okay then so task B now is is slightly slightly different and so what i've done
i've altered the board round so instead we're not using these lights anymore we're going to be using
these lights here so let's just turn turn them on so we turn the RCD on
and there you go you can see those lights have now come on so it's a slightly less load so we would
expect to see slightly less current so let's turn the ammeter to the current and this time we are
doing things in reverse so we're actually going to measure the secondary voltage and the secondary
current and then we work our way backwards we'll calculate the voltage and current in the primary
so again we can just take any of the uh either lead and we just put it round the clamp on ammeter
and this time it's reading 8.6 let's hold press hold and it's actually now reading 8.6
amps 8.6 amps and that's the secondary side so i'll just make a note of that
turn it um let's just turn it off and we'll also turn off the circuit breaker
and now i'll put it in to the voltage in parallel with the secondary side turn the voltage on
and read the voltage and measure and hold it and this time it's reading 12.53 oh it's 12.53 volts
so we've done that and now we can go back and do the calculations
so now we can do the calculations so we've got here we've measured the voltage in the
secondary and the current in the secondary this time we need to now um we need to calculate
the voltage and current in the primary so again we're going to use the same formula i'll just
write it up here so we've got the voltage in the primary over voltage in the secondary
is equal to the number of turns in the primary over number of turns in the secondary which is
equal to the number to the voltage of the current in the secondary over the current in the primary
so to calculate the voltage we're just going to use that part of the formula so
again i like to write it all out again so VP over VS is equal to NP over NS
okay so again we write these down the the figures down on the on the right hand side
just so it's all clear so the voltage in the primary so that's what we want to calculate
the voltage in the secondary is 12.53 volts the number of turns in the primary and number
of turns in the secondary so again it's exactly the same so that was 600
to 30 which brings it down to the ratio of 20 over 1 which makes it much more simple so this time
let's work on this so VP is what we want to find so again I put X there over VS which was 12.53
is equal to 20 over one and this time all we have to do again to get everything on the top line
is we just cross multiply so one times X is just X so X is equal to 12.53
times 20. so therefore X is equal
so that's twelve point five three times twenty
at equal two hundred and fifty point 6
volts so that's what we've calculated it to be which looks slightly high but
we won't worry about that just yet and then we will measure the current
in the primary side as well so again we can use go back to the original formula
we can either use VP over VS is equal to IS over IP which we did previously or we can
do this one and i think because we did it the other way around we'll do this one for this one
so we'll write this down NP over NS is equal to IS over IP
and this is it's IP that we want to find so we have a ratio NP over so 20 over one
is the number of turns in the primary in the secondary is equal to IS which we know IS is 8.6
over X which is IP
so let's write the figures down here so we know that NP is 20 we know NS is one we know is
is 8.6 amps and we know IP question mark just quickly i'll just notice there that
i forgot to write the answer there so that's 250.6
volts let's back back to this one so this time then we now just cross multiply
exactly the same as previously so that's one times 8.6 is equal to 20X so that's 8.6
is equal to 20X
okay so now we need to do a little bit further calculation so we want X to be on its own so
if we divide this side by 20 we must divide this side by 20. by dividing this side by 20 it cancels
out the 20. so X is equal to 8.6 over 20. when i wrote this out here i should have wrote it so
it should have really been 20X is equal to 8.6 but so now all i've done i've just
flipped it around the other way so X is on the left-hand side so X is equal to 8.6 over 20.
so X is equal to again we'll press the fraction button and that was 8.6
over 20. press equals
and that's 0.43 amps X is equal to 0.43 amps
0.43 amps now that sounds about right because of previously when we had the higher um wattage
lamps it was i think it was was it 0.69 amps or something so we would expect that one to be lower
anyway so so what we've done we have calculated this one and we have calculated that one
so now all that's left to do is go and re-measure the primary side and compare the results
okay then so back on the rig so let's turn turn this back on
so this time then so previously we measured the um initially we measured the current and
the voltage in the secondary side so now we're going to measure the voltage and the current
in the primary side so simple measuring the current we just go around one of the loops
let it settle we take the rms value and this time it's reading 0.53 amps
so let's just make a note of that so that's 0.53 amps
we turn that off and this time we measure the voltage
again the voltage is measured in parallel to the supply turn it on
let it settle press hold and you can see there 243.6
let's make a note of that 243.6
so that's that so let's just go and look at the results and see
see what we can make from that this one is task B so let's just go through what we did we've got
the secondary that we measured the both the voltage and the current we then calculated
the voltage and the current in the primary and as you can see it's laid out nice and clearly
everything all labeled nice and clear and you can see exactly what we've done now as far as i'm
concerned that is the end of the task but i took it one step further how we would be doing it in
on the assessment because that's what we would be doing
and in the video so we've actually now measured and we've got the current and the volts 0.53
243 now as you can see they are considerably out so the the calculated voltage was 250
the measured voltage was 243 and and the same with the current 0.43.53 now the reason because of that
on task a we did it in the correct order we measured the primary voltage first
and primary current and then we worked our way through and we got some reasonable
results and we expected to see what we got because of the um because of the losses
this time however we are calculating the secondary side on task b we've calculated the secondary side
and then we've worked our way backwards doing the calculations which don't seem to always
work out well they don't work out and that's why um because we haven't been we haven't taken
into account the losses properly so we've got a higher voltage and a higher current than um
a higher a higher voltage and a lower current compared to what we should have had really but
it doesn't really matter um the the task for this exercise isn't isn't to actually is to basically
show that you understand the calculations and you understand the process of how to do it
and it's also how to use um a an ammeter for measuring the current and the voltage so for
that for the sake of that i think that's a good um that's a good task that we've done