Circuit with an uncharged capacitor - desperate

AI Thread Summary
When the switch S is closed, the uncharged capacitor begins to charge, and initially, all the current flows into the capacitor rather than through Resistor 2. This occurs because the capacitor acts as a short circuit at the moment of closing, allowing the current to build up until the capacitor is fully charged. The potential difference from the battery causes the current to flow into the capacitor, and since the capacitor is initially uncharged, there is no voltage across it, resulting in no current through Resistor 2. As the capacitor charges, the current gradually decreases until it stops when the capacitor is fully charged. Understanding the capacitor's behavior and the role of the resistors is crucial for analyzing the circuit dynamics.
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circuit with an uncharged capacitor -- desperate

Homework Statement


http://img91.imageshack.us/img91/7533/physicsimage.gif
Switch S has been open for a long time. It is closed suddenly at t=0.
A long time later, it is opened.
Determine the potential across and the current through each component at the following times.

Immediately after S closes. <-- one i don't understand...
A long time after S closes.
Immediately after S is reopened.

Homework Equations


Q=CV

The Attempt at a Solution


So immediately after S closes, the capacitor charge is neutral, right? (Since it is uncharged.) The solution says that the current will not go through Resister 2.
So I do not understand why the current will not go through Resistor 2 and go through Capacitor instead.
I thought current charges cannot jump to the other side of the capacitor...There is potential difference between battery, so I thought the current will try to go through resister 2 due to the emf of the battery...
But that is not the case.

I am wondering why all currents going to R2 and C is used to charge the capacitor...
and i also do not understand why that results in no current through R2...

Thank you SO much in advance.

I am sso stuck and very desperate...

If you cannot answer all my questions, my biggest most important question is... Why does current NOT go through R2?

Again, thanks so much for your time.
 
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For the time being forget R2.
Close the switch .What happens?
 


You need another equation for the capacitor.

i = C \frac{dv}{dt}
 


rl.bhat said:
For the time being forget R2.
Close the switch .What happens?

If there is no R2, then the battery charges the capacitor.
Btw, that raises another question.
Since electrons are the charge carriers and electrons cannot ever travel to positive side of the capacitor, does the resistor 1 connected to the positive plate of the capacitor play a role? I am so confused. Thanks for your help.@Phrak:
Thank you so much for your response. I am wondering if you could elaborate a little more. ^^
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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