Few electricity (b) Berkeley questions

[JP2740]

8.6b, 8.8b, 9.1b, 9.5b

8.6b Is about which dipole creates the most electrostatic potential energy.

8.8b about the oscilattion frequency of an electron-had no idea here.

9.1b- why not temperature or length? this is about the fixed potential difference from a resistive wire.

9.5b- I guess I don't know this loop rule stuff. I only know how to do this from finding the equivalence resistance and working backwards, but apparently this doesn't work.

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[Gryhu]

I actually just finished these chapters. I'm answering from what I understand so I might not be 100% correct.

8.6b: The greatest electrostatic PE is where you would have the charges furthest from equilibrium. So you would want opposite charges to be furthest apart and like charges to be closest. Choice C will have the highest PE since the like charges are closest together.

8.8b: Remember v=f(lambda) if you decrease wavelength, then you're going to increase frequency @ a constant v. So you can cross out B, C, D.

9.1b: If you want to increase Power loss @ a fixed voltage you need to increase current. Current is the flow of charge/time. Temperature and length won't affect the current (I think) and if you increase the conductivity of a wire you'll increase the charge which will increase the current.

9.5b: You right, if you find equivalent resistance, you get that the potential difference for the circuit is 12V. Now using the loop rule you know that the current going into the loop is going to equal the current going out of the loop. The current will be distributed to each resistor in opposite ratios to their resistances. Since the ratio is 2:1 (R1+R2:R3) you can say that 6 amps will be split in a 1:2 fashion so R3 will get 4 amps and R1+R2 will get 2amps. From that you can choose choice C since thats the only one with 4 amps. On the MCAT if you find part of the choice and its the only one with that has it thats the right answer. Don't waste time and move on.

Hope this helped

 

[JP2740]

I actually just finished these chapters. I'm answering from what I understand so I might not be 100% correct.

8.6b: The greatest electrostatic PE is where you would have the charges furthest from equilibrium. So you would want opposite charges to be furthest apart and like charges to be closest. Choice C will have the highest PE since the like charges are closest together.

8.8b: Remember v=f(lambda) if you decrease wavelength, then you're going to increase frequency @ a constant v. So you can cross out B, C, D.

9.1b: If you want to increase Power loss @ a fixed voltage you need to increase current. Current is the flow of charge/time. Temperature and length won't affect the current (I think) and if you increase the conductivity of a wire you'll increase the charge which will increase the current.

9.5b: You right, if you find equivalent resistance, you get that the potential difference for the circuit is 12V. Now using the loop rule you know that the current going into the loop is going to equal the current going out of the loop. The current will be distributed to each resistor in opposite ratios to their resistances. Since the ratio is 2:1 (R1+R2:R3) you can say that 6 amps will be split in a 1:2 fashion so R3 will get 4 amps and R1+R2 will get 2amps. From that you can choose choice C since thats the only one with 4 amps. On the MCAT if you find part of the choice and its the only one with that has it thats the right answer. Don't waste time and move on.

Hope this helped

Thanks a bunch. But for 8.8b, why did you assume constant v, and using that equation I thought IV might be included.

Thanks for the other ones though.

 

[PiBond]

Thanks a bunch. But for 8.8b, why did you assume constant v, and using that equation I thought IV might be included.

Thanks for the other ones though.

i think it's in a vacuum so it's going 3e8 m/s and can't be sped up. i might be wrong though

 

[Gryhu]

Yeah you're right. In a vacuum, all light travels at the same constant speed. 3e8 m/s.