Capacitor & Resistor: Connecting in Parallel w/ Battery

[kahwawashay1]

My physics teacher said the other day that, if you connect a capacitor and resistor in parallel with a battery, no current will go through the resistor, since the current doesn't like resistance and so will follow path of least resistance (to the capacitor). But then the next day, I think he was trying to say that he was wrong. But idk. So now I am confused, would someone please tell me the correct result of this situation? Using loop rule, you would get a value for the current through the resistor. But it also makes sense that the current would want to avoid resistance...

 

[skeptic2]

Suppose you had a water pump that was pumping water at pressure. The pipe the pump was pumping water into had a T with one side going to a small orifice and the other side going to an empty tank with no outlet.

How do you think the water would flow.

 

[phinds]

Do you think the state of the circuit will be identical at the time the switch is closed and then at later times?

 

[kahwawashay1]

Do you think the state of the circuit will be identical at the time the switch is closed and then at later times?

Well I know that for a resistor and capacitor in series, the circuit gradually reaches final values, but in parallel, idk? My physics teacher at first was saying that when in parallel, the resistor acts as a broken wire. If that is true, then yes, the state of the circuit is identical at the time the switch is closed and at later times. If my teacher was wrong, then I have no clue, and would appreciate an explanation of the situation

 

[kahwawashay1]

Suppose you had a water pump that was pumping water at pressure. The pipe the pump was pumping water into had a T with one side going to a small orifice and the other side going to an empty tank with no outlet.

How do you think the water would flow.

I think some of it would flow to the small pipe and some to the empty tank, until the tank became full, and then more would begin to flow through the small pipe and none to the tank...so then you are saying that at the beginning, some current would flow through both the resistor and to the capacitor, until the capacitor became fully charged, and then more current would flow through the resistor than before?

 

[phinds]

Well I know that for a resistor and capacitor in series, the circuit gradually reaches final values, but in parallel, idk? My physics teacher at first was saying that when in parallel, the resistor acts as a broken wire. If that is true, then yes, the state of the circuit is identical at the time the switch is closed and at later times. If my teacher was wrong, then I have no clue, and would appreciate an explanation of the situation

How much current flows through an uncharged capacitor when a voltage is applied? How much current flows through the capacitor once it is fully charged?

 

[Dadface]

In order to understand this you need to know how a capacitor charges.I leave you to look up the details but in a nutshell the current through the capacitor starts at a high value and then falls exponentially approaching zero as the capacitor charges.
Whoops you beat me to it phinds.

 

[kahwawashay1]

How much current flows through an uncharged capacitor when a voltage is applied? How much current flows through the capacitor once it is fully charged?

Well with resistance,
q = Q(1-e^(-t/τ)) where τ is RC
so the current will be the derivative of that with respect to time

Without resistance, the charge on the capacitor would appear instantaneously

Either way, once fully charged, there would be no current..

So then all of the current will go through the resistor when the capacitor has become fully charged...but when the capacitor is still charging and there is room for more current to go through it, will some of the current still go through the resistor?

 

[kahwawashay1]

I think some of it would flow to the small pipe and some to the empty tank, until the tank became full, and then more would begin to flow through the small pipe and none to the tank...so then you are saying that at the beginning, some current would flow through both the resistor and to the capacitor, until the capacitor became fully charged, and then more current would flow through the resistor than before?

Actually I am more confused with this water analogy now, as I don't think it fully carries over to electricity? Because I just found a problem in my book, not with a capacitor, but with a resistor connected in parallel with a resistanceless wire and battery. The solution was that none of the current went through the resistor while the resistanceless wire was connected across it, so using same reasoning on the capacitor-resistor in parallel, I would assume that none of the current goes through the resistor until the capacitor would become fully charged. But in the water analogy, you would still expect water to flow through the smaller pipe even though it would be easier for it to go through the larger pipe into empty tank while the tank would still not be full

 

[kahwawashay1]

In order to understand this you need to know how a capacitor charges.I leave you to look up the details but in a nutshell the current through the capacitor starts at a high value and then falls exponentially approaching zero as the capacitor charges.
Whoops you beat me to it phinds.

