Physics Circuit Question (AAMC Self Assessment #92)

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moto_za

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I'm having trouble understanding this question. How do you know it is in series or parallel? And how do you know in what direction the current passes through the resisters?

Pic attached.

The question is:

If all the resisters in figure 3 are 200 ohm, what is the current from the battery when V0 = 12?

Answer: 60 mA

TIA!

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I'm having trouble understanding this question. How do you know it is in series or parallel? And how do you know in what direction the current passes through the resisters?

Pic attached.

The question is:

If all the resisters in figure 3 are 200 ohm, what is the current from the battery when V0 = 12?

Answer: 60 mA

TIA!

The upper two which are in series are in parallel with the bottom two which are also in series. Whenever there's a junction and a possibility of splitting current, the resistors on opposite sides of the junction become connected in parallel. The two little extensions coming out of the upper and bottom wires are disconnected, so no current goes through them, and thus can not be considered junctions.

With that said, the upper two will add up to 400ohms and the bottom will add up to 400ohms. Now to add the top and bottom to each other, do the "product over the sum", and that gives you 200ohms for the equiv. resistance. Plug and chug using the battery voltage.
 
The upper two which are in series are in parallel with the bottom two which are also in series. Whenever there's a junction and a possibility of splitting current, the resistors on opposite sides of the junction become connected in parallel. The two little extensions coming out of the upper and bottom wires are disconnected, so no current goes through them, and thus can not be considered junctions.

With that said, the upper two will add up to 400ohms and the bottom will add up to 400ohms. Now to add the top and bottom to each other, do the "product over the sum", and that gives you 200ohms for the equiv. resistance. Plug and chug using the battery voltage.

Thanks for the help. So does the current go to the left starting from V0 then split at R2 & R3?
 
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Thanks for the help. So does the current go to the left starting from V0 then split at R2 & R3?

Yup, current flows from the cathode in a clockwise fashion until it hits the first junction, at which point it then splits up at R2 and R3 (into top branch and bottom branch) which have equivalent resistances as Ibn mentioned of (200+200=400 ohms) since the top and the bottom branches are connected in series respectively.
 
BTW TBR physics is so spot on. There was a wheatstone circuit in one of the passages I just did...smh.
 
Yup, current flows from the cathode in a clockwise fashion until it hits the first junction, at which point it then splits up at R2 and R3 (into top branch and bottom branch) which have equivalent resistances as Ibn mentioned of (200+200=400 ohms) since the top and the bottom branches are connected in series respectively.

Thanks a lot guys. I thought the bottom and top were connected in parallel though?
 
Thanks a lot guys. I thought the bottom and top were connected in parallel though?

They are. By "respectively" I meant the (top 2 resistors in series,R1 & R2) are parallel with (the bottom 2 resistors in series R3 and RSG).

-------200+200--------(eq resistance=400)
-------200+200--------(eq resistance=400)

So essentially 2 resistors of 400 resistance in parallel

(400*400)/(400+400) =200ohms

12V/200 ohms= .06A=60 mA
 
I got most of that passage right and I was still upset at how ridiculous of a passage it was. God I hate TBR sometimes.

Agreed. I wasn't sure what was going on with the whole changing of the length of the resistor but i just assumed the smaller side length equals the smaller resistance and somehow managed to get the entire passage right lol. Did you find the circuits chapter particularly more easy than the rest of the other physics II chapters?
 
I know I'm wrong, but why aren't all of the resisters in series?? Can't the current go from resister 2 to 1 to SG back to 3? That's why I'm thinking that they are in series. Am I missing something here?
 
I know I'm wrong, but why aren't all of the resisters in series?? Can't the current go from resister 2 to 1 to SG back to 3? That's why I'm thinking that they are in series. Am I missing something here?

Trace the flow of current (from the larger terminal cathode) clockwise starting at the larger line in the battery. As you do this, you will come to the first "junction"... if you are an electron in this tide of current as soon as you come to the "fork in the road" you can either take the top branch or the bottom branch, but you will not take both! In a parallel circuit, the voltage drop is the same for both the top and the bottom branch, but the current will vary based on the path of least resistance, whereas in a series circuit, the current is the same everywhere in the circuit, however the voltage depends on the resistor in question.

If all the resistors were connected in series then there essentially would be no junction and no "fork in the road" in the circuit. If you trace the flow of electrons starting at the positive terminal in any series circuit there are no points in your "journey" where you ever posit the question "can i take more than one path?"

If there is a point in any circuit where the answer to "if i were an electron, could I take more than one path" is yes, then it is a parallel circuit.
 
Agreed. I wasn't sure what was going on with the whole changing of the length of the resistor but i just assumed the smaller side length equals the smaller resistance and somehow managed to get the entire passage right lol. Did you find the circuits chapter particularly more easy than the rest of the other physics II chapters?

I thought the passages of the circuits were for the most part easier but the content review sucked.
 
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