# Back to the Future Physics Questions (Power and charge) 🚗

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#1
Hello, I have some rather fun questions below which I was wondering if anyone could help me with in terms of improving or correctly my answers. I would be very thankful of any help or responses👌
Question i Currents of approximately 10000 A are common for lightning bolts. Lightning only lasts for approximately 1 millisecond.

1. Calculate the charge passing during this time.

I=Q/t
Therefore, Q=I*t
Q=10,000 A * (1.0*10^-3) s
Q= 10 C

2. In the film Back to the Future, a conducting rod was used to power the DeLorean time machine. Assuming the potential difference between the charged cloud and the conducting rod is 100 million volts (a standard result, but values can vary considerably), calculate:
a. The energy transferred along the conducting rod.
100 million volts = 100 MV
Voltage = energy transferred/charge passing
Energy transferred = voltage * charge
E=100,000,000 * 10 C
E= 10,000,000 J or 1.0 * 10^7J

b. The theoretical power generated.
Power = E/ t
Power = 10,000,000 J /0.001 s
Power =10,000,000,000 W or 1.0 * 10^10 W

3. Doc Brown from the film said 1.21 Giga Watts were required. Does your answer suggest there was sufficient power? What else might need to be considered?
Yes, there is sufficient power since 10,000,000,000 W = 10 GW ~ 10 times the required power. However, I am not sure what else might need to be considered?

Question ii In an experiment with an electric motor, an object of mass 550 g was lifted a height of 1.5 m in 12.1 s. A voltmeter read the potential difference across the motor as 5.1 V and an ammeter read the current to the motor as 1.41 A.
1. How much work was done lifting the load?

Work done= mgh
Work done = 0.5*9.81*1.5
Work done =7.3575 ~ 7.40 J

2. What was the power output?
Power=work done/time taken
P=7.3575/12.1
P=0.60805785 ~ 0.608 W

3. What was the power input to the motor?
Energy input = electrical energy used = VIt
VIt=5.1*1.41*12.1=87.0111 ~ 87 W

4. What was the efficiency?
Efficiency =useful energy output/total energy input
Efficiency =0.608 /87.0111
Efficiency = 0.00698 * 100% = 0.698%
I think this is wrong would it rather be:

Efficiency =mgh/total energy input
Efficiency =7.3575/87.0111
Efficiency =0.08455*100% =8.455%

5. Suggest where energy might have been wasted.
Energy may have been wasted as thermal energy resulting from the friction of the rope raising the mass and slightly as sound.

Question iii.
In Question i, a calculation was done on lightning being conducted via a lightning rod to power a DeLorean time machine. If the length of wire was 8.5 m long, with a diameter of 2.4 mm and had a resistance of 9 Ω:
a. What was the resistivity of the material?
R=pL/A
A = π r^2 or πd^2/4
Rearranging:
p=RA/L
p=R*( πd^2/4)/L
p=9*(π*2.4^2/4)/8.5
p=4.79 Ωm

b. How much power was dissipated due to the resistance of the wire for a current of 10000 A? Does this affect your answer to Question ii 3?
Power dissipation equation:
P=I^2R
P=10000^2*9
P=900,000,000 W or 9.0 * 10^8 W

I am not sure how this effects the power input to the motor?
Last edited by Alexandramartis; 1 year ago
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1 year ago
#2
Made a mistake in 2, you divided by 10 instead of multiplying
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#3
(Original post by Sinnoh)
Made a mistake in 2, you divided by 10 instead of multiplying
Thank you I missed that! Correcting this :

2. a. Energy transferred = voltage * charge
E=100,000,000 * 10 C
E= 1,000,000,000 J or 1.0 * 10^9J

b. The theoretical power generated.
Power = E/ t
Power = 1,000,000,000 J /0.001 s
Power =1,000,000,000,000 W or 1.0 * 10^12 W

