1 / 43

Topic 5: Electric currents 5.2 Electric circuits

Topic 5: Electric currents 5.2 Electric circuits. 5.2.1 Define electromotive force. 5.2.2 Describe the concept of internal resistance. . Topic 5: Electric currents 5.2 Electric circuits. Define electromotive force.

hyatt-jensen
Download Presentation

Topic 5: Electric currents 5.2 Electric circuits

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 5: Electric currents5.2 Electric circuits 5.2.1 Define electromotive force. 5.2.2 Describe the concept of internal resistance.

  2. Topic 5: Electric currents5.2 Electric circuits Define electromotive force. The electromotive force (or emf) of a cell is the amount of chemical energy converted to electrical energy per unit charge. Since energy per unit charge is volts, emf is measured in volts. This battery has an emf of  = 1.6 V. Because the battery is not connected to any circuit we say that it is unloaded. 01.6 00.0

  3. Topic 5: Electric currents5.2 Electric circuits Define electromotive force. EXAMPLE: How much chemical energy is converted to electrical energy by the cell if a charge of 15 C is drawn by the voltmeter? SOLUTION: From ∆V =∆EP/q we have  = ∆EP/q. Thus ∆EP=q =(1.6)(1510-6) = 2.410-5 J. 01.6

  4. Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. If we connect a resistor as part of an external circuit, we see that the voltage read by the meter goes down a little bit. We say that the battery is loaded because there is a resistor con- nected externally in a circuit. This battery has a loaded potential difference (p.d.) of V = 1.5 V. 01.5 01.6

  5. cell heat rate P = I2r r = internal resistance Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. All cells and batteries have limits as to how rapidly the chemical reactions can maintain the voltage. There is an internal resistancer associated with a cell or a battery which causes the cell’s voltage to drop when there is an external demand for the cell’s electrical energy. The best cells will have small internal resistances. The internal resistance of a cell is why a battery becomes hot if there is a high demand for current from an external circuit. Cell heat will be produced at the rate given by

  6. Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. PRACTICE: A battery has an internal resistance of r = 1.25 . What is its rate of heat production if it is supplying an external circuit with a current of I = 2.00 A? SOLUTION: Rate of heat production is power. P = I2r P = (22)(1.25) = 5.00 J/s (5.00 W.) FYI If you double the current, the rate of heat generation will quadruple because of the I2 dependency. If you accidentally “short circuit” a battery, the battery may even heat up enough to leak or explode!

  7. r Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. If we wish to consider the internal resistance of a cell, we can use the cell and the resistor symbols together, like this: And we may enclose the whole cell in a box of some sort to show that it is one unit. Suppose we connect our cell with its internal resistance r to an external circuit consisting of a single resistor of value R. All of the chemical energy from the battery is being consumed by the internal resistance r and the external resistance R. r  R

  8. R r emf relationship  = IR + Ir = I(R + r) Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. But from EP = qV we can write: Electrical energy being created from chemical energy in the battery: EP = q. Electrical energy being converted to heat energy by R: EP,R = qVR. Electrical energy being converted to heat energy by r: EP,r = qVr. From conservation of energy EP = EP,R + EP,r so that q = qVR + qVr. Note that the current I is the same everywhere. From Ohm’s law VR = IR and Vr = Ir so that q = qIR + qIr.

  9. R 1.6 V 330  1.5 V   r r Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. • EXAMPLE: • A cell has an unloaded voltage of 1.6 V and a loaded voltage of 1.5 V when a 330  resistor is connected as shown. • Find . • (b) Then from the second circuit find I. • (c) Finally, find the cell’s internal resistance. • SOLUTION: • (a) From the first schematic we see that = 1.6 V. (Unloaded cell.) • (b) From the second diagram we see that the voltage across the 330  resistor is 1.5 V. • V = IR so that 1.5 = I(330) and I = 0.0045 A.

  10. R 1.6 V 330  1.5 V   r r Topic 5: Electric currents5.2 Electric circuits Describe the concept of internal resistance. • EXAMPLE: • A cell has an unloaded voltage of 1.6 V and a loaded voltage of 1.5 V when a 330  resistor is connected as shown. • Find . • Then from the second circuit find I. • (c) Finally, find the cell’s internal resistance. • SOLUTION: • (c) Now we can use the emf relationship = IR + Ir. • 1.6 = (0.0045)(330) + (0.0045)r • 1.6 = 1.5 + 0.0045r • 0.1 = 0.0045r • r = 22 .

