Physics Oscilloscope Question Watch

QuantumBoi
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Hi, can anyone help me with part b of this question? I'm not sure what equations I'm supposed to use.

https://imgur.com/a/7jo2BNv
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Ecdysiastt
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It's a power question. You know R is 100 and in the first 1ms that V is -5, so you can use P=(V^2)/R
Do the same for the other time divisions
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QuantumBoi
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(Original post by Ecdysiastt)
It's a power question. You know R is 100 and in the first 1ms that V is -5, so you can use P=(V^2)/R
Do the same for the other time divisions
It's not a power question. It's asking for the amount of energy dissipated.
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BobbJo
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(Original post by QuantumBoi)
It's not a power question. It's asking for the amount of energy dissipated.
Energy = Power x time
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Ecdysiastt
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(Original post by BobbJo)
Energy = Power x time
Ty, forgot to add this part on. You can work out the power using V and R and then multiply by t to get the energy
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QuantumBoi
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(Original post by Ecdysiastt)
Ty, forgot to add this part on. You can work out the power using V and R and then multiply by t to get the energy
Ah thanks, so what would I use as my value for power in a whole cycle? The total change in power (10V - (-5V))?
Last edited by QuantumBoi; 1 month ago
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QuantumBoi
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What would you take power to be for iii and iv? (and why)?
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QuantumBoi
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(Original post by BobbJo)
Total power dissipated in 1 cycle = Power dissipated in 1 cycle

(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle
Yeah but how do you find the power dissipated in 1 cycle. Do you add up the power (in which case power would equal 0)? Do you multiply power by time? Do you find an average for power by dividing power by the number of horizontal lines? What do you do?
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BobbJo
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(Original post by QuantumBoi)
Yeah but how do you find the power dissipated in 1 cycle. Do you add up the power (in which case power would equal 0)? Do you multiply power by time? Do you find an average for power by dividing power by the number of horizontal lines? What do you do?
Power will not be equal to 0.

P = V^2/R = I^2 R which are never negative.

Mean power = Energy in 1 cycle / Time for 1 cycle
Use the time base and y-gain.

Energy dissipated in 1 cycle = Energy during 0 to 1 ms + Energy during 1 to 1.5 ms because first 1.5ms is one cycle

so ans to (b)(iii) = ans to (b)(i) + ans to (b)(ii)

Energy dissipated in 1 cycle = energy dissipated in time for 1 cycle / time for 1 cycle = energy dissipated in T seconds / T

Mean power dissipation = energy dissipated in T seconds / Time for 1 cycle = energy dissipated in T seconds / T

ans to (b)(iv) = energy dissipated in 1 cycle / time for 1 cycle = ans to (b)(iii) / time for 1 cycle = ans to (b)(iii) / period of signal = ans to (b)(iii) x frequency of signal = ans to (b)(iii) x ans to (a)(iii)
Last edited by BobbJo; 1 month ago
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Eimmanuel
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(Original post by QuantumBoi)
Hi, can anyone help me with part b of this question? I'm not sure what equations I'm supposed to use.

https://imgur.com/a/7jo2BNv
To attempt (b), it may be useful to transform the voltage versus time to V2 versus time graph. For example, for the first 1.0 ms, the voltage is -5.0 V, just square it and it becomes 25.0 V and then do it for the rest of the time intervals.

To find for the energy dissipated in the resistor of resistance R for any time interval Δt, find the area under the graph of V2 versus time for that time interval Δt, and then divided by R.

Note that I find there are some confusing phrases used above. Don’t really know what info it is conveying.

Total power dissipated in 1 cycle = Power dissipated in 1 cycle
(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle
Mean power is just mean power. In kinematics, we can say average velocity which is found by dividing the total displacement by total time taken. Based on the above writing,

(Average) velocity in 1 second = velocity occur in 1 cycle / Time for 1 cycle

(IMO) It is a confusing way of writing.
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Eimmanuel
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I just moved this to the physics forum where some of the experts may provide you with more info if the need arises.
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BobbJo
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(Original post by Eimmanuel)
becomes 25.0 V^2 and
Mean power is just mean power. In kinematics, we can say average velocity which is found by dividing the total displacement by total time taken. Based on the above writing,

(Average) velocity in 1 second = velocity occur in 1 cycle / Time for 1 cycle

(IMO) It is a confusing way of writing.
They just asked how to find to find answer to (b)(iv) in post 9. I'm very sorry you find it confusing. It is the mean power dissipated = total power dissipated in 1 cycle / time for 1 cycle.

