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[471.92s -> 480.85s] how far can we go now okay so if we look at what we've got so we've got we've completely solved all of our unknowns on our secondary winding |
[481.36s -> 494.67s] and so now what we need to do is we now need to start concentrating on our primary winding okay so at the moment we've only got 240 volts on the primary okay |
[495.09s -> 510.06s] so what we need to do is we can't use our formula wheel okay so we're going to have to go across and we're going to have to use this formula here okay so we're going to use |
[510.38s -> 513.74s] This formula, okay |
[514.80s -> 529.39s] to get the marks okay so the easier if the information is there okay the assessor can can mark it if you just put the answer then you know and you get it wrong obviously then you get score zero marks a lot of the the points for the exam |
[529.39s -> 539.55s] actually in the working so showing your workings out so actually providing the right answer is quite low well I will only produce like one or two points but demonstrating you can actually |
[539.55s -> 544.98s] Carry out the task to do the correct formula layout accurately okay is |
[545.49s -> 558.67s] That's where the points are won and lost, if you know what I mean. So with this formula here, so VP over VS equals NP over NS equals IS over IP. |
[558.67s -> 571.39s] Now, we don't necessarily have to use VP over VS equals NP over NS. We can use VP over VS equals IS over IP. We can use NP. |
[571.39s -> 585.07s] over NS over NS NP over NS or equals IS over IP I'm tripping myself up here okay so we don't necessarily need to use it from left to right we can actually you know we can use those two |
[585.07s -> 592.59s] and those two we can use those two and those two or we can use these two and these two okay |
[592.88s -> 605.92s] So the first thing I would do is write the formula down, and then underneath that or next to it, write in the values that you've got. And this way we can actually look at what we have. |
[605.92s -> 619.73s] and what we need to find okay and this will actually help us in selecting the formula that we need to actually find our unknowns so we're going to look for |
[619.73s -> 633.14s] NP so we've got because we know VP we know VS and we know NS so we're going to find NP okay so but you could but you could if you wanted to you could go |
[633.14s -> 635.82s] VP over VS, okay? |
[636.69s -> 649.74s] over is over ip okay so you could use these two and find ip but at the moment we're going to find np okay so i've transposed this formula okay |
[649.74s -> 663.71s] Transposing sounds like a horrible word, but all it is is just moving the formula around. I'm not going to go into it now because it's a whole other hour of PowerPoint, if you know what I mean. So what we're going to do is I'm just going to give you the formula. |
[663.71s -> 676.30s] So this is the formula we're going to use so effectively that goes up there, so it's going to be VP times NS over VS equals NP so then what we can do with that is we can then punch in our values |
[677.17s -> 691.10s] Okay, like so, so 240 times 120, divide by 20, okay, and that gives us 1,440, okay, and that's n, so that's the number of turns. So that tells me... |
[691.10s -> 698.91s] okay that the primary coil has 1440 turns on it okay |
[698.91s -> 713.90s] And the secondary has 120. So you can tell that that is going to be a step-down transformer. Or we can say that anyway, but it's a step-down transformer. So the primary now has 144. |
[714.22s -> 720.88s] 144, 140, I'm trying to get my words out, 1440 turns on it |
[723.50s -> 729.74s] So the next step is to then look at we're still back on this formula. Okay, but now we've got |
[730.03s -> 744.48s] All we've got left is the IP to find okay, so That's the formula we're going to use so it's going to be NS times is over NP and that equals IP So then we punch in our room |
[744.48s -> 756.53s] the data that we have so it's going to be 120 times 2.5 over 1440 and that gives us a current on the |
[756.72s -> 768.85s] primary coil of 0.2083 okay so it's quite low current so that's like 208 milliamps so as you can see the higher the voltage |
[769.01s -> 776.59s] Okay, so if the if the voltage is higher on the primary than on the secondary the the current will be lower |
[777.26s -> 789.97s] on the primary okay and higher on the secondary okay so lower voltages generally higher currents higher voltages generally lower currents okay so that's our |
[790.06s -> 794.64s] current on our primary |
[795.76s -> 809.82s] So now what we've got is we've now exhausted this formula here So we can't use that anymore because we know all of the values So what we're then going to have to do is turn back to our formula world. Okay, and we're going to use |
[809.82s -> 823.12s] our formula will and only the information that we have on the primary to solve for our other two unknowns okay so we need to find the power and the resistance okay so |
[823.25s -> 836.86s] We're going to look for the resistance. OK, so resistance from our even from our own law triangle is V divided by I equals our resistance. OK, so voltage divided by the current equals our resistance. |
[836.86s -> 839.44s] And from there we punch in our values that we know. |
[839.73s -> 854.03s] So it's going to be 240 volts divided by 0.2083 and that gives us a resistance on the primary of 1152.18 ohms. |
[854.03s -> 864.69s] quite high resistance okay and that's another value found okay so punch that into your master table so you can keep track of all your values |
[866.16s -> 877.10s] then the next step okay we've got the last thing to do is the power okay as a rule okay the power on the primary |
[877.10s -> 882.61s] okay equals the power on the secondary and vice versa so whatever the power is on the primary |
[882.86s -> 894.50s] Will be on the secondary. Okay. So in our case that we have got 50 watts on the secondary Okay, we will have 50 watts on the primary. Okay, but we're going to |
[894.50s -> 904.61s] We're just going to prove that theory. So we're now going to do voltage times the current, and that equals power in watts. |
[904.61s -> 910.26s] okay so then we punch in what we know so 240 volts times 0.2083 |
[910.51s -> 924.24s] and that equals 49.992 watts okay and the chances are the only reason why it's not bang on 50 is because we rounded our current |
[925.07s -> 938.67s] okay so that's where that comes from so then now you can punch that into our master table and we have completed the problem that is it done nothing more to do okay |
[938.83s -> 948.59s] We found all our unknowns, okay, and we can see We can see what values we have on either the primary or the secondary |
[948.91s -> 958.13s] There is one more step that we can do and it's more of a sanity check just to make sure that our values are correct. |
[958.86s -> 971.79s] What we've got here so 240 divided by 20 will have a value 1440 divided by 120 will have a value and 2.5 divided by zero point |
[972.69s -> 984.53s] two zero eight three will have a value as well and they should all equal the same okay so if you divide that by that actually that will equal 12 now by that that should equal 12 |
[984.53s -> 988.78s] And that by that should equal 12. So they all equal the same. |
[989.55s -> 1003.62s] And so we're just going to just check that mathematically. Okay, so I've done 240 divided by 20 is 12. Okay, 1440 divided by 120 is 12. Okay, and |
[1003.62s -> 1017.31s] 2.5 divided by 0.2083 is 12.00192 and again the reason why that doesn't add up exactly to 12 is because we would have rounded this one |
[1017.31s -> 1021.55s] Okay, so we would have rounded our current on the primary |
[1022.99s -> 1034.77s] That's fine. That is fine if you come up with that and you've got that and you're getting these values It's all good. Okay, so I would be happy with that and it's that is purely just down to |
[1034.77s -> 1046.02s] I'm not rounding error but just the way we've rounded okay so thanks for watching I hope this helps and if you have any questions now please see me in class have a good |
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