The relationship of the current before and after a change.

In summary, the current before frequency duplication and the current after frequency duplication are both proportional to the voltage. The impedance of the circuit before and after frequency duplication are also proportional to the voltage.
  • #1
Asmaa Mohammad
182
7

Homework Statement


An AC source was connected to a capacitor and an inductor (ignoring its ohmic resistance) in series. If the inductive reactance was equal to twice the capacitive reactance, and then the frequency duplicated, determine:

1. the relationship between the the current before and after duplication of frequency. (i₁/i₂)
2. The relationship between the impedance of the circuit before and after duplication of frequency. (Z₁/Z₂)

Homework Equations


i=V/Z
Xc=1/2πfC
V=NBAω

The Attempt at a Solution


Uauvh.jpg

Note: In my solution I considered the maximum values of both the current and the voltage, is this correct?
Is my solution correct?
 
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  • #2
Where does your V1 come from? The problem statement doesn't say anything about a coil rotating in a magnetic field, and certainly the constants A and N are not given. It only says that there is an AC source. An for your determination of I1 you've only considered the capacitive reactance. Why is that?

You can assume that the AC source is an ideal voltage source with a fixed amplitude. Only the frequency of the source changes.
 
  • #3
Asmaa Mohammad said:

Homework Statement


An AC source was connected to a capacitor and an inductor (ignoring its ohmic resistance) in series. If the inductive reactance was equal to twice the capacitive reactance, and then the frequency duplicated, determine:

1. the relationship between the the current before and after duplication of frequency. (i₁/i₂)
2. The relationship between the impedance of the circuit before and after duplication of frequency. (Z₁/Z₂)

Homework Equations


i=V/Z
Xc=1/2πfC
V=NBAω

The Attempt at a Solution


Uauvh.jpg

Note: In my solution I considered the maximum values of both the current and the voltage, is this correct?
Is my solution correct?
The statement doesn't mention that the AC source will change the amplitude of voltage as the frecuency changes. Are you assuming that or is missing in the statement?. Anyway when you calculate current why did you use only the Xc and not the series impedance?
 
  • #4
Diegor said:
The statement doesn't mention that the AC source will change the amplitude of voltage as the frecuency changes. Are you assuming that or is missing in the statement?. Anyway when you calculate current why did you use only the Xc and not the series impedance?
Edit. Sorry I see that your using the xc as impedance because is the result of 2xc-xc. That is correct.
 
  • #5
gneill said:
Where does your V1 come from? The problem statement doesn't say anything about a coil rotating in a magnetic field, and certainly the constants A and N are not given. It only says that there is an AC source.
Yes, it didn't say anything about it, but I am assuming it, because in my physics assignment in general this year, all the AC sources are considered to be Dynamos, and since it didn't mention that the A and N will vary I assumed that they are constant.
gneill said:
An for your determination of I1 you've only considered the capacitive reactance. Why is that?
Gneil, in the problem statement it says :"the inductive reactance is twice the capacitive reactance" so Xl=2Xc, and when determining the equivalent impedance of both the capacitive reactance and the inductive reactance, I did this:
Z=Xl-Xc=2Xc-Xc=Xc
so the impedance of the circuit in the first case equals the capacitive reactance, doesn't it?
 
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  • #6
Diegor said:
The statement doesn't mention that the AC source will change the amplitude of voltage as the frecuency changes. Are you assuming that or is missing in the statement?
I am assuming this because in my textbook the only AC source is a Dynamo, and in dynamo:
V=NBA (2πf), and the frequency change here must happen in the dynamo, so I am assuming that the amplitude of the voltage will change too.
 
  • #7
Asmaa Mohammad said:
Yes, it didn't say anything about it, but I am assuming it, because in my physics assignment in general this year, all the AC sources are considered to be Dynamos, and since it didn't mention that the A and N will vary I assumed that they are constant.
I see. Well, that's an approach I've not seen before. So your source voltage is going to double when its frequency doubles. That will have an effect on the current.
Gneil, in the problem statement it says :"the inductive reactance is twice the capacitive reactance" so Xl=2Xc, and when determining the equivalent impedance of both the capacitive reactance and the inductive reactance, I did this:
Z=Xl-Xc=2Xc-Xc=Xc
so the impedance of the circuit in the first case equals the capacitive reactance, doesn't it?
Okay.
 
  • #8
gneill said:
I see. Well, that's an approach I've not seen before. So your source voltage is going to double when its frequency doubles. That will have an effect on the current.

Okay.
OK, and what is your opinion about the solution as a whole?
 
  • #9
Asmaa Mohammad said:
I am assuming this because in my textbook the only AC source is a Dynamo, and in dynamo:
V=NBA (2πf), and the frequency change here must happen in the dynamo, so I am assuming that the amplitude of the voltage will change too.
I agree with gneill about the voltage source i think you should use the same amplitude V in both cases only the Xc and Xl will chanche with the frecuency. For me the rest of your approach is correct.
 
