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GR9677 #1
Problem
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Electromagnetism$\Rightarrow$}RC Circuit

One can immediately eliminate plots A, C, E, since one would expect an exponential decay behavior for current once the switch is flipped.

More rigorously, the initial circuit with the switch $S$ connected to $a$ has the following equation,

$V-\dot{Q} r - Q/C =0 \Rightarrow \frac{dQ}{dt}=\frac{1}{r}\left(V-Q/C\right)
$

Once integrated, the equation becomes,

$Ve^{-\frac{t}{rC}} = V-Q/C \Rightarrow Q/C = V\left(1-e^{-\frac{t}{rC}} \right)
$

The charge stored on the capacitor after it is fully charged is $Q=CV$ (at $t=0$).

When the switch $S$ is switched to $b$, the equation becomes,

$Q/C = \dot{Q}R \Rightarrow Q(t) = Q_0 e^{-\frac{t}{RC}},
$

where $Q_0=CV$ from the initial connection.

Current is the negative time derivative of charge, and thus,

$I(t)=-\dot{Q}=\frac{Q_0}{RC} Q(t) = \frac{V}{R}Q(t)
$

The initial current IS $I(0)=V/R$, and thus choice (B) is right.

Alternate Solutions
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Enigmation
2016-10-06 21:52:37
 no_go2016-10-11 22:59:41 at the bottom of Alex Lang\\\\\\\'s page:\\\\r\\\\nhttp://www.alexhunterlang.com/physics-gre
 no_go2016-10-11 23:01:06 at the bottom of Alex Lang\'s page:\r\nhttp://www.alexhunterlang.com/physics-gre
 no_go2016-10-11 23:02:01 ignore the \\r\\n
dberger8
2016-08-07 00:24:00
Eliminate the three obvious ones. Then, we are stuck between B and D. Think of $r$ as an internal resistor to the emf charging the capacitor until it has voltage $V$. Then the current in second circuit is independent of $r$. Hence the answer is B.
Solarmew
2013-09-23 12:24:43
Awwwww. How come there are no problems posted yet for the last two exams? :<
rigustaf
2011-09-14 12:00:56
Where is the problem statement?
testingtest
2009-05-18 03:08:59
ok
zaijings
2009-03-24 19:38:55
The EMF after switch to B must start with V/R, so only A and B are possible right. The current can not be the same all the time, eliminate A, so the right answer is B.

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