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Electric Circuits 5 Advanced Circuit Analysis

Electric Circuits 5 Advanced Circuit Analysis. Advanced Circuit Analysis Question 1. Because 1A = 1000mA 5 mA = 0.005 A 150 mA = 0.15 A 0.3 A = 300 mA 0.042 A = 42 mA. ??. ??. ??. ??. Advanced Circuit Analysis Question 2. Because 1k  = 1000  3.5k  = 3500 

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Electric Circuits 5 Advanced Circuit Analysis

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  1. Electric Circuits 5Advanced Circuit Analysis

  2. Advanced Circuit AnalysisQuestion 1 Because 1A = 1000mA • 5 mA = 0.005 A • 150 mA = 0.15 A • 0.3 A = 300 mA • 0.042 A = 42 mA ?? ?? ?? ??

  3. Advanced Circuit AnalysisQuestion 2 Because 1k= 1000 • 3.5k = 3500 • 25000 = 25 k • 0.45k = 450 • 22  = 0.22k ?? ?? ?? ??

  4. V IR Advanced Circuit Analysis 3. If the current through the circuit opposite is 600mA, what is the EMF of the battery? V = ? I = 600mA= 0.6A R = 20 V = IR V = 0.6  20 V = 12V P = VI = I2R = =

  5. Advanced Circuit Analysis 4. What power is being dissipated in the resistor opposite? P = ? V = 6V R = 10 P = P = P = 3.6W

  6. Advanced Circuit Analysis 5. What is the resistance of a globe rated at 200mW when it is being driven by a 60V battery? R = ? P = 200mW = 0.2W V = 6V P = R = R= R = 180

  7. Advanced Circuit Analysis 6. What is the power (in mW) being used by an LED that has 0.70V across it and a current of 5.0mA flowing through it? P = ? V = 0.70V I = 5.0mA = 0.005A P = VI P = 0.7  0.05 P = 0.035W P = 35mW

  8. Advanced Circuit Analysis 7. How much heat is dissipated in the resistor opposite in 3 minutes? W = ? V = 6V R = 100 t = 3  60 = 180s E = E = E = 64.8J E  65J

  9. Advanced Circuit Analysis 8. What is the total resistance of the parallel resistors opposite? = + Rtot= 1.5

  10. Advanced Circuit Analysis 9. In the circuit opposite: (a) what is the current in ammeter 1? I = ? V = 6V R = 16 I I I = 0.375A I  0.38A (b) what is the current in ammeter 2? I = ? V = 6V R = 60 I I I = 0.10A 16 = + Rtot= 15

  11. Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (a) What is the total resistance of the circuit? Rtot = 2 + 4 Rtot = 6 4 = + Rtot= 4

  12. Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (b) What current will flow? I = ? V = 12V R = 6 I I I = 2.0A 4

  13. Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (c) What will be the power dissipated in the 20 ohm resistor? P = ? I = 2AR = 2 P = I2R P = 22 2 P = 8W 4 From q2b I = 2A

  14. Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (d) What will be the power drain on the battery? P = ? I = 2A V = 12V P = VI P = 12  2 P = 24W 4 I = 2A

  15. Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (e) What will be the potential drop across the parallel components? V = ? I = 2A R = 4V V = IR V = 2 4 V = 8V Or use one of the voltage divider techniques. 4 I = 2A

  16. From q8f 8V Advanced Circuit Analysis 10. Assuming the battery opposite is ideal? (f) What current will flow through the 60 ohm resistor? Method 1 using Vparallel I = ? V = 8V R = 6 I I I = 1.333A Method 2 –I divider formula 2 4 I = 2A Method 3 – Ratios I6: I12 : since I  12:6 2:1 So I6will be 2 part out of 1 of Itot 2  3  1 = 1.333 A

  17. Advanced Circuit Analysis 11. What is Vout? Method 1 – Every reliable formula Vout= ?Vin= 16V RD= 6000 Ω Rout= 2000Ω 4.0V Method 2 – Formula in your head Rout is of the total resistance so 4.0V Method 3 – Ratios Rout : RD 2:2 1:3 So Voutwill be 1 part out of 4 of Vin 16  4  1 = 4.0V