I know how capacitor charges. My question is that while it is charging, would some of the current still go through the resistor connected in parallel to it

 

[skeptic2]

The current through a resistor is dependent upon the voltage across the resistor. At the first instant when the switch is closed there is no voltage across the capacitor and therefore none across the resistor, so no current flows through the resistor. As the capacitor charges, the voltage across it increases. In problems like these you can never assume a resistanceless wire and battery. Since current = V/R, it is the same as dividing by zero. Provided however that the resistance of the wire and battery are much less than that of the parallel resistor, the parallel resistor will have negligible effect on the charging rate.

The charging rate will be primarily determined by the resistance of the battery and wires and will follow the curve V*(1 - e^-t) where R is the resistance of the battery and wire and t is time in RC time constants. t = 1 for R*C. V is the voltage across the capacitor when it is fully charged. A capacitor is generally assumed to be charged after 5 time constants. As the battery charges, the current through the resistor increases proportionally.

 

[kahwawashay1]

The current through a resistor is dependent upon the voltage across the resistor. At the first instant when the switch is closed there is no voltage across the capacitor and therefore none across the resistor, so no current flows through the resistor. As the capacitor charges, the voltage across it increases. In problems like these you can never assume a resistanceless wire and battery. Since current = V/R, it is the same as dividing by zero. Provided however that the resistance of the wire and battery are much less than that of the parallel resistor, the parallel resistor will have negligible effect on the charging rate.

The charging rate will be primarily determined by the resistance of the battery and wires and will follow the curve V*(1 - e^-t) where R is the resistance of the battery and wire and t is time in RC time constants. t = 1 for R*C. V is the voltage across the capacitor when it is fully charged. A capacitor is generally assumed to be charged after 5 time constants. As the battery charges, the current through the resistor increases proportionally.

But you're describing formulas for a capacitor and resistor in series...I'm talking about a capacitor and resistor in parallel...

 

[skeptic2]

Yes, I said:

Provided however that the resistance of the wire and battery are much less than that of the parallel resistor, the parallel resistor will have negligible effect on the charging rate.

If the resistance of the wire and battery are not much less than the parallel resistor, then we have to calculate or measure the total amount of current and multiply the current by the resistance of the wire and battery. The resistance of the wire and battery together with the resistance of the parallel resistor make a voltage divider with a constantly changing divide ratio. The value of resistance the capacitor sees is the parallel resistance of the series and parallel resistors. The charging voltage is the voltage seen at the divider output.
Engineers believe their equations approximate the real world. Physicists believe the real world approximates their equations.

 

[kahwawashay1]

Yes, I said:


If the resistance of the wire and battery are not much less than the parallel resistor, then we have to calculate or measure the total amount of current and multiply the current by the resistance of the wire and battery. The resistance of the wire and battery together with the resistance of the parallel resistor make a voltage divider with a constantly changing divide ratio. The value of resistance the capacitor sees is the parallel resistance of the series and parallel resistors. The charging voltage is the voltage seen at the divider output.

But I am saying let the wire and battery be ideal with no resistance. And I'm actually not really asking about what the capacitor sees and how it charges, I'm just asking is there any current through the resistance if it is connected in parallel with a capacitor, with resistanceless battery and resistanceless wire. I know that amounts in reality to dividing by zero, but books make the assumption of zero resistance all the time, so I am just asking about this ideal case, so there is absolutely no resistance in series with the capacitor

 

[skeptic2]

Then the instant the switch is closed, there is no current through the resistor. The next instant, an infinitesimal time later, there is no current through the capacitor and all the current is through the resistor.

 

[kahwawashay1]

Then the instant the switch is closed, there is no current through the resistor. The next instant, an infinitesimal time later, there is no current through the capacitor and all the current is through the resistor.

thank you! :D