- C Solve Examples 23 7 Calculating The Self Inductance And 23 8 Calculating The Energy Stored In Section 23 9 Indu 1 (17.97 KiB) Viewed 43 times
- C Solve Examples 23 7 Calculating The Self Inductance And 23 8 Calculating The Energy Stored In Section 23 9 Indu 2 (39.25 KiB) Viewed 43 times
- C Solve Examples 23 7 Calculating The Self Inductance And 23 8 Calculating The Energy Stored In Section 23 9 Indu 3 (79.17 KiB) Viewed 43 times
- C Solve Examples 23 7 Calculating The Self Inductance And 23 8 Calculating The Energy Stored In Section 23 9 Indu 4 (78.31 KiB) Viewed 43 times
- C Solve Examples 23 7 Calculating The Self Inductance And 23 8 Calculating The Energy Stored In Section 23 9 Indu 5 (96.89 KiB) Viewed 43 times
EXAMPLE 23.7 Calculating the Self-inductance of a Moderate Size Solenoid Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils. Strategy This is a straightforward application of L = WNA, since all quantities in the equation except L are known. Solution Use the following expression for the self-inductance of a solenoid: Access for free at opensta.org 23.9. Inductance 1015 LE HONA The cross-sectional area in this example is A = * =(3.14...X0.0200 m) = 1.26 x 10 m, is given to be 200, and the length 1 0.100 m. We know the permeability of free space is 80 = 4* x 10-7Tm/A. Substituting these into the expression for L. gives (x10TIVAX2007 (1.26x10 L = 100 23.42 = 0.632 mH Discussion This solenoid is moderate in size. Its inductance of nearly a millihentry is also considered moderate. One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. An electrical circuit with an inductor is placed in the road under the place a waiting car will stop over. The body of the car increases
Generators convert mechanical energy into electrical energy s motors convert electrical energy into mechanical energy. Furthermore, motors and generators have the same construction. When the coil of a motor is turned, magnetic flux changes, and an emf (consistent with Faraday's law of Induction) is induced. The motor thus acts as a generator whenever its coil rotates. This will happen whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, an emfis generated. Lenz's law tells us the emf opposes any change, so that the input emf that powers the motor will be 02 Chapter 23. Electromagnetic Induction. AC Circuits, and Electrical Technologies opposed by the motor's self-generated emf, called the back emf of the motor (See Figure 23.25.) 0-120 V Could be greater it overspun w R Back em 1120 V Driving em Figure 23.25 The coll of a DC motor is represented as a resistor in this schematic. The back of is represeryed as a variable em that opposes the one driving the motor. Back emf is zero when the motor is not turning, and it increases proporfonally to the motor's angular velocity Back emf is the generator output of a motor, and so it is proportional to the motor's angular velocity ). It is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster and faster the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This effect is noticeable in a number of situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lines by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchait going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overhear it (via resistive power in the coil, P = 1R), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it will increase its angular velocity until the back emfis nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction Consider, for example, the motor coils represented in Figure 23.25. The coils have a 0,400 22 equivalent resistance and are 1 driven by a 48. Vemf. Shortly after being turned on the draw a current/= V/R =(48.0 V V(0.40022)= 120 A and, thus, dissipate P = FR = 5.76 kW of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emfis 40.0 V. Then at operating speed, the total voltage across the coils is 8.0V (48.0 V minus the 400 V back em), and the current drawn is I = V/R =(8.0 V) / 0.400 2)= 20 A. Under normal load, then, the power dissipated is P=/=(20 AV(8,0 V)= 160 W. The latter will not cause a problem for this motor, whereas the former 5.76 kW would burn out the coils if sustained.
23.6 Back Emf It has been noted that motors and generators are very similar. Generators convert mechanical energy into electrical energy, whereas motors convert electrical energy into mechanical energy. Furthermore, motors and generators have the same construction. When the coil of a motor is turned, magnetic flux changes, and an emf (consistent with Faraday's law of induction) is induced. The motor thus acts as a generator whenever its coil rotates. This will happen whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, an emf is generated. Lenz's law tells us the emf opposes any change, so that the input emf that powers the motor will be Chapter 23 - Electromagnetic Induction, AC Circuits, and Electrical Technologies opposed by the motor's self-generated emf, called the back emf of the motor. (See Figure 23.25.) 0-120 V Could be greater if overspun W R Back em 120v 120 V 1 Driving em Figure 23.25 The coil of a DC motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes the one driving the motor Backem is zero when the motor is not turning, and it increases proportionally to the motor's angular velocity Back emf is the generator output of a motor, and so it is proportional to the motor's angular velocity a), Ir is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster and faster the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This effect is noticeable in a number of situations. When a vacuum cleanet, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lincs by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low 2 speed, the larger current can overheat it (via resistive power in the coil, P = 1R), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it will increase its angular velocity es until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction Consider, for example, the motor coils represented in Figure 33.25. The coils have a 0.400 Q equivalent resistance and are driven by a 48. Vemf. Shortly after being turned on, they draw a current I = V/R =(48,0 V (0.400 22)- 120 A and, thus, dissipate P=FR = 5.76 kW of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0V 48.0 Y minus thro DA
Driving emt Figure 23.25 The coil of a DC motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes the one driving the motor. Back emf is zero when the motor is not turning, and it increases proportionally to the motor's angular velocity Back emfis the generator output of a motor, and so it is proportional to the motor's angular velocity 0). It is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it i on but not turning. As the motor turns faster and faster, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This effect is noticeable in a number of situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lines by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overheat it (via resistive power in the coil, P = PR), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it will increase its angular velocity a until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction. Consider, for example, the motor coils represented in Figure 23.25. The coils have a 0.400 22 equivalent resistance and are driven by a 48. Vemf. Shortly after being turned on, they draw a current I = V/R =(48.0 V)/(0.400 22)= 120 A and, thus, dissipate P=1R = 5.76 kW of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0V (48.0 V minus the 40.0 V back emf), and the current drawn is I = V/R =(8.0 V)/(0.400 82)= 20 A. Under normal load, then, the power dissipated is P = IV =(20 A)/(8.0 V)= 160 W. The latter will not cause a problem for this motor, whereas the former 5.76 kW would burn out the coils if sustained. 23.7 Transformers Transformers do what their name implies—they transform voltages from one value to another (The term voltage is used rather than emf, because transformers have internal resistance). For example, many cell phones, laptops, video games, and power tools and small appliances have a transformer built into their plug-in unit (like that in Figure 23.26) that changes 120 V or 240 V AC into whatever voltage the device uses. Transformers are also used at several points in the power distribution systems such as illustrated in Fimur