3. Yes, there is sufficient power since 1,000,000,000,000 W = 1000 GW ~ 1000 times the required power. However, I am not sure what else might need to be considered?
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1 year ago
#4
(Original post by Alexandramartis)
However, I am not sure what else might need to be considered?
Well you're assuming a lot of things in this - that the current is constant, that all the energy is transferred to the generator, which probably wouldn't be true - there would definitely be some dissipation of energy through heat
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#5
(Original post by Sinnoh)
Well you're assuming a lot of things in this - that the current is constant, that all the energy is transferred to the generator, which probably wouldn't be true - there would definitely be some dissipation of energy through heat
So how could I improve my answer? I was trying to follow the information given as closely as possible .
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1 year ago
#6
(Original post by Alexandramartis)
So how could I improve my answer? I was trying to follow the information given as closely as possible .
I meant that as a hint for "what else needs to be considered"
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#7
(Original post by Sinnoh)
I meant that as a hint for "what else needs to be considered"
Oh, sorry I thought you were referring to the entirety of the question.
3. So, considering the variables given account that the current is constant and all of the energy is transferred to power the time machine. In actuality, energy would be dissipated as heat so the theoretical power generated would be less than 1000GW, but still sufficient to fulfil the 1.2GW mark.
How could I elaborate further?
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1 year ago
#8
(Original post by Alexandramartis)
Oh, sorry I thought you were referring to the entirety of the question.
3. So, considering the variables given account that the current is constant and all of the energy is transferred to power the time machine. In actuality, energy would be dissipated as heat so the theoretical power generated would be less than 1000GW, but still sufficient to fulfil the 1.2GW mark.
How could I elaborate further?
How many marks is it? Sounds like enough to me
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#9
2 marks, but I thought that they were sort of similar points so maybe I should include something else? Thank you for your help I am very grateful 😁👍
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#10
Sinnoh, please see my original post. I have included some further related questions, would you be able to advice on these at all? Sorry if it is too much to ask and thank you for your help so far 👍😁
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1 year ago
#11
For power input to the motor, it looks like you just calculated an energy?
P = IV. No need to use the time taken. Then for efficiency that eliminates any ambiguities because you can just do power out / power in.
And it looks like you didn't convert mm to m in your resistivity equation. Other than that everything looks good.

My guess for the last part is that you would have to subtract the wasted power from the input and see if you still have the required power. May be wrong but I'm not sure of what else to do with that info.
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#12
(Original post by Sinnoh)
For power input to the motor, it looks like you just calculated an energy?
P = IV. No need to use the time taken. Then for efficiency that eliminates any ambiguities because you can just do power out / power in.
And it looks like you didn't convert mm to m in your resistivity equation. Other than that everything looks good.

My guess for the last part is that you would have to subtract the wasted power from the input and see if you still have the required power. May be wrong but I'm not sure of what else to do with that info.
Thank you very much again for all of your help 👍In hindsight, I am not sure why I wrote Power=VIt, I think I read it in my textbook with a similar investigation and it must have stuck in my mind. 😂

3. Energy input = electrical energy used = VI
VI=5.1*1.41=7.191 W

4. What was the efficiency?
Efficiency =useful energy output/total energy input
Efficiency =0.608 /7.191
Efficiency = 0.08455 * 100% = 84.55%

Question iii.
In Question i, a calculation was done on lightning being conducted via a lightning rod to power a DeLorean time machine. If the length of wire was 8.5 m long, with a diameter of 2.4 mm and had a resistance of 9 Ω:
a. What was the resistivity of the material?
R=pL/A
A = π r^2 or πd^2/4
Rearranging:
p=RA/L
p=R*( πd^2/4)/L
p=9*(π*0.0024^2/4)/8.5
p=4.79 * 10^-6 Ωm

b. So power dissipated = 9.0 * 10^8
Power input= 7.191 - 9.0 * 10^8 = -899999992.809

I think that I have made a mistake here?
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1 year ago
#13
(Original post by Alexandramartis)
b. So power dissipated = 9.0 * 10^8
Power input= 7.191 - 9.0 * 10^8 = -899999992.809

I think that I have made a mistake here?
I'd have used the value of 1010 W from earlier
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#14
(Original post by Sinnoh)
I'd have used the value of 1010 W from earlier
Sorry, I overlooked that value, for some reason I keep assuming they are seperate questions. In which case.
Question iii.
b. Power dissipation equation:
P=I^2R
P=10000^2*9
P=900,000,000 W or 9.0* 10^8 W

Power input = 1.0 * 10^10 W
So, 1.0 * 10^10 W-9.0* 10^8 W= 9,100,000,000 W or 9.1 * 10^9 W

