A

12  Task  Using what you know about Kirchoff’s second law work out the internal resistance of the battery, Kirchoff’s second law

13 I =1.5A 12 V 2Ω2Ω4Ω4Ω I x R = 6V I x R = 3V r= 2Ω This time we are taking the internal resistance of the battery into consideration The sum of the pd’s across the two resistors does not equal the EMF of the cell. There is a 3 V across over the internal resistance

14 Kirchoff’ second law

15 Kirchoff’s second law Example: A circuit consists of a cell of emf 1.6 V in series with a resistance 2.0 Ω connected to a resistor of resistance 3.0 Ω in parallel with a resistor of resistance 6.0 Ω. Determine the total current drawn from the cell, the potential difference across the 3.0 Ω resistor and the current I 1

16 Solution Total resistance of the parallel resistors = (R1 x R2)/R1 +R2 (3 x 6)/ /9 = 2Ω This is in series with the 2Ω internal resistance Total resistance 4Ω

17 Total current = V/R = 1.6/4 = 0.4A Pd from cell (1.6v) = Pd across parallel set + Pd across internal resistance = (total current x R parallel set ) + (total current x R int ) = (total current x 2) + (total current x 2) = 1.6v (total current x R parallel set ) = 0.8v Pd across the 3Ω = 0.8v

18  Current through the 3 resistor  = V/R  = 0.8/3 = 0.267A

19 Kirchoff’s second law M A B C D 20 Ω 40 Ω 30 Ω X 4 V Current directions I AB, I AD and I BD Develop expressions for the following meshes ABC, supply voltage A ABDA BDCB The resistance of monitoring device M = 80Ω

20 1, 4 = 20I AB + X(I AB – I BD ) 2, 0 = 20I AB + 80I BD – 40I AD 3, 0 = 80I BD + 30(I BD +I AD ) – X(I AB -I BD ) 80 BD + 30I BD + 30 AD – XI AB + XI BD = 110I BD + XI BD + 30 I AD - XI AB Kirchoff’s second law

21 1, 4 = 20I AB + 60(I AB – I BD ) 4 = 20I AB + 60I AB – 60I BD 4 = 80I AB – 60I BD 2, 0 = 20I AB + 80I BD – 40I AD 3, 0 =110I BD + 60I BD + 30 I AD – 60I AB 0 =170I BD + 30 I AD – 60I AB Multiply 2, by 3 and 3, by 4 If X = 60Ω calculate the current flowing through the monitoring device

22 0 = 60I AB + 240I BD – 120I AD Add them together 0 =920I BD – 180I AB 920I BD = 180I AB 5.1I BD = I AB Substitute for I AB in 1, Kirchoff’s second law

23 4 = 408I BD – 60I BD 4 = 348I BD I BD = A Current through monitoring device 11.5 mA Kirchoff’s second law

24 Kirchoff’s second law example 2 50Ω A B C D 10 Ω 15 Ω 30 Ω 25Ω 6 V Current directions I AB, I AD and I BD Develop expressions for the following meshes ABC, supply voltage A ABDA BDCB Find the current through the 50Ω resistor

25 Kirchoff’s second law example 2 50Ω A B C D 10 Ω 15 Ω 30 Ω 25Ω 6 V Pd from A – C = 6V Pd across the 10 Ω resistor = current x resistance = 10x I AB (10I AB ) Pd across the 25Ω resistor = 25(I AB – I BD ) = 25I AB –25I BD The pd across both the resistors (ABC) is 10I AB + 25I AB –25I BD = 35I AB –25I BD = 6V

26 Kirchoff’s second law 50Ω A B C D 10 Ω 15 Ω 30 Ω 25Ω 6 V Pd from ABDA= 0V Pd across the 50Ω resistor = current x resistance = 50x I BD (30I BD ) Pd across the 15Ω resistor = - 15I AD The pd across both the 10Ω resistor is 10I AB PD of the mesh ABDA 50I BD +10I AB -15I AD = 0V

