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Transformers Tutorials | Transformers Physics Problems – Voltage, Current & Power Calculations – Electromagnetic Induction

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Transformers Physics Problems - Voltage, Current & Power Calculations - Electromagnetic Induction

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In this video, we’re gonna talk about Transformers, So we’re not talking about Optimus Prime or anything like that, But an electrical device that’s commonly found in an AC plug or an adapter and things like that. So what’s the purpose of a transformer? A transformer can be used to increase or decrease an AC voltage. A step-up transformer increases the AC voltage. A step-down transformer decreases the AC voltage. So let me give an example of one. So this is the circuit diagram or the electrical symbol for a transformer and so there’s two sets of coils on the Left. We have the primary coil on the right the secondary coil so notice that the right side has more coils than the left side. This indicates that we have a step-up transformer. Now let’s say that the number of coils on the left side is a hundred, so that’s the primary coils and the number of coils on the right side is a thousand, so that’s the secondary coil. So the secondary coil has a thousand turns and a primary coil has a hundred turns. Now, let’s say if the primary voltage is 12 volts. What do you think the secondary voltage will be? Its proportional to the number of coils so notice that the number of coils increases by a factor of 10 a thousand divided by 100 is 10 so the secondary voltage is going to be 10 times Higher 12 times 10 is 120 volts. Now, let’s say the primary current, which we’ll call. IP, let’s say it’s 10 amps. What do you think, the primary? I mean, the secondary current will be in the second coil. Now, as the number of turns increases as you go from left to right primary to secondary, the voltage will increase, but the current will decrease, so the current is going to decrease by a factor of 10 ten amps divided by ten will give us a secondary current of one amp now. What is the power on the left side? Power is voltage times current. So if we multiply the primary voltage by the primary current 12 times 10 is 120 so the input power or the power in the primary coil is 120 watts now for the secondary coil or the output power, it’s going to be a 120 times 1 which is 120 so notice that the input power is equal to the output power. The power absorbed by the transformer is equal to the power delivered by the transformer, and this makes sense because energy must be conserved. The rate at which energy is transferred into the transformer has to equal the rate at which energy is transformed or transferred out of the transformer so we can come up with an equation that can relate these variables together, so we said that as the number of coils increases, the voltage increases the current decreases, but the power remains the same, so NS divided by N P. That’s the ratio between the turns in the secondary coil, compared to the turns in a primary coil that’s equal to vs over VP because these two are proportional, but this is equal to IP over is because the current is inversely related to N and V. And then the input power is equal to the output power, which means that the primary voltage times the primary current is equal to the secondary voltage times the secondary current so for transformers, These are the main equations that you need to be familiar with. Now let’s work on some problems. You can pause the video. If you want to and try the problem. A transformer has 400 primary turns and 1800 secondary turns. The input voltage is 12 volts, and the output current is 3 amps when connected across a resistor. What is the output voltage? So let’s start with a picture. So this is going to be the primary coil and here. We’re gonna have the secondary coil now. The secondary coil is connected across a resistor and the primary coil is connected to a source voltage now. The primary coil has 400 turns and the secondary coil has 1200 turns, so VP represents the input voltage and V S represents the output voltage or the secondary voltage. IP represents the current flown in the primary coil or the input. Current is represents the output current or the current in the secondary coil. Now we’re given the input voltage or the primary voltage test 12v we don’t know the output voltage, and we don’t know the input current. However, we do know the output current and so that is 3 amps, so let’s start by finding the output voltage so we can use this formula and P divided by N. S is equal to VP over V S now. I need to make one small correction For some reason. I put 1200 as opposed to 1800 so in this example, NP is 400 and S is 1800 VP is 12 volts and let’s calculate V s, so lets cross multiply 1800 times. Twelve is twenty one thousand six hundred and that’s equal to 400 times V S, so let’s divide both sides by 400 so twenty one thousand six hundred divided by 400 will give us a source voltage or rather a secondary voltage of 54 volts. So that’s the voltage across the coil, the secondary coil and it’s also the voltage across the resistor. Now let’s move on to Part B. Determine the input current so we can use this formula. Np over and S is equal to is over IP Since the current and the number of turns in the coils are inversely related, so NP is 400 and S is 1800 is s3 and that was calculate so 1800 times 3 That’s 5,400 and so that’s going to equal to 400 times IP, so let’s divide both sides by 400 and this will give us a current of 13.5 amps. See