  11. Topic 5: Electric currents5.2 Electric circuits 5.2.3 Apply the equations for resistors in series and parallel. 5.2.4 Draw circuit diagrams. 5.2.5 Describe the use of ideal ammeters and ideal voltmeters.

  12. equivalent resistance in series R = R1 + R2 + R3 Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. Resistors can be connected to one another in series, which means one after the other. Note there is only one current I and I is the same for all components of a series circuit. Conservation of energy tells us  =V1 + V2 + V3. Thus  =IR1 + IR2 + IR3 (from Ohm’s law)  =I(R1 + R2 + R3) (factoring out I)  =I(R), where R = R1 + R2 + R3. R3 R2 R1 

  13. Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V bat- tery in series as shown. (a) What is the circuit’s equivalent resistance? (b) What is the current in the circuit? (c) What is the voltage on each resistor? SOLUTION: (a) In series, R = R1 + R2 + R3 so that R = 330 + 330 + 330 = 990 . (b) Since the voltage on the entire circuit is 6 V, and since the total resistance is 990 , from Ohm’s law we have I = V/R = 6/990 = 0.0061 A. R3 R2 R1 

  14. Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V bat- tery in series as shown. (a) What is the circuit’s equivalent resistance? (b) What is the current in the circuit? (c) What is the voltage on each resistor? SOLUTION: (c) The current I we just found is the same everywhere. Thus each resis- tor has a current of I = 0.0061 A. From Ohm’s law, each resistor has a voltage given by V = IR = (.0061)(330) = 2.0 V. R3 R2 R1 FYI In general the V’s are different if the R’s are.

  15. Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. Resistors can also be connected in parallel. Each resistor is connected directly to the cell. Thus each resistor has the same voltage V and V is the same for all components of a parallel circuit. We can then write  =V1 = V2 = V3V. But there are three currents I1, I2, and I3. Since the total current I passes through the cell we see that I = I1+ I2+ I3. If R is the equivalent or total resistance of the three resistors, then I = I1+ I2+ I3 becomes /R =V1/R1 + V2/R2 + V3/R3 (Ohm’s law I = V/R)  R2 R3 R1

  16. equivalent resistance in parallel 1/R = 1/R1 + 1/R2 + 1/R3 Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. Resistors can also be connected in parallel. Each resistor is connected directly to the cell. Thus each resistor has the same voltage V and V is the same for all components of a parallel circuit. We can then write  =V1 = V2 = V3V. From  =V1 = V2 = V3V and /R =V1/R1 + V2/R2 + V3/R3 we get V/R =V/R1 + V/R2 + V/R3. Thus the equivalent resistance R is given by  R2 R3 R1

  17. R2 R3 R1 Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V cell in parallel as shown. (a) What is the circuit’s equivalent resistance? (b) What is the voltage on each resistor? (c) What is the current in each resistor? SOLUTION: (a) In parallel, 1/R = 1/R1 + 1/R2 + 1/R3 so that 1/R = 1/330 + 1/330 + 1/330 = 0.00909. Thus R = 1/0.00909 = 110 . (b) The voltage on each resistor is 6 V, since the resistors are in parallel. (Or each resistor is clearly directly connected to the battery).

  18. R2 R3 R1 Topic 5: Electric currents5.2 Electric circuits Apply the equations for resistors in series and parallel. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V cell in parallel as shown. (a) What is the circuit’s equivalent resistance? (b) What is the voltage on each resistor? (c) What is the current in each resistor? SOLUTION: (c) Using Ohm’s law (I = V/R): I1 = V1/R1 = 6/330 = 0.018 A. I2 = V2/R2 = 6/330 = 0.018 A. I3 = V3/R3 = 6/330 = 0.018 A. FYI In general the I’s are different if the R’s are different.

  19. Topic 5: Electric currents5.2 Electric circuits Draw circuit diagrams. A complete circuit will always contain a cell or a battery. The schematic diagram of a cell is this: A battery is just a group of cells connected in series: If each cell is 1.5 V, then the battery shown above is 3(1.5) = 4.5 V. Often a battery is shown in a more compact form: Of course a fixed-value resistor looks like this: cell battery battery resistor

  20. Topic 5: Electric currents5.2 Electric circuits Draw circuit diagrams. EXAMPLE: Draw schematic diagrams of each of the following circuits: SOLUTION: For A: For B: solder joints A B

  21. 1.06 Topic 5: Electric currents5.2 Electric circuits Describe the use of ideal ammeters and ideal voltmeters. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI Be sure to position the voltmeter across the correct resistor inparallel.

