Latop Power, Efficiency and Dissipation Questions help 💻

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Alexandramartis
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Hello, I have a question regarding electricity topics focusing on finding efficiency, power dissipation and power. I would be very grateful if someone could help me to improve my solutions below, or just offer friendly advice 👍😁

a. An adaptor for a laptop has the following information: input: 230 V, 1.6 A, output: 19.5 V, 3.34 A.
Calculate:
i. The input power
Input power = V * I
Input power = 230*1.6
P=368 W

ii. The output power
Output power = V * I
Output power = 19.5*3.34
P=65.13 W

iii. The efficiency
Efficiency =useful energy output/total energy input
Efficiency =65.13/368
Efficiency = 0.1769836957 * 100% = 17.69836957% ~17.7% (3.s.f)


b. When in use, the adaptor gets very hot. How is this related to the low efficiency?
I am not sure, but would this be because at high temperatures significant power is dissipated as heat. Since efficiency is defined as the ability of a device to convert energy usefully, this would mean that heat is not a useful power output of a laptop and results in decreasing the overall efficiency.
How can I improve this answer?

c. It takes 5 hours to charge the laptop battery.
i. Calculate the total charge transferred from the adaptor to the laptop battery.
Q=I*t
Q=1.6 (5*60*60)
Q=1.6A*18,000s
Q=28,800 C

ii. Calculate the work done by the adapter.
Work done = V * Q
W=230*28,800
W=6,624,000 J


Thank you to anyone who responds ✌️
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tmr19
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Just looking briefly at the calculations, you've used the right equations and approaches so they're fine assuming you've entered the values right

For b, you've explained it correctly but maybe in an overly verbose way. My main points would relate to - the adaptor getting hot indicates electrical energy is being converted into heat energy, which in this application is not considered a useful output. A large amount of waste heat energy would mean that relatively little useful electrical energy is outputted - indicating low efficiency.

I don't know how many marks b is worth, but basically everything you've done is fine
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Stonebridge
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To calculate the charge transferred from the adapter to the laptop battery, you need to use the current supplied by the adapter to the laptop.
Use the 3.34A value. Not the input current. That would calculate the charge transferred from the main supply to the adapter.
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Alexandramartis
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(Original post by Stonebridge)
To calculate the charge transferred from the adapter to the laptop battery, you need to use the current supplied by the adapter to the laptop.
Use the 3.34A value. Not the input current. That would calculate the charge transferred from the main supply to the adapter.
Thank you for your reply. Sorry, I overlooked this of course you are correct.

So for c i. Calculate the total charge transferred from the adaptor to the laptop battery.
Q=I*t
Q=3.34 (5*60*60)
Q=3.34A*18,000s
Q=60,120 C


ii. Calculate the work done by the adapter.
Work done = V * Q
W=230*60,120
W=13,827,600 J


Is this right now? Have I made any other mistakes? 😊
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Alexandramartis
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(Original post by Stonebridge)
To calculate the charge transferred from the adapter to the laptop battery, you need to use the current supplied by the adapter to the laptop.
Use the 3.34A value. Not the input current. That would calculate the charge transferred from the main supply to the adapter.
Thank you for your reply. Sorry, I overlooked this of course you are correct.

So for c i. Calculate the total charge transferred from the adaptor to the laptop battery.
Q=I*t
Q=3.34 (5*60*60)
Q=3.34A*18,000s
Q=60,120 C


ii. Calculate the work done by the adapter.
Work done = V * Q
W=230*60,120
W=13,827,600 J


Is this right now? Have I made any other mistakes? 😊
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Stonebridge
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(Original post by Alexandramartis)


ii. Calculate the work done by the adapter.
Work done = V * Q
W=230*60,120
W=13,827,600 J


Is this right now? Have I made any other mistakes? 😊
By the same token as my earlier reply, the work done by the adapter would require you to use the output voltage of the adapter, not the input voltage.
I presume they mean the total energy output from the adapter here.
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Alexandramartis
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(Original post by Stonebridge)
By the same token as my earlier reply, the work done by the adapter would require you to use the output voltage of the adapter, not the input voltage.
I presume they mean the total energy output from the adapter here.
Sorry so;
c i. Calculate the total charge transferred from the adaptor to the laptop battery.
Q=I*t
Q=3.34 (5*60*60)
Q=3.34A*18,000s
Q=60,120 C


ii. Calculate the work done by the adapter.
Work done = V * Q
W=19.5*60,120
W=1,172,340 J ?
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Stonebridge
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Yes.
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Alexandramartis
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(Original post by Stonebridge)
Yes.
Great thank you for your help 😁
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