RC Circuit with parallel resistor

[CrazyNeutrino]

Homework Statement

A circuit contains a capacitor of capacitance C, a power supply of emf E, two resistors of resistances R1 and R2 , and a switch S2 . Resistor R1 is in series with the power supply and R2 is in parallel with the capacitor and the power supply. S2 switches the branch that contains R2 (in parallel with the capacitor). Initially, the capacitor is CHARGED and S1 is closed while S2 is open. Switch S2 then gets closed at time t=0. Sketch the currents in resistance R1 and R2 as a function of time t.

Homework Equations

V=IR
Kirchoffs Loop and Junction Rule
Capacitor Differential equations

The Attempt at a Solution

I know that the solution is two exponential curves that asymptotically approach some current value from the top and from 0. I figured that the current in R1 is 0 when the switch is immediately closed while the initial current in R2 is E/R2 since the voltage across the capacitor and R2 must both equal E. My reasoning was that as t approaches infinity, the current in R1 and R2 will tend towards E/(R1+R2). However, I am not really sure why. I tried writing individual differential equations for the currents in R1, R2 and the capacitor but ended up confusing myself further. Could somebody please explain conceptually, what happens to the voltage across the capacitor and the currents in both resistors over time? I sort of understand what happens but don't particularly understand why.

 

[kuruman]

If I understand the situation correctly from your description without a diagram, when you open the switch, the capacitor becomes isolated and the charge on its plates has nowhere to go. What does that imply about the voltage across its plates?

 

[CrazyNeutrino]

Here's a picture of the circuit. It is still in parallel with the resistor and the power supply. I don't understand why it is isolated, although, if it was I suppose the voltage across it would be the same as it was when it was initially fully charged, i.e = E.

 

[kuruman]

OK, I see, thanks. I imagined a different circuit that's why I said the capacitor becomes isolated. In this case the voltage across the capacitor when S1 is closed for a long time is E. Now immediately after S2 is closed, call that t = 0, it is still E. A long time after S2 is closed, the voltage across the capacitor becomes the same as the voltage across resistor $R_2$ because the two are in parallel. Find what that voltage is and connect the dots from t = 0 to t = "a long time later" with an increasing or decreasing exponential as the case might be.

 

[CrazyNeutrino]

Fair enough I understand that part. What seems counterintuitive to me is that a current starts flowing in R1 just by introducing a new resistor. Why?

 

[kuruman]

Fair enough I understand that part. What seems counterintuitive to me is that a current starts flowing in R1 just by introducing a new resistor. Why?

Because, initially at t = 0, when S2 is closed, the voltage across the capacitor, $V_C$, is instantaneously $V_C = E$. A long time after S2 is closed, $V_C$ is only a fraction of $E$. Now remember that $Q=CV_C$ which means that charge has to leave the plates when the voltage is decreased. "Charge leaving the plates" means that current is flowing in the circuit, as long as $V_C$ is decreasing. Eventually, $V_C$ matches the voltage across $R_2$ at which point current stops flowing and a new steady state situation is reached.

 

[CrazyNeutrino]

Thank you so much! Couldn't visualise what was going on for the longest time but it really does make sense now.