TBR Physics Diagnostic Passage #2.4 Circuits Question

[arc5005]

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Question <---

answer:

C) 2A

A long time after both switches are closed, he current will travel exclusively through R1 because the capacitor passes no current once it is fully charged. V = IR ...

10 mV = I(5 m-ohms)

10 mV / 5 m-ohms = I = 2 amps

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I'm confused how the current (I) through R1 can be calculated without taking into account a parallel resistor. Why is the current exclusively going to R1 and ignoring R2? Wouldn't the current be split between the two resistors, R1 and R2?

 

[mk04447]

That's about the easiest circuit you'll get. I suggest studying more.

 

[arc5005]

That's about the easiest circuit you'll get. I suggest studying more.

not helpful. in every other setup i've seen with resistors in parallel, the current is split. why not in this scenario?

 

[mk04447]

not helpful. in every other setup i've seen with resistors in parallel, the current is split. why not in this scenario?

Because the capacitor is also in parallel? I can't help you learn the material. I can explain areas of confusion, but there is none here. You just need to reread the section.

 

[Feste85]

Your solution is wrong. Right answer, wrong solution.

With both switches closed, current will flow and the capacitor will charge. In a steady-state condition (long elapsed time), the capacitor will be fully charged. At that point, the current is *effectively* flowing through only the resistors. The amount of current depends on the arrangement. Using Kirchoff's laws, we know that the current into the junction splits: I=I1+I2. We also know that the resistors are equivalent, so I1=I2. Now, the easy way to determine how much current is flowing through the circuit after a long time, we use the concept of equivalent resistance. R_eq=R1*R2/(R1+R2) = (1/2) R. V = I*R_eq = I *(1/2) R. Plugging in the values and solve for I:
10 mV = I * (1/2)*(5 mOhms)
I=(10/2.5) A = 4 A.
Now, that's the total current flowing, which means that each resistor will get half of that current...or, 2 A.

 

[arc5005]

Your solution is wrong. Right answer, wrong solution.

With both switches closed, current will flow and the capacitor will charge. In a steady-state condition (long elapsed time), the capacitor will be fully charged. At that point, the current is *effectively* flowing through only the resistors. The amount of current depends on the arrangement. Using Kirchoff's laws, we know that the current into the junction splits: I=I1+I2. We also know that the resistors are equivalent, so I1=I2. Now, the easy way to determine how much current is flowing through the circuit after a long time, we use the concept of equivalent resistance. R_eq=R1*R2/(R1+R2) = (1/2) R. V = I*R_eq = I *(1/2) R. Plugging in the values and solve for I:
10 mV = I * (1/2)*(5 mOhms)
I=(10/2.5) A = 4 A.
Now, that's the total current flowing, which means that each resistor will get half of that current...or, 2 A.

thank you. this is what I thought. I was so frustrated trying to understand their solution.

 

[PlsLetMeIn21]

Your solution is wrong.

Actually, their solution is right, and it's the best (fastest) one shown here so far. Your solution is correct as well, but it takes too much time.

With both switched closed for a long time, the capacitor drops out of the circuit (because it's full.) There are two parallel resistors and nothing else, so both resistors get 10mV, because circuit elements in parallel share the same voltage drop. Each resistor gets 2A, because 10mV/5m-ohms = 2A for both parallel pathways.

I like your solution for a physics class, but theirs is better for the MCAT given the limited time per question. The best solution is the simplest and fastest solution.

 

[BerkReviewTeach]

Really nicely worded PlsLet!

@arc5005, I want to add to the solution above. When looking at any circuit, you should consider the various segments. In this case, there is a segment before the junction (with just a wire and no circuit elements), between the junctions (a capacitor, a resistor, and another resistor), and after the junction (also with just a wire and no circuit elements.) As mentioned above, after a long enough period of time (less than a second as a general rule), the capacitor will be full. So no current will run through that segment. Current will flow through each of the two resistors. The key thing here is that circuit elements in parallel share the same voltage drop. That can be thought of by the comment "path of least resistance", which tells us that the smaller resistor (the path of least resistance) will get more current, and it happens that they get the same value for I x R, which is the voltage drop.

So both resistors in the question will get the same voltage drop. Because no voltage was dropped before or after the junction, all of the voltage is dropped in the middle segment, when current passes through each resistor. Both restores will experience the entire voltage of the circuit, in this case 10 mV. Each resistor has a resistance of 5 mΩ, so the current (found using I = V/R) is 10 mV/5mΩ = 2 A.

That is the fastest and easiest solution.

Faste85's method is also very good, where you first determine that the equivalent resistance for the middle segment of the circuit is (5 x 5)/(5 + 5) = 2.5 mΩ. This shortcut for getting the equivalent resistance for two parallel resistors can be found on page 72 of your book.

The total current for the circuit (the current leaving the cathode and arriving at the anode) is 10mV/2.5mΩ= 4 A. The two resistors have equal resistance, so they will split the 4A evenly, each getting 2A. This is the Kirchoff's junction rule.

That solution, if you use the BR shortcut for resistors in parallel is also fast and easy.

It is important to keep in mind that there is often more than one way to solve a given question, and that the one that is most instinctive for you is the one you should choose.

Best of luck!