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Calculations on Power [Help please]

I'm stuck on two questions but I have attempted some parts within the question and I'd like some guidance on the questions that I'm unsure with please.

Question 1: An electric car has a maximum power output of 102 kW. The power supply is a bank of lead acid batteries arranged to give an output voltage of 312 V.
a) What is the maximum current through the batteries?
~ My Answer: I got 326.9 A (1dp)

b) If the batteries store 6.7*10^7 Joules of energy. How long will they last at maximum power?
~ My Answer: I got 656.9 seconds (1dp)

ci) The manufacturers claim that for a normal city/highway driving the car needs 9.3*10^7 J to cover 100 miles. What would the expected range be?
~ I am very stuck on this. Hints would be appreciated in working this out.

cii) The manufacturer quotes the range being 55 to 90 miles. Comment on this.
~ Stuck on this too because I don't know the answer to ci).

Question 2: A remote farm community generates its own electricity using a wind generator. The designer claims that it has an output of 720 W and 12 V in a 20mph breeze.

a) calculate the output current in a 20mph breeze.
~ My Answer: 60 A

b) A circuit known as an inverter converts 12 V d.c. (input) to 230 V a.c. (output). Assuming that the system is 100% efficient show that the output current from the inverter is approximately 3 A.
~ Stuck on this :frown:

Thank you for your time in reading this /hugs

Reply 1

wofldog

Original post by minibuttons

ci) The manufacturers claim that for a normal city/highway driving the car needs 9.3*10^7 J to cover 100 miles. What would the expected range be?
~ I am very stuck on this. Hints would be appreciated in working this out.

cii) The manufacturer quotes the range being 55 to 90 miles. Comment on this.
~ Stuck on this too because I don't know the answer to ci).

So the car batteries can store

6.7 \times 10^7 J

of energy, and it takes

9.3 \times 10^7 J

to travel 100 miles. You could say that it takes

6.7 \times 10^7 J

to travel R miles. This lets you write the problem as

\frac{R}{100 \ miles} = \frac{6.7 \times 10^7 J}{9.3 \times 10^7 J}

This gives you a range of

\approx 72

miles.

I'll leave the commenting part to you!

Reply 2

wofldog

b) A circuit known as an inverter converts 12 V d.c. (input) to 230 V a.c. (output). Assuming that the system is 100% efficient show that the output current from the inverter is approximately 3 A.
~ Stuck on this :frown:

Thank you for your time in reading this /hugs

Sorry, didn't spot this one!

You can solve this two ways. For the first, you need to use your result from the previous question,

60 A

. The inverter scales the voltage up by a scale factor of

\frac{230 V}{12 V} \approx 19

.

Because the voltage has increased, the current must have decreased by the same scale factor to maintain a constant power output of

720W

(this is because of the formula

P = IV

).

This means that the output current will be

\frac{60A}{19} \approx 3

Alternatively, you can do it the easier way, which is to realise that if the efficiency is 100%, you can use the formula

P = IV

using

P = 720 W

(because the efficiency is 100%) and

V = 230 V

(from the question). Simply rearrange and solve!

If I've made any mistakes or you have any more questions, feel free to send me a message!

Reply 3

minibuttons

Original post by wofldog

Sorry, didn't spot this one!

You can solve this two ways. For the first, you need to use your result from the previous question,

60 A

. The inverter scales the voltage up by a scale factor of

\frac{230 V}{12 V} \approx 19

.

Because the voltage has increased, the current must have decreased by the same scale factor to maintain a constant power output of

720W

(this is because of the formula

P = IV

).

This means that the output current will be

\frac{60A}{19} \approx 3

Alternatively, you can do it the easier way, which is to realise that if the efficiency is 100%, you can use the formula

P = IV

using

P = 720 W

(because the efficiency is 100%) and

V = 230 V

(from the question). Simply rearrange and solve!

If I've made any mistakes or you have any more questions, feel free to send me a message!

Omg thank you so much, but honestly you didn't have to tell me the answers straightaway. Rules on TSR say you cannot give answers straightaway but instead you have to give guidance - if a moderator comes and tells you off about it then apologies. You're too kind giving me answers though. I will definitely give you rep tomorrow for your generousity in helping me because I've reached a limit already. Thank you once again ^__^

Reply 4

wofldog

Original post by minibuttons

Oh, really? Sorry! I only joined yesterday, I'll remember that for next time :smile:

Reply 5

minibuttons

Original post by wofldog

So the car batteries can store

6.7 \times 10^7 J

of energy, and it takes

9.3 \times 10^7 J

to travel 100 miles. You could say that it takes

6.7 \times 10^7 J

to travel R miles. This lets you write the problem as

\frac{R}{100 \ miles} = \frac{6.7 \times 10^7 J}{9.3 \times 10^7 J}

This gives you a range of

\approx 72

miles.

I'll leave the commenting part to you!

So for the commenting part would you say that the expected range is within what the manufacturer quoted? Could you just put it as simply as that?

Original post by wofldog

Alternatively, you can do it the easier way, which is to realise that if the efficiency is 100%, you can use the formula

P = IV

using

P = 720 W

(because the efficiency is 100%) and

V = 230 V

(from the question). Simply rearrange and solve!

If I've made any mistakes or you have any more questions, feel free to send me a message!

Now that I look at this, it's actually really simple. I wish I could just answer the question without complicating it so much :/ the joys of being me *sigh* Thank you once again.

Original post by wofldog

Oh, really? Sorry! I only joined yesterday, I'll remember that for next time :smile:

Don't apologise it's not your fault :smile:

Ahh you joined yesterday? Well, I give you a warm welcome in joining the TSR community :biggrin:

Reply 6

wofldog

Original post by minibuttons

So for the commenting part would you say that the expected range is within what the manufacturer quoted? Could you just put it as simply as that?

I'm not sure how you'd answer that one to be honest. I think that because the expected range is slap bang in the middle of the manufacturer's quote, and there is a large margin for error either way, you could say that their quote is accurate.

Original post by minibuttons

Now that I look at this, it's actually really simple. I wish I could just answer the question without complicating it so much :/ the joys of being me *sigh* Thank you once again.

Something that always helps me is to write down everything important to the question as simply as possible. List all the variables that you know the value of, list the variables you are trying to find, and list any equations you think are relevant. That way, it's much easier to see which equations you need to use. The hardest part of worded questions is turning the words into useable mathematical expressions!

Don't apologise it's not your fault :smile:

Ahh you joined yesterday? Well, I give you a warm welcome in joining the TSR community :biggrin:

Thank you very much :smile:

(edited 10 years ago)

Reply 7

Stonebridge

Original post by wofldog

Oh, really? Sorry! I only joined yesterday, I'll remember that for next time :smile:

The guidelines are here.
http://www.thestudentroom.co.uk/wiki/study_help_guidelines

As minibuttons says, we ask that, if you do help out with answering questions, that you avoid, if possible, giving the full answer to numerical problems. Just give gentle hints.

Thanks for your cooperation.

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