ANSWERS: 5
  • The longer the cord, the more voltage is lost along the cord to the resistance of the wires in the cord, so the less current will be drawn by the appliance.
  • current will indeed be lost but so little be changed that it wont affect the appliance being used unless you were using 1000 mile long extension cord. the power of the current dimishes very very slowly.
  • An 18ga copper wire will have a resistance of about 6.4ohms per 1000ft. You can figure out how much voltage you will lose using the E=IR formula (voltage drop = current * resistance). remember that the wire distance is twice the length of the cord because the current has to go both ways - out and back. If the drop is 5v, your device would be trying to run on 105v vs 110v. Lower voltage will deliver less current to your device. See http://www.powerstream.com/Wire_Size.htm for more info. It has a pretty cool voltage drop calculator.
  • Generally, current delivered to static loads is lowered by increased extension cord length. Resistance goes up; current goes down (I = E / R). Resistance increases with extension cord length, thus voltage drop across the line lowers the voltage available at the end of the cord for use by the appliance. A heavier extension cord (one with more copper in it) has less resistance than a lighter cord, just as in plumbing, a larger pipe has a lessened restriction to the flow of water. Extension cords are rated by an AWG (American Wire Gauge) number; the higher the number, the smaller the cross-section (roughly: diameter) of the (usually copper) conductors in the cord. Stranded wires of the same AWG rating have higher resistance per linear foot than do solid wires; thus solid copper (romex) drops less voltage over longer runs. Extension cords are made of stranded copper (or other metals, perhaps) to maintain flexibility, for temporary applications that can tolerate some amount of voltage drop in order to gain positional advantage (where it would be impractical to run solid copper). In electrical circuits, current (in Amperes) is shown in formulas as represented by the letter I; voltage (in Volts) is represented by the letter E; and resistance (in Ohms) takes the letter R as its representation in a formula known as Ohm's Law: E = I * R (Voltage equals Current times Resistance) Power is calculated by this law: P = E * I (Power equals Voltage times Current) Power is expressed in Watts in electrical circuits. Anything that uses power in an electrical circuit is generally referred to as the load on the circuit. The length of an extension cord, along with the wire gauge of its conductors, influences the current by lowering it (I = E/R; R is increased by extension cord length). For example, ten 60-watt light bulbs connected directly to a 120 volt source draw 5 amperes of current, and present a 24-ohm load to the voltage source and use 600 watts of electrical power between them. However, if a 100-foot, 16-AWG extension cord is placed between the 120 volt source and the light bulbs, a small resistance (0.8344 ohms) is added in series with the 24-ohm load. This affects the circuit in the following ways: 1. Total load is now 24.8344 ohms (24 + 0.8344; R total = R load + R line) 2. Total current is now 4.832 amperes (120 / 24.8344; I = E / R) 3. I-R drop of the load is now 115.97 volts (4.832 * 24; E = I * R) 4. I-R drop of the line is now 4.032 volts (4.832 * 0.8344; E = I * R) 5. Power dissipated by the load is now 560.36 watts (115.97 * 4.832; P = E * I) 6. Power dissipated by the line is now 19.48 watts (4.032 * 4.832; P = E * I) 7. Power dissipated by the system is now 579.84 watts (560.36 + 19.48; P total = P load + P line) Commentary: 1. The circuit gained resistance and will therefore draw less current. 2. The current in the circuit dropped from 5 amps to 4.832 amps. 3. The voltage available to the light bulbs dropped to 115.97 volts. 4. The extension cord dropped the line voltage at the far end by 4.032 volts. 5. The light bulbs use 560.36 watts instead of 600 watts (and are less bright). 6. The extension cord uses 19.48 watts of power (dissipated as heat). 7. The system uses less electric power (579.84 watts not 600 watts). Now, increase the conductor diameter of the extension cord to 12 AWG and shorten it from 100 feet to 25 feet: 1. Total load is now 24.0825 ohms (24 + 0.0825) 2. Total current is now 4.983 amperes (120 / 24.0825) 3. I-R drop of the load is now 119.59 volts (4.983 * 24) 4. I-R drop of the line is now 0.411 volts (4.983 * 0.0825) 5. Power dissipated by the load is now 595.89 watts (119.59 * 4.983) 6. Power dissipated by the line is now 2.048 watts (0.411 * 4.983) 7. Power dissipated by the system is now 597.945 watts (595.89 + 2.048) Commentary (25 foot 12 AWG compared with 100 foot 16 AWG, same work-load): 1. The circuit with the heavier, shorter extension cord lost resistance, and will therefore draw more current with the same 600 watt bulb array (load) attached. 2. The current in the circuit increased a bit, to 4.983 amps. 3. The voltage available to the light bulbs increased to 119.59 volts, closer to nominal 120 volts (design voltage for these bulbs) than the 115.97 volts supplied to the bulbs using the longer, thinner cord. 4. The extension cord dropped the line voltage at the far end by only 0.411 volts, a tenth the (4.032 volt) loss of the other extension cord. 5. The light bulbs use 596 watts -- close to the 600 watts nominal expected of a perfect circuit, and give off more light. 6. The extension cord is heated by only 2.048 watts of power -- much less loss due to heat and provides a margin of safety as well as a stability (resistance increases with temperature); this is more than nine times less heat dissipated by this extension cord as compared with the longer, thinner cord. However, it is dissipated over a shorter run (25 feet) and so the power dissipated per foot is 2.048 watts / 25 feet or 0.082 watts/foot as compared with the longer cord: 19.48 watts / 100 feet or 0.1948 watts/foot. The thinner cord runs hotter, even though it is dissipating its power loss over a longer run. Its resistance increases as it warms and this creates even more heat loss through the cord, over time. 7. The system uses more power (597.945 watts versus 579.84 watts with the other cord). So, the more conservative cord (12 AWG and shortened to 25 feet) will cost more to operate the ten bulbs at the same supply voltage; we get more light, less heating of the extension cord, and if this were a single-phase AC motor load (instead of light bulbs) there'd be less risk to the motor due to brownout (lowered line voltage at the motor). Motors heat rapidly during brownout, dramatically shortening service life, due to longer acceleration times of the rotor as it spins up to operating speed (the longer spin-up times increase heating). Wire Constants (varies by manufacture): 12 AWG stranded: 1.65 Ohms / 1000 feet (606.06 feet per Ohm) 14 AWG stranded: 2.624 Ohms / 1000 feet (381.1 feet per Ohm) 16 AWG stranded: 4.172 Ohms / 1000 feet (239.69 feet per Ohm) 12 AWG solid: 1.5883 Ohms / 1000 feet (629.6 feet per Ohm) 14 AWG solid: 2.5252 Ohms / 1000 feet (396.0 feet per Ohm) 16 AWG solid: 4.0160 Ohms / 1000 feet (249.0 feet per Ohm) Formulas: Power: P = E * I Voltage: E = I * R Resistance: R = E / I Current: I = E / R The algebraic sum of the voltages in a series circuit is zero, thus Source Voltage minus IR drop (line) minus IR drop (load) equal zero. (Note: the terms used throughout this discussion are mine and may not match standard terminology; other inaccuracies may be present. -tetonca)
  • that was shit

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