TBR physics Circuits (capacitor question) - driving me crazy

[09grad]

---

Members don't see this ad.

For anyone who has TBR physics, I'd appreciate your help with Question 34 in Section 9 (Electric Circuits).

The question asks about heat dissipated, giving values for current and resistance. The answer in the back uses P = (I^2)*R, but I don't understand why we can't use P=IV and still get the same answer. We know that the voltage across the capacitor after a long time will be 10mV (the emf of the battery), and since the resistors are in parallel with the capacitor, they should have the same V. Yet, P=IV doesn't seem to work.

Thanks in advance!

 

[lorenzomicron]

For anyone who has TBR physics, I'd appreciate your help with Question 34 in Section 9 (Electric Circuits).

The question asks about heat dissipated, giving values for current and resistance. The answer in the back uses P = (I^2)*R, but I don't understand why we can't use P=IV and still get the same answer. We know that the voltage across the capacitor after a long time will be 10mV (the emf of the battery), and since the resistors are in parallel with the capacitor, they should have the same V. Yet, P=IV doesn't seem to work.

Thanks in advance!

First of all, capacitors dont develop power because they have no resistance. Can you please post the whole problem so I understand what you're talking about?

 

[dingyibvs]

For anyone who has TBR physics, I'd appreciate your help with Question 34 in Section 9 (Electric Circuits).

The question asks about heat dissipated, giving values for current and resistance. The answer in the back uses P = (I^2)*R, but I don't understand why we can't use P=IV and still get the same answer. We know that the voltage across the capacitor after a long time will be 10mV (the emf of the battery), and since the resistors are in parallel with the capacitor, they should have the same V. Yet, P=IV doesn't seem to work.

Thanks in advance!

P = (I^2)R is the formula for heat dissipation, P = IV is power delivered.

They're not always the same, which is why electricity is transferred over long distance in high voltage, low resistance power lines. With V being high, I can be small to obtain the same IV and therefore the same power delivered. But with a low I and R, the heat dissipated along the way will be less.

If you want a specific explanation, then you need to write out the problem in more detail.

 

[lorenzomicron]

P = (I^2)R is the formula for heat dissipation, P = IV is power delivered.

They're not always the same, which is why electricity is transferred over long distance in high voltage, low resistance power lines. With V being high, I can be small to obtain the same IV and therefore the same power delivered. But with a low I and R, the heat dissipated along the way will be less.

If you want a specific explanation, then you need to write out the problem in more detail.

I'm not sure what you mean by "they're not always the same", because those are equal. The only manner of concern is that we must look specifically at which currents, resistances, and voltages for the specific element. But, I really could not agree that they are not the same.

 

[plsfoldthx]

how hard is it to write out the whole question?
​

 

[09grad]

Sorry, it's just that the question involves a circuit. I've now uploaded a pic.h

V = 10mV, C=2uF, R=5mOhms

question is: If a steady current of 1mA passes through resistor R1 for 10 sec, how much heat is generated.

The answer uses P=I^2*R but I am wondering, since we know that after a time, V across the resistor will be 10mV, and we know the current, why can't we use P=IV?

Thanks

 

[Handinhand]

Because the V of the battery is not the voltage across the resistor. If you work it out with the current and the resistor, you shouldn't get the same volate reading. It's because wires themselves have an inherent resistance that will also dissipate some energy.

 

[lorenzomicron]

Sorry, it's just that the question involves a circuit. I've now uploaded a pic.h

V = 10mV, C=2uF, R=5mOhms

question is: If a steady current of 1mA passes through resistor R1 for 10 sec, how much heat is generated.

The answer uses P=I^2*R but I am wondering, since we know that after a time, V across the resistor will be 10mV, and we know the current, why can't we use P=IV?

Thanks

You can use P=VI, but first you must find V of R1 which uses Ohm's Law to be V=IR= 1mA*5mOhms. (also then multiply P by t, time, to find the heat energy disspitated)

When the switches circuit is open, it is not a parallel circuit, and in fact, there is a short circuit.

Also, since the circuit is "ideal", one need not worry about wire dissipation effects or anything like that.

However, if the current is said to be 1 mA, then the voltage of the source must be 5 and not 10 as you previously stated.

where did you get the value of 10 mV from?

 

[09grad]

hey thanks for your help lorenzo. I used 10mV because after the capacitor is charged, the voltage across it will equal the voltage across the resistor, since they are in parallel.

However, maybe it was wrong of me to assume the capacitor is already fully charged.

 

[dingyibvs]

I'm not sure what you mean by "they're not always the same", because those are equal. The only manner of concern is that we must look specifically at which currents, resistances, and voltages for the specific element. But, I really could not agree that they are not the same.

Technically true, but I like to keep it simple for the MCAT. In order to reconcile that concept with my high voltage wire example, I'd have to explain alternators and their impedances, etc. For the MCAT, it's safer to just use P = I^2 * R for heat dissipation.

Sorry, it's just that the question involves a circuit. I've now uploaded a pic.h

V = 10mV, C=2uF, R=5mOhms

question is: If a steady current of 1mA passes through resistor R1 for 10 sec, how much heat is generated.

The answer uses P=I^2*R but I am wondering, since we know that after a time, V across the resistor will be 10mV, and we know the current, why can't we use P=IV?

Thanks

Clearly the voltage over the resistor V = IR = 1 x 5 = 5 mV, and not 10 mV. So using V = 10mV will definitely give you a wrong answer. According to the drawing and the information you gave, there is no way for that to happen, so it must be some sort of theoretical situation.

If you close the circuit, then yes, eventually the capacitor will have 10 mV across it, and eventually, the current through R1 will be 2 mA since I = V/R = 10/5 = 2 mA.

 

[lorenzomicron]

hey thanks for your help lorenzo. I used 10mV because after the capacitor is charged, the voltage across it will equal the voltage across the resistor, since they are in parallel.

However, maybe it was wrong of me to assume the capacitor is already fully charged.

Yes, the 10 mV assumption is not right. But, the voltages would be equal due to parallel, just in this case, the voltage is 5mV.

great.