Transcript transformer
Electromagnetic Induction Current Induction Current Induction Basic Generator Current Induction Current Generator Transformers Transformers Transformers Transformer is just wire coiled around metal • • • • • Magnetic field is generated Secondary Voltage is V2 = (N2/N1) V1 Secondary Current is I2 = (N1/N2) I1 But Power in = Power out – negligible power lost in transformer Works only for AC, not DC An Example • My laptop computer requires about 20 Volts AC, which comes from an adapter (transformer) that is plugged into the wall socket. • What is the approximate ratio of the number of turns on this transformer, given 120 VAC from wall socket? Which coil has more turns, the primary or the secondary? Current Induction between Coils Electric Guitars Microphones Speakers Electrical Power Distribution Household AC Power Transformers Electronic Components • Resisters: Electrical Friction, energy released as heat, but not stored. • I = V/R • Capacitors: Store Electrical Energy • P=V*I • Electric companies bill us for Energy: E = P * t measured in units of kilowatt-hours • D - cell battery stores about 27 watt-hours • Car Battery stores 120 watt-hours Electricity is a Medium for Transporting Energy Where does the energy come from? Is energy “lost” in the transmission wires? What’s the goal, in terms of energy transfer? Putting Electricity to Work • Recall the power consumed by an electrical device is given by the product of the current through it times the voltage drop across it P = VI • Many ways to get the same useful work done, i.e., same power output from electrical device Voltage Current or Voltage Current Segment of an Electrical Power Transmission Cable • Recall Power dissipated in a resistor is P = I2R • How can we minimize power dissipated in the cable? – Minimize R • Short cables with large cross sections • Use high conductivity materials (silver is good!) • Economic considerations limit the cross section and materials • Distribution requirements establish needed lengths – Minimize I • In order to do this, while keeping power delivered to the household appliances (P = VI) the same, must raise the voltage difference between the 2 transmission lines • Which as a bigger impact, halving I or R? Power Dissipated in an Electricity Distribution System 150 miles 120 Watt Light bulb Power Plant • • • • • 12 Volt Connection Box Estimate resistance of power lines: say 0.001 Ohms per meter, times 200 km = 0.001 W/m 2105 m = 20 Ohms With one light bulb on, we can figure out the current it draws using P = VI so I = P/V = 120 Watts/12 Volts = 10 Amps Power in transmission line is P = I2R = 102 20 = 2,000 Watts!! “Efficiency” is e = 120 Watts/2120 Watts = 0.6%!!! What could we change in order to do better? Answer: Must reduce either resistance or the current 1. Reduce resistance in the power lines – Already we’re using pretty hefty copper lines, not very cost-effective to do anything else (superconductors?). 2. Raising the voltage in the system reduces current! – Repeating the above calculation with 12,000 Volts delivered to the house draws only I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving P = I2R = (0.01)220 = 2010-4 Watts, so P = 0.002 Watts of power dissipated in transmission line Efficiency in this case is e = 120 Watts/120.002 = 99.998% Need a Way to Convert! • We need a way to transform from a high voltage electrical distribution system to a low voltage electricity within a household..... • So, use a transformer! High voltage in Low voltage out... A way to provide high efficiency, safe low voltage: step-up to 500,000 V step-down, back to 5,000 V ~5,000 Volts step-down to 120 V High Voltage Transmission Lines Low Voltage to Consumers Examples Class Problem Compared to the huge force that attracts an iron tack to a strong magnet, the force that the tack exerts on the magnet is: a) relatively small b) equally huge Class Problem The pair of forces between the tack and the magnet comprises a single interaction and both are equal in magnitude and opposite in directionNewton's Third Law. Class Problem When current flows in the wire that is placed in the magnetic field shown, the wire is forced upward. If the wire is made to form a loop as shown below, the loop will tend to: a) rotate clockwise b) rotate counter-clockwise c) remain at rest Class Problem The left side is forced up while the right side is forced down as shown. If you make the loop rotate against a spring and attach a pointer to it, you have a simple electric meter. At maximum, it can only make a half turn. But if you make the current change direction (alternate) at every half turn, it will rotate continuously as long as the alternating current persists. Then you have a motor. Class Problem What happens to the reading on the Galvanometer when the switch in circuit 1 is a) first closed b) kept closed c) opened again? Class Problem When the switch is first closed, a current is established in coil 1 and creates a magnetic field which extends to coil 2. This build-up of field in coil 2 induces current which is registered in the Galvanometer. The current is brief, however, because once the field is stabilized and no further charge takes place, no current is induced and the Galvanometer reads zero current. When the switch is opened, the current ceases in coil 1 and the magnetic field in the coil and the part that extends to coil 2 collapses. This change induces a pulse of current in the opposite direction which is registered on the Galvanometer.