what is the value of a resistor? So how can we calculate the value of the resistor? Well, we could use. V equals IR. In this case, we need to use the secondary voltage and the secondary current. So V s is 54 volts. The current in the secondary coil is 3 amps. So now we can calculate. R It’s 54 divided by 3 So the resistance is 18 ohms now. Part D how much power is dissipated by the resistor so we can use? I squared R. The current that flows in the resistor is 3 amps and the value of the resistance itself is 18 Ohms. So 3 squared is 9 9 Times 18 is 162 so 162 watts is dissipated by the resistor. Now, what is the power in the primary coil and in the secondary coil? So the input power or the power in the primary coil is equal to V P Times. I P So that’s 12 volts times 13.5 amps, so this is equal to 162 watts, so the power that was absorbed by the input coil or the primary coil is equal to the power dissipated by the resistor and the power delivered by the secondary coils must be the same -. So if we take V S and X is, we should get the same answer. So 54 volts. Times 3 amps is also equal to 162 watts number 2 An ideal 100 watt transformer has an input current of 20 amps and an output voltage of 12 volts. Determine the input voltage, so lets. Start with a picture, so right now we don’t know if we have a step up or step down transformer because we don’t know the relative amount of turns in the secondary coil compared to the primary coil, So I’m just going to draw a generic transformer and let’s say it’s connected across a resistor in the secondary coil. Now the input current. IP is equal to 20 amps MP. We know its 200 turns, according to Part C. We don’t have the value of NS, and we don’t know the value 4 is now we do have the value 4 vs. The output voltage. That’s 12 volts, but we don’t know VP. So how can we figure out VP? How can we determine the input voltage? Now keep in mind, we know it’s a 100 watt transformer, and we could use that fact to calculate the input voltage. So the power of the transformer is going to equal voltage times current. So 100 watts is equal to VP Times the current of 20 so we need to divide both sides by 20 so a hundred divided by 20 is 5 so the input voltage is 5 Now we could use the same formula to calculate the output current, so it’s going to equal vs. Times is so this is a hundred vs. Is 12 and the secondary current is going to be a hundred divided by 12 and so the secondary current is 8.33 amps. So now we can focus on Part C. How many secondary turns does the transformer have? If there are 200 primary turns, so let’s use this formula. Np / Ns is equal to VP over V S. So N Pianist, example, is 200 Our goal is to calculate N S. VP is 5 and V. S is 12 so 200 times 12 is 24 hundred, and that’s going to equal 5 times n s. So we need to divide this by 5 So N S is 24 hundred divided by 5 which works out to be 480 turns, so that’s. How many turns we have in the secondary coil? So would you say this is a step-up transformer or a step-down transformer? We have 200 turns in the primary coil and 480 turns in the secondary coil and also notice that the voltage increased from 5 to 12 so based on that fact, we know it’s a step-up transformer so anytime that NS is greater than NP, which means that V S will be greater than V P When these conditions are met, we have a step-up transformer, and if we have the reverse, let’s say if N P is greater than N S or if V P is greater than V S, then this is going to be a step-down transformer. So in this example, N S is greater than N P and V S is greater than V. P So we have this situation in his problem number. Three a transformer has an input voltage of 120 volts and an input current of five amps. The output voltage is eleven point nine volts and the output current is 49 amps. How much power was absorbed by the primary coil, so lets. Draw a picture by the way. Do we have a step up or a step-down transformer Notice that we’re going from one to one teats? We’ll let them point that. So this is a step-down transformer, so I need to put more coils on the left side than the right side, so the circuit should look something like this. So this is 120 volts on this side and the current is 5 amps on the right side. The voltage is 11 point nine volts and the current is 49 amps, so how much power was absorbed by the primary coil, so the input power is going to equal V P Times IP. So that’s 120 volts times 5 amps, now 120 times 5 is 600 Watts, so that’s. How much power was absorbed by the primary coil on the left side now? What about the power delivered by the secondary coil? So it’s going to equal vs. Times is so the secondary voltage is eleven point nine as the output voltage and the secondary current is 49 amp’s. Eleven point nine times 49 gives us a power of 580 three point One watts so notice that this transformer is not an ideal transformer. The power is not exactly the same. Most transform is a 99% efficient. Some are even better than that. But it should be close so to calculate the efficiency were gonna take the output power divided by the input power and then multiply by 100% so the output power is five. Eighty three point one watts. The input power is 600 and so if we divide those two numbers 580 3.1 divided by 600 That’s point nine, seven one eight and then multiply that by a hundred. So this particular transformer is ninety seven point, two percent efficient, which is not bad, but there is some amount of energy that’s lost in this example, so no transformers are 100 percent efficient, but there are some that are 99 percent efficient, which is good enough.

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