GR9677 #25


Problem


This problem is still being typed. 
Electromagnetism}Capacitors
Recall the following truths (held to be selfevident?) on the subject of capacitors: 1. series capacitors have equal charge (Equivalent capacitance of two capacitors is ); 2. parallel capacitance have equal voltage (); 3. ; 4. .
(A) Initially, before the switch is closed, only has a voltage across it, and hence it is charged. . But, afterwards, since the voltage stays the same, one has ; hence, .
(B) . Since , one has . This is true.
(C) By definition of circuit elements in parallel, one has each capacitor at the same potential. This is trivially true.
(D) Since one determined from (C) that the capacitors are at the same voltage, then because they have the same capacitance, they have the same energy as per . True.
(E) This is false, since , initially. In the final state, each capacitor has energy . The sum of energies is thus .


Alternate Solutions 
kroner 20091009 19:50:35  The charged up state of a capacitor depends only on its capacitance and the voltage across it. In this case and so all three are indistinguishable. So for every quantity that you might measure regardless of the formulas.   jmason86 20090930 17:35:59  With Q=CV and one can eliminate things pretty quickly. If C and V are the same, then Q and V must be the same as well (through the above equations).
The problem statements supplies C1 = C2.
If Q1 does happen to be equal to Q2 (i.e. answer (B)), then that means that (C) and (D) must also be true. Since we are looking for the statement that is incorrect, we can eliminate (B) (C) (D).
Charge damn well better be conserved, so you can't have Q1+Q2=Q0 or you'd be generating charge somewhere. Eliminate (A).
(E) remains.  

Comments 
jdbro 20141023 17:43:07  :: Careful, depends on . Use instead.
  kroner 20091009 19:50:35  The charged up state of a capacitor depends only on its capacitance and the voltage across it. In this case and so all three are indistinguishable. So for every quantity that you might measure regardless of the formulas.   dstahlke 20091008 21:35:34  Does anyone know what they mean by a "real capacitor"? Does this just mean that there is some resistance so that the wires don't melt due to the infinite amount of current that would flow otherwise? I scratched my head about this for a while before I took a look at the options and realized that the problem wasn't as hard as I was expecting it to be.   jmason86 20090930 17:35:59  With Q=CV and one can eliminate things pretty quickly. If C and V are the same, then Q and V must be the same as well (through the above equations).
The problem statements supplies C1 = C2.
If Q1 does happen to be equal to Q2 (i.e. answer (B)), then that means that (C) and (D) must also be true. Since we are looking for the statement that is incorrect, we can eliminate (B) (C) (D).
Charge damn well better be conserved, so you can't have Q1+Q2=Q0 or you'd be generating charge somewhere. Eliminate (A).
(E) remains.
Crandor 20100409 20:04:50 
Charge on the capacitors is not conserved; it is generated by the battery. (A) is true because C and V are the same for both capacitors, and (because V didn't change).

  a19grey2 20081029 15:14:46  So, the easy way I did this was not to check for everything being true, but to look quickly for clearly False statements.
E is false because the connection of a second capacitor in parallel must add energy to the system. This is seen because capacitors in parallel must have the same Voltage and so . Thus, the energy in the two capacitors is more than that in the one capacitor.
This solution is nice since it doesn't even require
chemicalsoul 20091017 01:06:19 
I don't understand just by closing a circuit we have gained on the total energy of the system. When the circuit was open the total energy of the system was equal only to the energy stored in the charged capacitor and there wasn't any energy in the uncharged one, where did this new energy come from ?

carle257 20100404 20:01:28 
The battery provides additional energy when the capacitance of the circuit changed. When the switch closed, we charged up

 

Post A Comment! 

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .

type this... 
to get... 
$\int_0^\infty$ 

$\partial$ 

$\Rightarrow$ 

$\ddot{x},\dot{x}$ 

$\sqrt{z}$ 

$\langle my \rangle$ 

$\left( abacadabra \right)_{me}$ 

$\vec{E}$ 

$\frac{a}{b}$ 





The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... 

