Electric Circuits Legacy Problem #18 Guided Solution

Having taught for several years, I'm very aware of the fact that when solving electric circuit problems, one of the most common questions which students have is what equation should I use? Now if you click the link that goes back to the overview page to this set of problems, you'll see a collection of equations there. And knowing which one to use centers around practicing the habits of an effective problem solver. That is as you read the problem very carefully that you write down the things that you know and you write down the things that you're looking for. As I read this problem, I read about a high voltage power transmission line and I happen to know the delta V or electric potential difference across the line. I write down delta V equal 4.0 times 10 to the fifth volts. And I also know the power, the rate of energy, energy transmission or the power of these lines. It's listed in megawatts, MW, capital MW, 1500 megawatts where a megawatt is a million watts. And so I can convert from 1500 megawatts to watts and so I write down P equal 1500 megawatts. Then I multiply by 1 times 10 to the sixth and I write down equal 1.5 times 10 to the ninth watts. Now what I'm looking for, my two unknowns are resistance and current. So I simply write R equal question mark and I equal question mark. And now I'm off to a list of equations trying to find the ones which relate P and delta V to I and P and delta V to R. Now when I do that, I find for calculating the current, one of the equations is P equal I times delta V. And so I can take my value for delta V for E to the fifth and for P 1.5 E to the ninth and substitute and solve for current. And when I do, I get 3,750 amps. And if needed, I can round that to two significant digits such as 3,800 amps. Now when it comes to calculating the resistance, I can look for another power equation that relates power to delta V and to R. One of the equations of choice might be P equal delta V squared over R. Substituting and solving would give me about 107 ohms which again can be rounded to two significant digits. Alternatively, there is an equation that relates delta V and I and R. And that equation is delta V equal I times R. And I could just as easily use that equation to solve for R by substituting the calculated value of 3,750 amps into the equation along with the value for delta V.