Lower resistance if larger resistor?

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monkeyMD

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This question stems back to the BR Physics book, example 9.5b.

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Depends on what you mean by larger, longer wires would have more total resistance, but in equally long wires the larger the cross section the less the resistance.
 
This question stems back to the BR Physics book, example 9.5b.

Which version of the physics book are you using?

In the new version 9.5b, you need to consider "larger" in the same context as the capacitor, given that it's in the same sentence. Larger in that context implies greater magnitude.

In the older version 9.5b, using a larger resistor at any of the resitor positions would result in a greater equivalent resistance for the circuit, but if it's R1 or R2 that becomes larger, then there would be no impact on I3.
 
Bump. I don't understand this either...I see that A has a larger capacitor,
And gains more q in a time t and thus has a greater current than B. But why does A have a larger (greater magnitude) resistor?? Maybe can you expand upon resistor/resistance/resistivity? I would think A would have a smaller resistor as it has a greater current, and resistors impede current flow. So wouldn't a bigger resistor resist current more??!!!!
 
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Resistance is proportional to Resistivity*Length of resistoy/Area of resistor


Increase Length, increase resistance
Increase Area, decrease resistance
increase resistivity, increase resistance


Think of a straw when drinking a milk shake. short, wide straw is the shiznit cuz its soo easy to get that drink through, but a long skinny straw is very hard to get that milkshake through.
 
Bump. I don't understand this either...I see that A has a larger capacitor,
And gains more q in a time t and thus has a greater current than B. But why does A have a larger (greater magnitude) resistor?? Maybe can you expand upon resistor/resistance/resistivity? I would think A would have a smaller resistor as it has a greater current, and resistors impede current flow. So wouldn't a bigger resistor resist current more??!!!!

looking at the OLD physics book:

in 9.5b, R1 and R2 are in series with each other, so you can kinda replace them with an equivalent resistor that is Req = R1 + R2. so now you have, Req (6 ohms) and R3 which are in parallel with each other. current leaving the battery is 6 amps, so when it reaches the node it splits off to R3 and Req.

the current going through Req + current going through R3 has to equal 6 by kirchoff's rule. since Req has two times more resistances (6 ohms) than R3 (3 ohms), R3 will have twice the current flowing through.

2I (flowing through R3) + I (flowing through Req) = 6 amps, so I = 2 amps. R3 has current of 4 amps going through. and since its in parallel with Req then you know the voltage across R3 is equal to the voltage across Req. but the current through Req is 2 amps, and since Req is two resistors in series, all resistors in series have the same current. so R2 has current 2 amps, and R1 has current 2 amps.

voltage drop across R2; V = IR = 2 amps * 3 ohms = 6 volts.
 
Sorry to bump an old thread, but I still don't understand example 9.5b in the newest BR physics book. Circuit A has an overall larger charge, so its capacitor must be larger than B . It also has a shorter charging time, so I would expect that it would have a smaller resistor. However, the answer says that A has a larger capacitor and large'r resistor. Can someone explain this? It's been several years since I've taken physics. Thanks!

Please excuse any spelling errors. This was posted from my phone.
 
Think of a straw when drinking a milk shake. short, wide straw is the shiznit cuz its soo easy to get that drink through, but a long skinny straw is very hard to get that milkshake through.

This is not only hilarious, but genius
 
The answer to the question is considering the actual physical size of the resistors. Since R is also equal to Length/Area, this tells us that as the area of the resistor increases the resistance decreases. So in the graph, capacitor A gains more total charge than capacitor B, so this means that capacitor A is connected to a physically larger resistor than capacitor B. Also, since the answer choices didn't mention any of the resistors decreasing in size, we can assume that the length of resistors were kept the same. Thus, its area is larger and its resistivity is lower. Therefore, allowing capacitor A (curve A) to have a greater total charge than capacitor B (curve b).
 

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