Can someone explain this A2 transformer question please? Watch

JamieOH
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Here's the context:

"When in standby mode, the transformer supplied an output current of 300mA at 9.0V to the internal circuits of the TV set."

Here's part 1:

"Calculate the power wasted in the internal circuits when the TV set is left in standby mode."

0.3*9 = 2.7W (I get this part)

And here is part 2, which I don't understand:

"If the efficiency of the transformer is 0.90, show that the current supplied by the 230 V mains supply under these conditions is 13mA."

I thought you would multiply 2.7W by 9 to get 90% (as 10% of power is wasted due the the transformer being 0.90 efficient) then divide this value by the 230V.
This isn't the case and instead you treat the 2.7W as the power that isn't wasted. So you're supposed to divide the 2.7W by 0.9 to get the total power and divide that by 230V.

I don't understand why the 2.7W isn't the 10% that is wasted, especially since part 1 calls it the wasted power.

Help! Thanks.
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atsruser
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(Original post by JamieOH)
Here's the context:

"When in standby mode, the transformer supplied an output current of 300mA at 9.0V to the internal circuits of the TV set."

Here's part 1:

"Calculate the power wasted in the internal circuits when the TV set is left in standby mode."

0.3*9 = 2.7W (I get this part)

And here is part 2, which I don't understand:

"If the efficiency of the transformer is 0.90, show that the current supplied by the 230 V mains supply under these conditions is 13mA."

I thought you would multiply 2.7W by 9 to get 90% (as 10% of power is wasted due the the transformer being 0.90 efficient) then divide this value by the 230V.
This isn't the case and instead you treat the 2.7W as the power that isn't wasted. So you're supposed to divide the 2.7W by 0.9 to get the total power and divide that by 230V.
If the power on the 230 V side is P_\text{in} and that on the 9 V side is P_\text{out} then we have:

2.7 = P_\text{out} = 0.9 P_\text{in}

so rearrange that to get the result in mark scheme.
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JamieOH
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(Original post by atsruser)
If the power on the 230 V side is P_\text{in} and that on the 9 V side is P_\text{out} then we have:

2.7 = P_\text{out} = 0.9 P_\text{in}

so rearrange that to get the result in mark scheme.
2.7 is the power wasted, so when talking about the efficiency why isn't the 2.7 the lost power to heat? And in that case 2.7 is 0.1 of the total power.

I know I am just completely misunderstanding something really fundamental but I just can't get my head around what is going on aha.
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JamieOH
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After the reading the question again properly I now understand! My inability to read the context will be my downfall. Thanks anyway.
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