  22. .003 Topic 5: Electric currents5.2 Electric circuits Describe the use of ideal ammeters and ideal voltmeters. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI Be sure to position the ammeter between the correct resistors in series.

  23. Topic 5: Electric currents5.2 Electric circuits Draw circuit diagrams. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI This circuit is a combination series-parallel. In a later slide you will learn how to find the equivalent resistance of the combo circuit.

  24. Topic 5: Electric currents5.2 Electric circuits Describe the use of ideal ammeters and idealvoltmeters. We have seen that voltmeters are always connected in parallel. The voltmeter reads the voltage of only the component it is in parallel with. The green current represents the amount of current the battery needs to supply to the voltmeter in order to make it register. The red current is the amount of current the battery supplies to the original circuit. In order to NOT ALTER the original properties of the circuit, ideal voltmeters have extremely high resistance() to minimize the green current.

  25. Topic 5: Electric currents5.2 Electric circuits Describe the use of ideal ammeters and ideal voltmeters. We have seen that ammeters are always connected in series. The ammeter reads the current of the original circuit. In order to NOT ALTER the original properties of the circuit, ideal ammeters have extremely low resistance (0) to minimize the effect on the red current.

  26. Topic 5: Electric currents5.2 Electric circuits 5.2.6 Describe a potential divider. 5.2.7 Explain the use of sensors in potential divider circuits. 5.2.8 Solve problems involving electric circuits.

  27. IN OUT Topic 5: Electric currents5.2 Electric circuits Describe a potential divider. Consider a battery of  = 6 V. Suppose we have a light bulb that can only use three volts. How do we obtain 3 V from a 6 V battery? A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage. The input voltage is the emf of the battery. The output voltage is the voltage drop across R2. Since the resistors are in series R = R1 + R2. potential divider R1 R2

  28. potential divider R1 IN OUT R2 potential divider VOUT = VIN[ R2/(R1 + R2) ] Topic 5: Electric currents5.2 Electric circuits Describe a potential divider. Consider a battery of  = 6 V. Suppose we have a light bulb that can only use three volts. How do we obtain 3 V from a 6 V battery? A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage. From Ohm’s law the current I of the divider is given by I = VIN/R = VIN/(R1 + R2). But VOUT = V2 = IR2 so that VOUT = IR2 = R2VIN/(R1 + R2).

  29. R1 IN R2 OUT Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: Find the output voltage if the battery has an emf of 9.0 V, R1 is a 2200  resistor, and R2 is a 330  resistor. SOLUTION: Use VOUT = VIN[ R2/(R1 + R2) ] VOUT = 9[ 330/(2200 + 330) ] VOUT = 9[ 330/2530 ] VOUT = 1.2 V. FYI The bigger R2is in comparison to R1, the bigger VOUT is in proportion to the total voltage.

  30. R1 IN R2 OUT Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: Find the value of R2 if the battery has an emf of 9.0 V, R1 is a 2200  resistor, and we want an output voltage of 6 V. SOLUTION: Use the formula VOUT = VIN[ R2/(R1 + R2) ]. Thus 6 = 9[ R2/(2200 + R2) ] 6(2200 + R2) = 9R2 13200 + 6R2 = 9R2 13200 = 3R2 R2 = 4400 

  31. R1 9.0 V electronic switch R2 Topic 5: Electric currents5.2 Electric circuits Explain the use of sensors in potential divider circuits. PRACTICE: A light-dependent resistor (LDR) has a resistance of 25  in bright light and a resistance of 22000  in low light. An electronic switch will turn on a light when its p.d. is above 7.0 V. What should the value of R1 be? SOLUTION: Use VOUT = VIN[ R2/(R1 + R2) ]. Thus 7 = 9[ 22000/(R1 + 22000) ] 7(R1 + 22000) = 9(22000) 7R1 + 154000 = 198000 7R1 = 44000 R1 = 6300  (6286)

  32. R1 9.0 V electronic switch R2 Topic 5: Electric currents5.2 Electric circuits Explain the use of sensors in potential divider circuits. PRACTICE: A thermistor has a resistance of 250  when it is in the heat of a fire and a resistance of 65000  when at room temperature. An electronic switch will turn on a sprinkler system when its p.d. is above 7.0 V. (a) Should the thermistor be R1 or R2? SOLUTION: Because we want a high voltage at a high temperature, and because the thermistor’s resistance decreases with temperature, it should be placed at the R1 position.

