TBR capacitor dependce on resistance

[Vicodeen]

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The maximum charge that can be stored on a capacitor
depends on:
I. the capacitance of the capacitor.
II. the emf of the battery.
III. the equivalent resistance of the circuit
IV. the power drain of each resistor in the circuit

A. I and III only
B. II and III only
C. I and II only
D. II only
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The answer is clearly C with process of elimination, but why III is untrue is unclear to me.

TBR says that "There is no dependence on resistance of any kind" in the explanations.

But yet just in a previous question they showed that the charge on a capacitor had resistance in the equations. Q=CV, and V is dependent upon the resistance in the circuit

Enlighten me please!

 

[sshah92]

The maximum charge that can be stored on a capacitor
depends on:
I. the capacitance of the capacitor.
II. the emf of the battery.
III. the equivalent resistance of the circuit
IV. the power drain of each resistor in the circuit

A. I and III only
B. II and III only
C. I and II only
D. II only
----------------------------------------------------------------------
The answer is clearly C with process of elimination, but why III is untrue is unclear to me.

TBR says that "There is no dependence on resistance of any kind" in the explanations.

But yet just in a previous question they showed that the charge on a capacitor had resistance in the equations. Q=CV, and V is dependent upon the resistance in the circuit

Enlighten me please!

Hey man,

For capacitors you will almost never use the V=IR equation because that's another type of circuit altogether. Therefore you have Q=CV, V=Ed, and C=EoKA/d. Also remember that the C in the Q equation is only a constant that isn't affected by Q or V, but rather A and d only.

So the reason that R is irrelevant is because it doesn't play a role in determining capacitance. (You could have a resistor in series with a capacitor but it would only slow down the rate at which the capacitor is charged, not change the amount of charge that the capacitor could hold.) And the reason V is important is because of Ed's equation, V=Ed!

Hope that helps 🙂

 

[99dmg]

Yes, the V in the equation C=Q/V does not depend on the resistance of the circuit.

Think of it like this:

If the capacitor is fully charged, there is no current in the circuit so any resistors in the circuit don't consume any potential (voltage).
But Kirchhoff's loop law states that the potential must be 0 for any loop in a circuit thus the capacitor has all the potential ( ie. emf battery = V capacitor)

 

[Vicodeen]

Yes, the V in the equation C=Q/V does not depend on the resistance of the circuit.

Think of it like this:

If the capacitor is fully charged, there is no current in the circuit so any resistors in the circuit don't consume any potential (voltage).
But Kirchhoff's loop law states that the potential must be 0 for any loop in a circuit thus the capacitor has all the potential ( ie. emf battery = V capacitor)

So that only holds true for a capacitor in series, correct? Say there was a circuit that looked like the attached file and the question that goes along with it. Explain what you would choose based on your logic posted above. (Circuit and Question are from TBR Circuits 25 question exam Passage III)

 

[sshah92]

So that only holds true for a capacitor in series, correct? Say there was a circuit that looked like the attached file and the question that goes along with it. Explain what you would choose based on your logic posted above. (Circuit and Question are from TBR Circuits 25 question exam Passage III)

Is it... B? That's what I would choose. I would think resistors only affect the timing of charge, not how much charge the capacitor holds.

 

[MTHFR]

Is it... B? That's what I would choose. I would think resistors only affect the timing of charge, not how much charge the capacitor holds.

Yeah, I think it's B. After charging, the capacitors charge should equal C•V according to the equation Q=CV. It seems like others have a greater intuition about your previous question, but for this, I think it's B.

 

[Vicodeen]

No, it's D.
Their reasoning as follows
1 Q=CV
2 Voltage across capacitor is same as resistor 3 since they are parallel
3 so Q = C (I* R3)
4 I = V / Req
5 Req = R1 + R3
6 Substituting 5 into 4 and then that into 3 gives answer choice D

This would go against the answer of the original question. So I'm assuming that the original question was badly worded and they intended to mean a capacitor that is not in parallel with anything does not have its total charge effected by the equivalent resistance of the circuit and therefore voltage drop of the circuit preceding it? Anyone knowledgeable enough to confirm or disprove what I said?

Also, sshah92 are you from Illinois? heh.

 

[Vicodeen]

.

 

[99dmg]

Yes, the V in the equation C=Q/V does not depend on the resistance of the circuit.

Think of it like this:

If the capacitor is fully charged, there is no current in the circuit so any resistors in the circuit don't consume any potential (voltage).
But Kirchhoff's loop law states that the potential must be 0 for any loop in a circuit thus the capacitor has all the potential ( ie. emf battery = V capacitor)

Sorry this explanation I gave was for a simple RC series circuit to try and explain the concept behind it.

The first question is still right even for a resistor in parallel even though the emf doesn't equal the V on the capacitor. The V on the capacitor will still DEPEND on the emf (ie. if emf increases the V on the capacitor increases) thus Q still DEPENDS on emf.

The second question's answer is D like you stated.

The easiest way to do this would be the parallel rule like TBR book does it. Components in parallel have the same V, so V for resistor 3 = V for capacitor.

The rational I stated before (Kirchhoff's loop law) still works since the resistor in series with the capacitor still experiences a V loss even in the loop with the capacitor. This is due to the fact that there is still current running through the resistor even when the capacitor is fully charged (through the first path)

 

[Vicodeen]

Sorry this explanation I gave was for a simple RC series circuit to try and explain the concept behind it.

The first question is still right even for a resistor in parallel even though the emf doesn't equal the V on the capacitor. The V on the capacitor will still DEPEND on the emf (ie. if emf increases the V on the capacitor increases) thus Q still DEPENDS on emf.

The second question's answer is D like you stated.

The easiest way to do this would be the parallel rule like TBR book does it. Components in parallel have the same V, so V for resistor 3 = V for capacitor.

The rational I stated before (Kirchhoff's loop law) still works since the resistor in series with the capacitor still experiences a V loss even in the loop with the capacitor. This is due to the fact that there is still current running through the resistor even when the capacitor is fully charged (through the first path)

Yeah its making more sense now, thank you.

 

[sshah92]

No, it's D.
Their reasoning as follows
1 Q=CV
2 Voltage across capacitor is same as resistor 3 since they are parallel
3 so Q = C (I* R3)
4 I = V / Req
5 Req = R1 + R3
6 Substituting 5 into 4 and then that into 3 gives answer choice D

This would go against the answer of the original question. So I'm assuming that the original question was badly worded and they intended to mean a capacitor that is not in parallel with anything does not have its total charge effected by the equivalent resistance of the circuit and therefore voltage drop of the circuit preceding it? Anyone knowledgeable enough to confirm or disprove what I said?

Also, sshah92 are you from Illinois? heh.

Haha nah I go to UT