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The concept of high voltage power transmission in electricity and the role of superconductors in reducing power loss. It includes discussions on the power transfer equation, superconductors, and their properties, as well as examples using light bulbs and circuits. The document also covers the importance of considering all elements in a circuit and the use of Ohm's Law.
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Lecture 28A "We live in a truly magical time. With a flick of a finger, the power of ten horses flows from a small wire in the wall of our homes to clean our carpets." --Steven Chu
Why do electric companies make power lines with very high voltages to deliver electricity to your house? Let’s make some estimations: R = 10 Ω, V = 100,000V to deliver 100kW of power. First, don’t make a common mistake: This equation can only be used with a potential difference (ΔV) across a resistor. <-no!
What if power companies used lower potential for their transmission lines (say 2,000V)? Now we can input this current in the proper power dissipation equations. 25% of power is lost to heat. That isn’t good business.
Supercondutors can help solve these problems. Superconductors are materials whose resistance falls to virtually zero below a certain critical temperature, TC. Once a current is set up in a superconductor, it persists without any applied potential difference.
Light bulbs are resistors that you can measure the power dissipated by them with their brightness. The brighter the bulb, the more power dissipation of the light bulb. I have two light bulbs, one 100W and one 60W. If I hook them up separately to the a given battery which one will be brighter? Correct, the 100W bulb was brighter! What if I hook them up in series to the same battery, which one will be brighter then?
I have two light bulbs, one 100W and one 60W. If I hook them up in series to the same battery, which one will be brighter? A) The 100W bulb. B) The 60W bulb. C) They will have the same brightness. D) It doesn’t matter what power rating they have it will always be the one that is closest to the positive terminal of the battery.
We will now start discussing how current, resistance, power, and devices work together as a circuit. First, become familiar with the following symbols that make circuit diagrams easy to draw.
There are three basic types of emf:
Technically, this is the only circuit you can use Ohm’s Law to solve for the current through the resistor. But we can take more complicated circuits and reduce them down this simple circuit and then apply Ohm’s Law. A typical task is to find the current in a circuit and, maybe, then find the power dissipated by various circuit elements. We will use the same ideas that we developed when finding out the equivalent capacitance for a given circuit.
Let’s say we have a circuit with two resistors in series. When these two resistors are hooked up to the battery, a current is then established. In this circuit, the amount of current moving through R 1 equals the amount of current moving through R 2 (and the same as the battery as well). Ibat = IR1 = IR2 = I
We can essentially replace the two resistors in series with one equivalent resistor. The battery sees current, I, passing through it and believes that the one equivalent resistor has a potential difference of ΔVbat. So, to the battery the equivalent resistance is:
This becomes: The equivalent resistance of resistors in series is greater than the individual resistors. In general, for series resistors:
We can essentially replace the two resistors in parallel with one equivalent resistor. The battery sees current I passing through it and believes that the one equivalent resistor has a potential difference of ΔVbat. As the current, I, approaches point a (a junction), it will break up into I 1 and I 2. I = I 1 + I 2 Req
In general, for parallel resistors: So, to the battery the equivalent resistance is: Req