  33. R1 V1 9.0 V electronic switch R2 V2 Topic 5: Electric currents5.2 Electric circuits Explain the use of sensors in potential divider circuits. PRACTICE: A thermistor has a resistance of 250  when it is in the heat of a fire and a resistance of 65000  when at room temperature. An electronic switch will turn on a sprinkler system when its p.d. is above 7.0 V. (b) What should the value of R2be? SOLUTION: In fire the thermistor is R1 = 250 . 7 = 9[ R2/(250 + R2) ] 7(250 + R2) = 9R2 1750 + 7R2 = 9R2 R2 = 880  (875)

  34. X Y Z 7.0 V V I Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (a) Sketch the variation of the p.d. V vs. the current I for a typical filament lamp. Is it ohmic? SOLUTION: Since the temperature increases with the current, so does the resistance. But from V = IR we see that R = V/I, which is the slope. Thus the slope should increase with I. ohmic means linear non-ohmic

  35. X Y Z 7.0 V Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (b) The potentiometer is adjusted so that the meter shows 4.0 V. Will it’s contact be above Y, below Y, or exactly on Y? SOLUTION: The circuit is acting like a potential divider with R1 being the resistance between X and Y and R2 being the resistance between Y and Z. Since we need VOUT = 4 V, and since VIN = 6 V, the contact must be adjusted above the Y.

  36. X Y Z 7.0 V Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (c) The potentiometer is adjusted so that the meter shows 4.0 V. What are the current and the resistance of the lamp at this instant? SOLUTION: P = 0.80 and V = 4.0. From P = IV we get 0.8 = I(4) so that I = 0.20 A. From V = IR we get 4 = 0.2R so that R = 20. . You could also use P = I2R for this last one.

  37. X Y Z 7.0 V Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (d) The potentiometer is adjusted so that the meter shows 4.0 V. What is the resistance of the Y-Z portion of the potentiometer? SOLUTION: Let R1 = X-Y resistance, R2 = Y-Z resistance. Then R1 + R2 = 24 so that R1 = 24 – R2. From VOUT = VIN[ R2/(R1 + R2) ] we get 4 = 7[ R2/(24 – R2+ R2) ]. R2 = 14 . (13.71) R1 R2

  38. X Y Z 7.0 V Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (e) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the Y-Z portion of the potentiometer? SOLUTION: V2 = 4.0 V because it is in parallel with the lamp. Then I2 = V2/R2 = 4/13.71 (use unrounded value) = 0.29 A R1 R2

  39. X Y Z 7.0 V Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (f) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the ammeter? SOLUTION: There are two currents supplied by the battery. The red current is 0.29 A because it is the I2 we just calculated in (e). The green current is 0.20 A found in (c). The ammeter has both so I = 0.29 + 0.20 = 0.49 A. R1 R2

  40. Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: Which circuit shows the correct setup to find the V-I characteristics of a filament lamp? SOLUTION: The voltmeter must be in paral- lel with the lamp. It IS, in ALL cases. The ammeter must be in series with the lamp and must read only the lamp’s current. The correct response is B. lamp current two currents no currents short circuit!

  41. equivalent ckt Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: A non-ideal voltmeter is used to measure the p.d. of the 20 k resistor as shown. What will its reading be? SOLUTION: There are two currents passing through the circuit because the voltmeter does not have a high enough resistance to prevent the green one from flowing. The 20 k resistor is in parallel with the 20 k voltmeter. Thus their equivalent resistance is 1/R = 1/20000 + 1/20000 = 2/20000. R = 20000/2 = 10 k. But then we have two 10 k resistors in series and each takes half the battery voltage, or 3 V.

  42. Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: All three circuits use the same resistors and the same cells. Which one of the following shows the correct ranking for the currents passing through the cells? SOLUTION: The bigger the resistance the lower the current. Highest I Middle I Lowest I 1.5R 2R 0.5R R 0.5R parallel series combo

  43. Topic 5: Electric currents5.2 Electric circuits Solve problems involving electric circuits. PRACTICE: The battery has negligible internal resis- tance and the voltmeter has infinite resistance. What are the readings on the voltmeter when the switch is open and closed? SOLUTION: With the switch open the greenR is not part of the circuit. Red and orange split the battery emf. With the switch closed the red and green are in paral- lel and are (1/2)R. E

More Related