A physics graduate student builds the circuit below, with a switch allowing the

Question

A physics graduate student builds the circuit below, with a switch allowing the circuit to close with either battery:

(a) First, the graduate student flips the switch to B, connecting the 2E battery. Assuming the capacitor starts uncharged, what will its charge go to if she waits a very long time?

(b) After waiting a long time, the graduate student declares it to be t = 0 and flips the switch from B to A, connecting the E battery. Find Q(t) by using Kirchhoff’s rules to get a differential equation and then plugging in the solution to that equation. (hint: what direction does current flow?)

(c) Check the t ? 0 and t ? ? limits of your answer, and plot Q as a function of time. Does your answer make sense?

Thanks in advance!

Explanation / Answer

(a) After a very long time, the capacitor behaves like an open circuit and charge on it is given by

Qo = CV = C (2E)    ==> Qo = 2CE

(b) In this case, the current starts growing through the circuit while capacitor discharges. So,

R dQ/dt + Q/C = E

R dQ/dt + (Q/C - E) = 0

The solution for this differential equation can be written as

Q(t) = EC + c1 e-t/RC where c1 is a constant of integration

(c) as t-> 0

Q(t) = EC + c1

as t->

Q(t) = EC = Qo/2

In both the cases, the graphs will show a straight line parallel to x-axis. This makes sense as at t=0, the capacitor is having charge Qo and is constant before it starts to discharge. At time t=, when capacitor has completely discharged, the charge in the circuit becomes independent of time and approaches a value of Qo/2.

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