I do not see what this has to do with the power input to the motor in question ii 3 though? The electric motor features in an experiment, not in the DeLorean questions? I am a little puzzled by the wording here.
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1 year ago
#15
(Original post by Alexandramartis)
Sorry, I overlooked that value, for some reason I keep assuming they are seperate questions. In which case.
Question iii.
b. Power dissipation equation:
P=I^2R
P=10000^2*9
P=900,000,000 W or 9.0* 10^8 W

Power input = 1.0 * 10^10 W
So, 1.0 * 10^10 W-9.0* 10^8 W= 9,100,000,000 W or 9.1 * 10^9 W

I do not see what this has to do with the power input to the motor in question ii 3 though? The electric motor features in an experiment, not in the DeLorean questions? I am a little puzzled by the wording here.
Yeah I thought they were unrelated
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#16
(Original post by Sinnoh)
Yeah I thought they were unrelated
I know! Maybe the question is written erroneously, yet it does emphasise specific mention of that question:

"How much power was dissipated due to the resistance of the wire for a current of 10000 A? Does this affect your answer to Question ii 3?"

Question ii 3 " What was the power input to the motor?"

Perhaps, they mean question i 3 " Doc Brown from the film said 1.21 Giga Watts were required. Does your answer suggest there was sufficient power? What else might need to be considered?"

That would be far more logical and would surely be a continuation of this working:
1.0 * 10^10 W-9.0* 10^8 W= 9,100,000,000 W or 9.1 * 10^9 W

Since the sufficient power is set at 1.21 GW,
9,100,000,000=9.1 GW which does affect my answer to question ii. 3 since the power is reduced from 10 GW to 9.1 GW, which although still provides sufficient energy, does mean there is less overall available energy and this is prior to considering energy losses e.g. heat dissipation. So although the DeLorean will still function effectively it does place a slight disadvantage in case of any adverse causes of external energy losses.
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1 year ago
#17
(Original post by Alexandramartis)
I know! Maybe the question is written erroneously, yet it does emphasise specific mention of that question:

"How much power was dissipated due to the resistance of the wire for a current of 10000 A? Does this affect your answer to Question ii 3?"

Question ii 3 " What was the power input to the motor?"

Perhaps, they mean question i 3 " Doc Brown from the film said 1.21 Giga Watts were required. Does your answer suggest there was sufficient power? What else might need to be considered?"

That would be far more logical and would surely be a continuation of this working:
1.0 * 10^10 W-9.0* 10^8 W= 9,100,000,000 W or 9.1 * 10^9 W

Since the sufficient power is set at 1.21 GW,
9,100,000,000=9.1 GW which does affect my answer to question ii. 3 since the power is reduced from 10 GW to 9.1 GW, which although still provides sufficient energy, does mean there is less overall available energy and this is prior to considering energy losses e.g. heat dissipation. So although the DeLorean will still function effectively it does place a slight disadvantage in case of any adverse causes of external energy losses.
He said jiggerWatts in the movie

have a look at the rating plate on a kettle or something... is there usually any other information given about the power supply it needs to be connected to apart from power/watts?
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#18
(Original post by Joinedup)
He said jiggerWatts in the movie

have a look at the rating plate on a kettle or something... is there usually any other information given about the power supply it needs to be connected to apart from power/watts?
Haha 😆 That unit should become commonplace.

Would the power supply need to convert the electrical current from a source to the correct voltage, current frequency to power the load?
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1 year ago
#19
(Original post by Alexandramartis)
Haha 😆 That unit should become commonplace.

Would the power supply need to convert the electrical current from a source to the correct voltage, current frequency to power the load?
Yeah those are all valid IMO

Also Watts is probably the wrong unit... especially if you don't specify a duration. e.g. one Gigawatt for a nanosecond isn't really an impressive amount of total energy transferred... one gigawatt continuous is pretty enormous - comparable to the full output of a large nuclear power station and probably enough to turn the delorian into a puddle of white hot metal in a second or two
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#20
(Original post by Joinedup)
Yeah those are all valid IMO

Also Watts is probably the wrong unit... especially if you don't specify a duration. e.g. one Gigawatt for a nanosecond isn't really an impressive amount of total energy transferred... one gigawatt continuous is pretty enormous - comparable to the full output of a large nuclear power station and probably enough to turn the delorian into a puddle of white hot metal in a second or two
Haha, so what unit should be used instead for it to be valid here? 😁
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