Commercial electricity is generated and transmitted as three-phas...

Question

Commercial electricity is generated and transmitted as three-phase electricity. Instead of a single emf, three separate wires carry currents for the emfs ε₁=ε₀ cos ωt, ε₂=ε₀ cos(ωt +120°), and ε₃=ε₀ cos(ωt−120°) over three parallel wires, each of which supplies one-third of the power. This is why the long-distance transmission lines you see in the countryside have three wires. Suppose the transmission lines into a city supply a total of 450 MW of electric power, a realistic value.b. In fact, transformers are used to step the transmission-line voltage up to 500 kV rms. What is the current in each wire?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Consider an experimental two phase power system with phase lines A and B. The ems generated by the system are epsilon A is equal to 45 volts multiplied by cosine of omega multiplied by T and epsilon B is equal to 45 volts multiplied by the cosine of omega multiplied by T minus pi divided by two respectively. Here, the value 45 volts represents the peak voltage. The power produced by the emfs transmitted by the two lines is 10 kilowatts. Each of these lines supplies half of the total power. A transformer is used to increase the R MS voltage of each EMF by a factor of 35 calculate the current transmitted by each wire at 45 volts. So that is our end goal is to calculate the current transmitted by each wire at 45 volts. OK. So we are given some multiple choice answers. They're all on the same units of S. So let's read them off to see what our final answer might be. A is 4.5 B is 6.3 C is 200 D is 220. Ok. So before we dive in, let's take a minute to make a mental note here. So in case we're confused, R MS stands for the root mean square. So we're gonna be using that abbreviation a lot in this video in the next few minutes. So let's keep that in mind. So the before we start solving, let's write down all of our known variables. So we're given in the problem that epsilon A epsilon subscript A is equal to 45 volts multiplied by the cosine of omega multiplied by T. We're also given that epsilon B epsilon subscript B is equal to 45 volts multiplied by the cosine of Omega multiplied by T minus pi divided by two. We're also given the total power. I'm gonna write it as capital T capital P. So the total power is equal to 10 kilowatts, which is the same as 10 multiplied by 10 to the power of three watts. And that's just a quick conversion. Now, at this stage, we need to recall and use the equation for the total power produced by two emfs. And let's call this equation one, we'll write it in blue. So equation one states that the power source, I'm gonna denote it as capital P subscript S the power source is equal to I which I represents the current. So Ir MS multiplied by E or I should say epsilon, epsilon R MS for phase line A is add it to Ir MS multiplied by epsilon R MS for phase line B. Awesome. So now we need to take note uh from the problem was the problem states that each of these wires supplies half of the total power. So using that piece of information, we can write that Ir MS multiplied by epsilon RMS for A is equal to Ir MS multiplied by epsilon RMS for line B is all equal to ims multiplied by epsilon RMS. So now we need to calculate the R MS voltage produced by the EMF values for epsilon A and epsilon B before it reaches the transformer. So we need to note that epsilon A represents the instantaneous voltage at any given time T. So since cosine of omega multiplied by T varies between minus one and positive one, the maximum amplitude of epsilon A is 45 volts. So to find the R MS value, we must divide the maximum amplitude by the square root of two. So let's write that down. So epsilon RMS A, so this is epsilon subscript RMS A. So this is focusing on phase line A is equal to 45 volts divided by the square root of two. So we also need to note this is a very important note that epsilon R MS A is equal to epsilon RMS B. Therefore, the current flowing through line A will be equal to the current flowing through line B. After it passes through the transformer, the R MS voltage of each EMF is increased by a factor of 35 as stated to us in the prom cell. Therefore, we can write that epsilon R MS is equal to 35 multiplied by 45 volts, all divided by the square root of two. So when we plug that into a calculator, we will get 1113.69 volts. Now we need to write equation one as the power source ps is equal to two multiplied by multiplied by Ir MS multiplied by. So on R MS. So now we need to take this equation and rearrange it and re rearrange it and isolate and solve for I the current R MS. So let me do that. We will get that Ir MS is equal to the power source divided by two multiplied by epsilon R MS. And we will call this equation two. So now we need to plug in all of our known variables and solve for our Ir MS value. So let's do that. So Ir MS is equal to the power source, which is, we're gonna write it as in watts instead of kilowatts is 10 multiplied by 10 to the power of three watts divided by two multiplied by our ems value which we determine to be 1113.69 volts. So when we plug that into a calculator, we will get four point five yours. Ok. So let's box our correct answer in green here. So our final answer is 4.5 amp here. So the current transmitted by each wire at 45 volts is 4.5 amp here. So looking at our multiple choice answers once again and using our handy dandy green highlighter. The final answer has to be the letter A 4.5 amp here. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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