27 Kirchoff’s second law 50Ω A B C D 10 Ω 15 Ω 30 Ω 25Ω 6 V =105I BD + 30I AD - 25I AB Pd from BDCB= 0V Pd across the 50Ω resistor = current x resistance =50 x I BD (50I BD ) The pd across both the 30Ω resistor is 30I AD + 30I BD Pd across the 25Ω resistor = - (25I AB –25I BD ) PD of the mesh BDCD 50I BD +30I AD +30I BD -25I AB + 25I BD = 0

28 We now have 3 equations 1, 35I AB –25I BD = 6V 2, 50I BD +10I AB -15I AD = 0V 3,105I BD + 30I AD – 25 AB =0V Multiply equation 2, by2 and call it 4, 4,100I BD +20I AB -30I AD = 0V add 3, and 4, 5, 205I BD - 5I AB = 0V 6, 205I BD = 5I AB

29 6, 205I BD = 5I AB I AB = 41I BD go back to equation 1and substitute 35I AB –25I BD = I BD –25I BD = I BD = 6 IBD = amps  0.4mA

30  Thevenin’s theorem  "Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor". In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below. Thevenins Theorem is especially useful in analysing power or battery systems and other interconnected circuits where it will have an effect on the adjoining part of the circuit. Thevenin’s theorem

31 R1 R2 R3 R LOAD A B V1 Consider a circuit consisting of a power source and resistors

32 Thevenin’s theorem R1 R2 R3 A B V1 V TH Thevenin’s voltage (V TH ) is the open circuit voltage

33 Thevenin’s theorem R1 R2 R3 A B V1 V TH Thevenin’s voltage (V TH ) = V1xR2/(R1 + R2). (R3 has no affect because there is no current through it) R1 and R2 act as a potential divider

34 Thevenin’s theorem R1 R2 R3 A B V1 Thevenin’s resistance (r) = R3 + (R1 xR2/(R1 + R2)). (All voltage sources are short circuited and all current sources open circuited)

35 Thevenin’s theorem Example 2Ω2Ω 4Ω4Ω 3Ω3Ω 10Ω A B 4 Volts Calculate V TH,r and the current through the 10Ω load

36 Thevenin’s theorem Example 2Ω2Ω 4Ω4Ω 3Ω3Ω 10Ω A B 4 Volts V TH = V1xR2/(R1 + R2). = 4 x 4/6 = 2.6 Volts

37 Thevenin’s theorem Example 2Ω2Ω 4Ω4Ω 3Ω3Ω 10Ω A B 4 Volts r = 3 + (8/6) = 4.3Ω

38 Thevenin’s theorem Example 2Ω2Ω 4Ω4Ω 3Ω3Ω 10Ω A B 4 Volts Current through the load = V/I = 2.6/( ) = 0.18A

39 Thevenin’s theorem with two power sources E1 E2 Load E1 = 8V with internal resistance 4Ω E2 = 6V with internal resistance 6Ω Load = 12 Ω Use Thevenin’s Theorem to find the current through the load A B

40 Thevenin’s theorem with two power sources Using the theorem with the load disconnected the current circulating E1 and E2 I E1E2 = (E1 – E2) / ( r1 +r2) = (8-6) / (4+6) = 0.2A E1 E2 Load A B

41 Thevenin’s theorem with two power sources Hence equivalent emf of sources =E1 – I 1 r 1 = 8 – ( 0.2 x 4) = 7.2 V E1 E2 Load A B

42 Thevenin’s theorem with two power sources Total internal resistance of sources in parallel 1/R= ¼ + 1 / 6 = 3 / / 12 = 5 / 12 R = 12/5 = 2.4Ω E1 E2 Load

43 Thevenin’s theorem with two power sources I L = 7.2 / ( ) =7.2/ 14.4 = 0.5 A E1 E2 Load

44  Hence equivalent emf of sources  =E1 – I 1 r 1 = 8 – ( 0.2 x 4) = 7.2 V  Total internal resistance of sources in parallel 1/R= ¼ + 1 / 6 = 3 / / 12 = 5 / 12  I L = 7.2 / ( ) =7.2/ 14.4= 0.5 A


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