TBR physics chapter 9 (circuits) passage 2 #12

[PanRoasted]

It says in the explanation to the problem that the charge on the capacitor is NOT equal to c*emf (derived from C=q/v), but is instead equal to c*emf*R3/(R1+R3). wtf? It says directly in the chapter that the voltage difference across a capacitor after it's finished charging always equals the voltage difference of the emf source, irrespective of any resistors that may be in series with the capacitor. Can someone explain this to me please? Thanks.

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[Jengreef]

Could you post the question? I have TBR, and in my version they have two different "Passage 2", and neither of them asks for the charge on the capacitor on #12.

 

[PanRoasted]

Essentially, there is a capacitor and a resistor in parallel with each other (C and R3). There is another resistor (R1) in series with the capacitor/resistor in parallel. The question asks what the charge stored on the capacitor is after it is finished charging.

 

[PanRoasted]

Ah...I re-read the explanation and I get it now. Sorry for the trouble everyone!

 

[V5RED]

In a circuit, the voltage across a capacitor is not automatically equal to the voltage of the battery. If it were a circuit in which the resistors were all in series with the capacitor, then it would be simply because there would not be a circuit once the capacitor had fully charged.

In the question, the capacitor is in parallel with R3 which is in series with R1. The voltage across the capacitor is equal to the voltage across R3.

To find the voltage across R3 you do circuit analysis.
note: current is equal through R1 and R3 because they are in series
V3=i*R3
i*(R1+R3)=Emf (again, this is because they are in series)
i=Emf/(R1+R3)

V3=Emf*R3/(R1+R3) (substitution)

q=CV

q=C*Emf*R3/(R1+R3)

 

[PanRoasted]

Thanks! That actually made it a bit clearer.