How Does Closing a Switch Affect RC Circuit Components?

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Closing a switch in an RC circuit creates a path of zero resistance, allowing current to flow and initiating the discharge of the capacitor through the resistor. At the moment the switch closes, the capacitor starts discharging with an initial current determined by the voltage across it and the resistance in the circuit. The left loop of the circuit remains unaffected by the switch closure, as it consists only of a voltage source and resistor, making it trivial compared to the right loop with the capacitor and resistor. The behavior of the circuit can be analyzed using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL), which show that the two loops operate independently after the switch is closed. Understanding these principles is essential for grasping the dynamics of RC circuits.
LongApple
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Homework Statement



See the imgur link
http://i.imgur.com/yZsyoHx.png

Homework Equations


See the imgur link

The Attempt at a Solution


I am just reading a problem in the textbook and am not solving the problem

Basically I am looking for a good explanation of how current behaves and components behave when you close a switch and maybe the real question is "What would you search on google to find a good explanation of the above imgur link questions."

The question "what happens after you close a switch is too general"
 
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Hello Long, and welcome to PF :)

Slightly irregular: a problem on the homework forum that the poster isn't solving. No wonder, because the solution is provided in the text.

In appreciation of your sense of wonder, a few questions/remarks to get you thinking a bit more:

"Why doesn't left loop happen and right loop be (do?) nothing ?" reduces to "Why doesn't right loop do nothing ?", because left loop really "happens".

The closing of S means there is zero resistance between its connection points, so zero voltage. At t=0 the capacitor is charged to a voltage ##\epsilon## and between its connection points it "sees" S and R2 in series. That means it starts to discharge ! Initially with a current ##\epsilon##/R2, but while it discharges, the voltage over the capacitor drops. (well, continue reading after "according to..".

So through switch S flow both the currents from the left and from the right circuit. Being a good conductor (zero resistance) it has no problem with that.What would you say to a classmate who is asking "Why doesn't right circuit happen and left circuit do nothing ?"

What about a second switch, S2, parallel to S, that is closed at the same time as S ? Would that make a difference in your perception ? And in reality ?
 
Hi LongApple, Welcome to Physics Forums.

You should include some text description of your question, or at least what it's about in general, in the Problem Statement part of the template; helpers shouldn't have to click through to another site in order to get the gist of your problem.

In your image you're wondering why after the switch closes only the right hand loop was considered. I'm afraid it's simply because it's the part of the circuit that contains the components (resistor with capacitor) that pertains to the course material you are studying at the moment -- the behavior RC circuits.

Note that the left hand loop still exists after the switch closes but it's rather trivial: just a voltage source and resistor in a loop.

An ideal switch, when it's closed, behaves like a perfect conductor (zero resistance), so no potential difference can be developed across it. This effectively isolates the two sides of the circuits from each other, since nothing that happens in one loop can affect the other loop (If you write KVL around each loop there are no shared potential drops, so the equations are completely independent of each other).

So the things to read up on are the basic circuit laws (KVL, KCL), and the basic RC circuit undergoing charging and discharging.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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