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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.

emerald-bryant
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Topic 5: Electric currents 5.2 Electric circuits

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

  2. 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. Topic 5: Electric currents5.2 Electric circuits 01.6 00.0

  3. 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. Topic 5: Electric currents5.2 Electric circuits 01.6

  4. 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. Topic 5: Electric currents5.2 Electric circuits 01.5 01.6

  5. cell heat rate P = I2r r = internal resistance 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 Topic 5: Electric currents5.2 Electric circuits

  6. 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.) Topic 5: Electric currents5.2 Electric circuits 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 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. Topic 5: Electric currents5.2 Electric circuits r  R

  8. R r emf relationship  = IR + Ir = I(R + r) 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. Topic 5: Electric currents5.2 Electric circuits

  9. R 1.6 V 330  1.5 V   r r 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. Topic 5: Electric currents5.2 Electric circuits

  10. R 1.6 V 330  1.5 V   r r 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 . Topic 5: Electric currents5.2 Electric circuits

  11. 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. Topic 5: Electric currents5.2 Electric circuits

  12. equivalent resistance in series R = R1 + R2 + R3 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. Topic 5: Electric currents5.2 Electric circuits R3 R2 R1 

  13. Apply the equations for resistors in series and parallel. Topic 5: Electric currents5.2 Electric circuits 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. Apply the equations for resistors in series and parallel. Topic 5: Electric currents5.2 Electric circuits 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. 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) Topic 5: Electric currents5.2 Electric circuits  R2 R3 R1

  16. equivalent resistance in parallel 1/R = 1/R1 + 1/R2 + 1/R3 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 Topic 5: Electric currents5.2 Electric circuits  R2 R3 R1

  17. R2 R3 R1 Apply the equations for resistors in series and parallel. Topic 5: Electric currents5.2 Electric circuits 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 Apply the equations for resistors in series and parallel. Topic 5: Electric currents5.2 Electric circuits 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. 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: Topic 5: Electric currents5.2 Electric circuits cell battery battery resistor

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

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

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

  23. Draw circuit diagrams. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: Topic 5: Electric currents5.2 Electric circuits 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. 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. Topic 5: Electric currents5.2 Electric circuits

  25. 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. Topic 5: Electric currents5.2 Electric circuits

  26. 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. Topic 5: Electric currents5.2 Electric circuits

  27. IN OUT 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. Topic 5: Electric currents5.2 Electric circuits potential divider R1 R2

  28. potential divider R1 IN OUT R2 potential divider VOUT = VIN[ R2/(R1 + R2) ] 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). Topic 5: Electric currents5.2 Electric circuits

  29. R1 IN R2 OUT 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. Topic 5: Electric currents5.2 Electric circuits FYI The bigger R2is in comparison to R1, the bigger VOUT is in proportion to the total voltage.

  30. R1 IN R2 OUT 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  Topic 5: Electric currents5.2 Electric circuits

  31. R1 9.0 V electronic switch R2 Explain the use of sensors in potential divider circuits. Topic 5: Electric currents5.2 Electric 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 Explain the use of sensors in potential divider circuits. Topic 5: Electric currents5.2 Electric 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 Explain the use of sensors in potential divider circuits. Topic 5: Electric currents5.2 Electric 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 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. Topic 5: Electric currents5.2 Electric circuits ohmic means linear non-ohmic

  35. X Y Z 7.0 V 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. Topic 5: Electric currents5.2 Electric circuits

  36. X Y Z 7.0 V 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. Topic 5: Electric currents5.2 Electric circuits

  37. X Y Z 7.0 V 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) Topic 5: Electric currents5.2 Electric circuits R1 R2

  38. X Y Z 7.0 V 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 Topic 5: Electric currents5.2 Electric circuits R1 R2

  39. X Y Z 7.0 V 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. Topic 5: Electric currents5.2 Electric circuits R1 R2

  40. 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. Topic 5: Electric currents5.2 Electric circuits lamp current two currents no currents short circuit!

  41. equivalent ckt 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. Topic 5: Electric currents5.2 Electric circuits

  42. 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. Topic 5: Electric currents5.2 Electric circuits Highest I Middle I Lowest I 1.5R 2R 0.5R R 0.5R parallel series combo

  43. 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 Topic 5: Electric currents5.2 Electric circuits

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