  18. Advanced Circuit Analysis 2V 12. What is R? 5 5 Method 1 – Every reliable formula Rout = ? Vin= 12V Vout= 10V RD = 100Ω 500 Method 2 – Ratios on the diagram Using the ratios on the diagram R = 100  2 = 500

  19. Advanced Circuit Analysis 25V 13. What is R? 5 5 Method 1 – Every reliable formula RD = ? Vin= 30V Vout= 5.0V Rout = 2000Ω 10000 10k Method 2 – Ratios on the diagram Using the ratios on the diagram R = 2k  5 = 10k

  20. Advanced Circuit Analysis Classical but Long Method Find R// = + Rtot= 1.5 Find V V = ? I = 12A R = 1.5 V = IR V = 12  1.5 V = 18V Find I I = ? V = 18V R = 6 I = 3.0A 14. What is I6.0? Method 1 –I divider formula 2 12A 6.0 2.0

  21. Circuit Design

  22. 100V Circuit DesignQuestion 1 The circuit below has a string of 5V 1W globes. (a) How many globes are there in the string if the globes are operating to their specifications? number of globes = = 20 globes

  23. 100V Circuit DesignQuestion 1 The circuit below has a string of 5V 1W globes. (b) What is the current flowing in the circuit? circuit current = current on 1 globe P = VI I I I = 0.20A I = ? V = 5V P = 1W

  24. Circuit DesignQuestion 2 The circuit for a small set of Christmas tree lights is shown below. The voltage supply (shown as a DC battery) is 240V. The complete circuit is designed to consume a total of 160W. Which of the following best describes the correct labelling for the lights in the circuit? • 20W; 30V • 20W; 240V • 160W; 30V • 30W; 20V

  25. Circuit DesignQuestion 3 Arrange the following items into a circuit designed to give a total light intensity of 200W. All items must be connected into the final circuit. 100V 100V 100V, 150W 50V 25W 50V 25W

  26. Circuit DesignQuestion 4 (b) How much current will be flowing through each battery? Find I in 150W Globe I = ? V = 100V P = 150W Find I in 25W Globe I = ? V = 50V P = 25W P = VI I I I = .5A P = VI I I I = 1.5A 100V 100V, 150W 100V 50V 25W Total Current = 1.5 + 0.5 = 2.0A So current through each battery will be 1.0A 0.5A 1.0A 2.0A 1.0A 1.5A 2.0A 50V 25W

  27. Examiners Comment Series and parallel circuits continue to cause difficulties for some students. Many were unable to work out how much current was flowing in each circuit. It was common for circuit A to have 4 A and circuit B to have 1 A. Other students tried unsuccessfully to obtain the total effective resistance of the circuits Circuit DesignQuestion 5 – Q18 2011 All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer? In Circuit A Circuit B uses more power In Circuit B P = VI P = 12 × 3 P= 36W P = VI P = 12 × 2 P= 24W P = ? V = 12V I = 2.0A P = ? V = 12V I = 3.0A 55%

  28. Circuit DesignQuestion 5 – Q18 2011 All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer? OR

  29. Circuit DesignQuestion 5 – Q18 2011 Examiners Comment Series and parallel circuits continue to cause difficulties for some students. Many were unable to work out how much current was flowing in each circuit. It was common for circuit A to have 4 A and circuit B to have 1 A. Other students tried unsuccessfully to obtain the total effective resistance of the circuits All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer? 55% Pglobe = VI P = 12 × 1 P= 12W Pglobe = VI P = 6 × 2 P= 12W Pglobe = ? V = 12V I = 1.0A Pglobe = ? V = 6V I = 2.0A Ptotal = 3 × 12 = 36W Ptotal = 2 × 12 = 24W

  30. Circuit DesignQuestion 6 – Q 13 2010 Complete the circuit opposite to show the connections when the heater is set to provide a heating power of 600 W. Note: draw your line in black pen so that it photocopies clearly. Do not use pencil or red pen. P = R = R = R= 96Ω Find R required for 600W R = ? P = 600 V = 240VRMS You could have also worked out the power for series and parallel connections to see which gives 600W For 96Ω the resistors need to be connected in series. 58%

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