Hi everyone!
I was having trouble with the explanation for this question.
I know that once a capacitor is fully charged, it acts like a "broken wire" in the circuit, and current doesn't flow through it (?). I thought that after a long time, all the current would be directed through the loop with R1 and no current would flow through the loop with R2 because the capacitor in that loop is charged after a long time. This lead me to choose A.
I guess my question is, will current still be sent into a loop with a fully charged capacitor from a junction?
Nathan
Discrete Science QBook, Physics #1545, pg. 256

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 Joined: Sat Mar 30, 2019 8:39 pm
Re: Discrete Science QBook, Physics #1545, pg. 256
It will not be "sent into" that loop once the capacitor is fully charged, although that phrasing is a bit dangerous (DC current simply cannot flow in this case).
However, what the question is saying is that under situation 1 (uncharged), equivalent resistance of the circuit is:
Under situation 2 (charged), equivalent resistance of the circuit is:
But since I = V/R, for situation 1 (uncharged) I1 = V/0.5R1 = 2V/R1. Since this current is split equally among two circuits due to their identical resistances, each one must be receiving I=V/R1 current.
Now for situation 2 (charged), I = V/R resolves even more simply: I2 = V/R1. Since there is only that loop present and the other is interrupted, the total current through the circuit must be the current through R1 as well.
Notice that the current through R1 is initially V/R1 and finally also V/R1. We know that our voltage is constant (given by the problem), so we know now that current through R1 is always the same.
This is honestly a really good problem.
However, what the question is saying is that under situation 1 (uncharged), equivalent resistance of the circuit is:
Code: Select all
1/rtotal=1/R1+1/R1=2/R1
Rtotal=R1/2
Code: Select all
Rtotal=R1
Now for situation 2 (charged), I = V/R resolves even more simply: I2 = V/R1. Since there is only that loop present and the other is interrupted, the total current through the circuit must be the current through R1 as well.
Notice that the current through R1 is initially V/R1 and finally also V/R1. We know that our voltage is constant (given by the problem), so we know now that current through R1 is always the same.
This is honestly a really good problem.

 Posts: 8
 Joined: Sun Aug 04, 2019 5:19 pm
Re: Discrete Science QBook, Physics #1545, pg. 256
Great, thank you! So would it be safe to say that given a problem with two resistors, we should calculate equivalent resistance of the whole circuit and use that to determine current if charge is flowing in all loops?

 Posts: 616
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Discrete Science QBook, Physics #1545, pg. 256
Right, but don't forget at least the intuitive interpretation of Kirchoff's (first) law: Current flowing into a junction must equal current flowing out of a junction. That means if a junction splits in two directions, and each direction has the same equivalent resistance, then current is simply split evenly between the two.
This is exactly why even though equivalent resistance drops by half when the capacitor is charged, current through the first loop stays exactly the same.
This is exactly why even though equivalent resistance drops by half when the capacitor is charged, current through the first loop stays exactly the same.