I don't understand what is meant by "velocity occur in 1 cycle".
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Eimmanuel
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(Original post by BobbJo)
They just asked how to find to find answer to (b)(iv) in post 9. I'm very sorry you find it confusing. It is the mean power dissipated = total power dissipated in 1 cycle / time for 1 cycle.

I don't understand what is meant by "velocity occur in 1 cycle".

Look at your writing, it originates from you.
(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle
(Average) velocity in 1 second = velocity occur in 1 cycle / Time for 1 cycle
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BobbJo
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(Original post by Eimmanuel)
Look at your writing, it originates from you.
(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle
(Average) velocity in 1 second = velocity occur in 1 cycle / Time for 1 cycle
Why not say displacement over 1 cycle?

I still do not understand velocity occur in 1 cycle. [I didn't say it]

I edited and explained it more. Also (-5V)^2 = 25V^2 ?
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Eimmanuel
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(Original post by BobbJo)
They just asked how to find to find answer to (b)(iv) in post 9. I'm very sorry you find it confusing. It is the mean power dissipated = total power dissipated in 1 cycle / time for 1 cycle.
(Original post by BobbJo)
Ok
If you read OP’s questions in post #9 again, the questions are raised in regard to your previous post #8. Don’t really know what you are trying to say in post #8. I believe in A-level physics, you have learned
Energy transferred = Power × time taken
and this should be used to find mean power
Mean power = total energy transferred in 1 period / 1 period
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BobbJo
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(Original post by Eimmanuel)
If you read OP’s questions in post #9 again, the questions are raised in regard to your previous post #8. Don’t really know what you are trying to say in post #8. I believe in A-level physics, you have learned
Energy transferred = Power × time taken
and this should be used to find mean power
Mean power = total energy transferred in 1 period / 1 period
Oh my mistake, I meant energy

I understand how to find an average

"Power dissipated in T seconds" should have been Energy dissipated in T seconds.
Last edited by BobbJo; 1 month ago
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QuantumBoi
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(Original post by BobbJo)
Power will not be equal to 0.

P = V^2/R = I^2 R which are never negative.

Mean power = Energy in 1 cycle / Time for 1 cycle
Use the time base and y-gain.

Energy dissipated in 1 cycle = Energy during 0 to 1 ms + Energy during 1 to 1.5 ms because first 1.5ms is one cycle

so ans to (b)(iii) = ans to (b)(i) + ans to (b)(ii)

Edited to explain mean power = energy dissipated in 1 cycle / time for 1 cycle:

Power dissipated in 1 cycle = energy dissipated in time for 1 cycle / time for 1 cycle = energy dissipated in T seconds / T

Mean power dissipation = energy dissipated in T seconds / Time for 1 cycle = energy dissipated in T seconds / T

ans to (b)(iv) = energy dissipated in 1 cycle / time for 1 cycle = ans to (b)(iii) / time for 1 cycle = ans to (b)(iii) / period of signal = ans to (b)(iii) x frequency of signal = ans to (b)(iii) x ans to (a)(iii)



Average velocity,  v \bar =  \displaystyle \dfrac{1}{t_2-t_1} \int_{t_1}^{t_2} v(x) dx
e.g v(t) = 2

Then average velocity for t = 3 to t = 5,  v = \displaystyle \dfrac{1}{5-3} \int_{3}^{5} 2 dx = 2

If motion is periodic and it is meaningful to talk of cycles, then v = displacement over 1 cycle / time for 1 cycle

I do not understand "velocity occur for 1 cycle".
Wha?? Why are you integrating??? . Power would equal 0 because:

V = -5 + -5 + 10 = 0 volts
P = V^2/R
P = 0^2/100
P = 0 Watts
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QuantumBoi
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Can someone just tell me how to find energy dissipated in one cycle and energy dissipated in one second. It's only one / two marks per part so I shouldn't need to integrate anything or drawn any separate graphs.
Last edited by QuantumBoi; 1 month ago
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Eimmanuel
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(Original post by QuantumBoi)
Can someone just tell me how to find energy dissipated in one cycle and energy dissipated in one second. It's only one / two marks per part so I shouldn't need to integrate anything.
See post #11
https://www.thestudentroom.co.uk/sho...8&postcount=11
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QuantumBoi
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You shouldn't need to draw a graph on top of the other graph as each part is only worth a mark (besides 25 won't fit on that graph). Also why find the area?? That's just going to give you V^2 x time which isn't useful at all. I'm trying to find energy dissipated.
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