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    Right I needed help on part b iii), but there are a few sub parts to this question so I will just provide the whole thing.
    Part b i) http://prntscr.com/bhyicv
    I just simply did P = IV, which had given me a value of 2.7 watts (300^10-3)*(9)
    Part b ii) http://prntscr.com/bhyj7u
    I used transformer efficiency equation e = (Is*Vs)/(Ip*Vp)
    We have e as 0.90, Is*Vs = 2.7 watts which we calculated in the first part, and also were told that Vp = 230 Volts. I simply rearranged for Ip which had given me a value of 13mA.
    Part b iii) This was the part I got stuck on and would appreciate any help, so I know that P = E/t, and we are told that 80% of the time the device is in standby, therefore I do 80% of (3.15*10^7 seconds) multiplied by 2.7 watts? This gives me an energy value of 6.31*10^7 Joules, however this is incorrect, as instead of using 2.7 watts (output power) they used 3.0 watts (input power), why was the input power used?

    Thanks in advance, also the correct answer to part b iii) is 7.56*10^7 Joules.
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    (Original post by 1017bsquad)
    Right I needed help on part b iii), but there are a few sub parts to this question so I will just provide the whole thing.
    Part b i) http://prntscr.com/bhyicv
    I just simply did P = IV, which had given me a value of 2.7 watts (300^10-3)*(9)
    Part b ii) http://prntscr.com/bhyj7u
    I used transformer efficiency equation e = (Is*Vs)/(Ip*Vp)
    We have e as 0.90, Is*Vs = 2.7 watts which we calculated in the first part, and also were told that Vp = 230 Volts. I simply rearranged for Ip which had given me a value of 13mA.
    Part b iii) This was the part I got stuck on and would appreciate any help, so I know that P = E/t, and we are told that 80% of the time the device is in standby, therefore I do 80% of (3.15*10^7 seconds) multiplied by 2.7 watts? This gives me an energy value of 6.31*10^7 Joules, however this is incorrect, as instead of using 2.7 watts (output power) they used 3.0 watts (input power), why was the input power used?

    Thanks in advance, also the correct answer to part b iii) is 7.56*10^7 Joules.
    Well every joule input to the transformer ends up on your electricity bill, including any losses in the transformer due to inefficiency.

    with the switch located to disconnect the transformer primary from the mains as shown in the diagram - the power when switched off is zero.
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    (Original post by Joinedup)
    Well every joule input to the transformer ends up on your electricity bill, including any losses in the transformer due to inefficiency.

    with the switch located to disconnect the transformer primary from the mains as shown in the diagram - the power when switched off is zero.
    Oh I see, thanks man!
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    (Original post by 1017bsquad)
    Oh I see, thanks man!
    it probably could have been worded more clearly,
    i.e. energy wasted by using standby mode instead of switching off properly
 
 
 
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