  • #10
Guys, I really can't imagine that the frequency through the capacitor and the inductor will change, but the frequency in the Dynamo will be constant?!
I mean the Dynamo is the AC source here, so when the frequency changes, that means the rate of the rotation of the coil changes, then the voltage going out of the dynamo has undergone a change in the frequency.
Right?!
 
  • #11
Asmaa Mohammad said:
OK, and what is your opinion about the solution as a whole?
Taking into account the odd (to me) way of dealing with the voltage source, your work looks good.
 
  • #12
Asmaa Mohammad said:
Guys, I really can't imagine that the frequency through the capacitor and the inductor will change, but the frequency in the Dynamo will be constant?!
I mean the Dynamo is the AC source here, so when the frequency changes, that means the rate of the rotation of the coil changes, then the voltage going out of the dynamo has undergone a change in the frequency.
Right?!
If you say that the AC source is a dynamo then taking into account the behavior of a dynamo is justified. But to me there is no evidence in the problem statement itself that a dynamo is implied. I've never seen a course where "AC source" must be interpreted as a dynamo by default. What I've seen is that when dynamos are being dealt with in a problem it is always explicitly stated.

Most often we treat AC sources like signal generators where voltage and frequency are both adjustable as desired, and changing one does not change the other.
 
  • #13
gneill said:
If you say that the AC source is a dynamo then taking into account the behavior of a dynamo is justified. But to me there is no evidence in the problem statement itself that a dynamo is implied. I've never seen a course where "AC source" must be interpreted as a dynamo by default. What I've seen is that when dynamos are being dealt with in a problem it is always explicitly stated.

Most often we treat AC sources like signal generators where voltage and frequency are both adjustable as desired, and changing one does not change the other.
OK, gneil, you gave me a new concept to think about when dealing with AC problems. It's a new information for me that the AC sources are like signal generators. This concept is not mention in my textbook, though. And I don't actually know if my course gives any significance to this point.
Anyway, let's say your point of view is the correct one, would the solution be like this (correcting some missing things and incorrect mathematical calculations):
Qqig5.jpg
 
  • #14
Asmaa Mohammad said:
Anyway, let's say your point of view is the correct one, would the solution be like this (correcting some missing things and incorrect mathematical calculations):
Sure. Only I'd probably disregard the dynamo details entirely and just assume that the voltage was "V" for both cases, form the ratio and cancel it out early on:

##\frac{I_1}{I_2} = \left( \frac{V_1}{Z_1} \right) \cdot \left( \frac{Z_2}{V_2} \right) = \frac{Z_2}{Z_1}~~~~## since ##V_1 = V_2##

Then go ahead and work with the known details about the impedances.
 
  • #15
gneill said:
Sure. Only I'd probably disregard the dynamo details entirely and just assume that the voltage was "V" for both cases, form the ratio and cancel it out early on:

##\frac{I_1}{I_2} = \left( \frac{V_1}{Z_1} \right) \cdot \left( \frac{Z_2}{V_2} \right) = \frac{Z_2}{Z_1}~~~~## since ##V_1 = V_2##

Then go ahead and work with the known details about the impedances.
OK, and that's what I got. In the solution attached in #1, in the final step I had Z1/Z2 = 1/3.5.
And then In the solution attached in #13, I had I1/I2 = 3.5, which equals Z2/Z1.
That means my solution is correct, right?
 
  • #16
It means that you got the impedance ratio correct, yes. Whether or not you need to account for a "dynamo factor" I leave up to you :smile:
 
Last edited:
  • #17
gneill said:
It mean that you got the impedance ratio correct, yes. Whether or not you need to account for a "dynamo factor" I leave up to you :smile:
Thank you very much, gneil! You all the time help me a lot, I'm so grateful to you.
 

Related to The relationship of the current before and after a change.

1. How does the current change before and after a change?

The current is the flow of electric charge through a circuit. Before a change, the current will remain constant as long as the external factors such as resistance and voltage remain unchanged. After a change, the current may increase or decrease depending on the nature of the change and its effect on the circuit's resistance or voltage.

2. What factors can cause a change in current?

There are several factors that can cause a change in current, including changes in resistance, voltage, temperature, and the number of components in a circuit. Any external influence that alters the flow of electrons can result in a change in current.

3. How is the current affected by a change in resistance?

The current is inversely proportional to the resistance, meaning that an increase in resistance will result in a decrease in current and vice versa. This is known as Ohm's Law, which states that the current is equal to the voltage divided by the resistance.

4. Can a change in voltage affect the current?

Yes, a change in voltage can affect the current. According to Ohm's Law, an increase in voltage will result in an increase in current, and a decrease in voltage will lead to a decrease in current. However, the resistance of the circuit must remain constant for this relationship to hold true.

5. How does a change in temperature impact the current?

A change in temperature can affect the current in a circuit by altering the resistance. In some materials, as the temperature increases, the resistance decreases, resulting in an increase in current. In others, the resistance may increase with temperature, leading to a decrease in current. This relationship depends on the type of material and its temperature coefficient of resistance.

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