Solving Discharging Circuit Homework

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The discussion revolves around calculating the time it takes for the charge on capacitor C2 to decrease to 36.8% of its fully charged value after switch S2 is reopened. Participants clarify that the term "drop to 1/e" refers to the charge decreasing to a specific percentage of its initial value, which is linked to voltage changes. To determine the time constant, participants emphasize the need to find the equivalent resistance in the circuit after S2 is opened. There is uncertainty about whether the resistors are in parallel at this stage. Understanding the configuration of the circuit is crucial for solving the problem accurately.
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Homework Statement


http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam2/fa06/fig17.gif
After a very long time with both switches in the closed position, switch S2 is reopened. How long does it take for the charge on capacitor C2 to drop to 1/e (36.8%) of its fully-charged value (i.e. of the value it had just before S2 was reopened)?


Homework Equations


Time constant = RC
capchg7.gif


The Attempt at a Solution


I try to find the total resistance, but am not sure how. I am not even sure what it means by "drop to 1/e (36.8%)". Does it mean 36.8% of the total charge?
 
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driedupfish said:

Homework Statement


http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam2/fa06/fig17.gif
After a very long time with both switches in the closed position, switch S2 is reopened. How long does it take for the charge on capacitor C2 to drop to 1/e (36.8%) of its fully-charged value (i.e. of the value it had just before S2 was reopened)?


Homework Equations


Time constant = RC
capchg7.gif


The Attempt at a Solution


I try to find the total resistance, but am not sure how. I am not even sure what it means by "drop to 1/e (36.8%)". Does it mean 36.8% of the total charge?

yes, since charge is a function of voltage here. As the voltage drops, less charge is accumulated on the plates of the capacitor.
 
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Okay, to find the time constant, RC, I have to find the equivalent resistor at that point. But I don't know how? With only S2 reopened, are all the resistors now in parallel?
 
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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