Capacitor in parallel with a resistor

[IlyaR]

The answer says that since R3 and C are in parallel, the voltage is 0 for both (since there is no charge on the C yet). It also says that all of the charge will flow to the C initially since it offers no resistance. What I don't understand though is how the current flows through R3.

I'm assuming as SOON as the current hits C, there will be a small potential on it, thus R3 will have the same small voltage, and I=V/R. As V gets larger in C, the current gets smaller through C and larger through R3, is that correct?

Thanks!

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[milski]

Yes, you have it right. When the switch is closed, there are no charges at all on C. It's innately easier for the current to go through C, piling up charges on it. As more charges accumulate on C, it becomes harder to bring any more charges and some of the current starts to flow through R₃, just in the way you described.

 

[sps27]

The answer says that since R3 and C are in parallel, the voltage is 0 for both (since there is no charge on the C yet). It also says that all of the charge will flow to the C initially since it offers no resistance. What I don't understand though is how the current flows through R3.

I'm assuming as SOON as the current hits C, there will be a small potential on it, thus R3 will have the same small voltage, and I=V/R. As V gets larger in C, the current gets smaller through C and larger through R3, is that correct?

Thanks!

The way I understand this is that current always chooses the path of least resistance. So if switch 2 is open no current will flow through R3. C will continue to get charged till the time the potential between the plates of capacitor can match the potential drop (I*R3) if current were to flow through R3. The potential drop across R3 is static. But the potential drop across the plates of capacitor is slowly building up as it gets charged. So there is no current flow through R3 and it will be like that till the potential in C can build up to a value that represents the same potential drop as current flowing across R3........is the ans A?.......or I could be wrong.........

 

[milski]

The way I understand this is that current always chooses the path of least resistance. So if switch 2 is open no current will flow through R3. C will continue to get charged till the time the potential between the plates of capacitor can match the potential drop (I*R3) if current were to flow through R3. The potential drop across R3 is static. But the potential drop across the plates of capacitor is slowly building up as it gets charged. So there is no current flow through R3 and it will be like that till the potential in C can build up to a value that represents the same potential drop as current flowing across R3........is the ans A?.......or I could be wrong.........

The answer is A. The potential across R₃ and C is always the same - the terminals of each are directly connected to each other. That leads you to another way of explaining the answer. The potential across C in the beginning is 0, so the potential across R₃ also must be 0, thus I=V/R=0/R=0.

 

[sps27]

The answer is A. The potential across R₃ and C is always the same - the terminals of each are directly connected to each other. That leads you to another way of explaining the answer. The potential across C in the beginning is 0, so the potential across R₃ also must be 0, thus I=V/R=0/R=0.

Ok, gotcha. So essentially I can measure the potential across the capacitor plates by connecting a voltmeter across R3. Meaning, as the capacitor is getting charged the current flow across R3 increases.

 

[milski]

Ok, gotcha. So essentially I can measure the potential across the capacitor plates by connecting a voltmeter across R3. Meaning, as the capacitor is getting charged the current flow across R3 increases.

Yes, that's all correct. They are in parallel, so connecting a voltmeter across one is the same as connecting a voltmeter across the other.