Potential difference in serie

In summary, the conversation discusses a physics problem involving a current flowing through a resistor, capacitor, and inductor, and the potential difference between two points. The solution is calculated using Ohm's law and the loop rule, and it is concluded that the potential difference is zero due to the influence of the inductor. The conversation also touches on the fact that the solution is only valid for a certain time interval and the importance of understanding the underlying principles in solving physics problems.
  • #1
bulbanos
11
0
I've been thinking while studying physics and came up with the following problem:

A current I, starting at a point A, flows through a resistor with R=25 kΩ, a capacitor with C=4milliF, and an inductor with L=7 Henry to finish in point B. At a time when Q=8 milliC and the current I=0.2 milliAmps and is changing at a rate dI/dt=-1.0 A/s, what is the potential difference VB-VA (inVolts)?

Solution: Moving in the direction of the current flow we have
VB-VA=-IR-Q/C-L dI/dt
IR =5 V
Q/C = 2 V
L dI/dt = -7 V

Thus, VB-VA = -5V -2V +7V = 0V (?)

how is this possible? There must be some loss due to the resistor, otherwise we could be able to make perfect circuits with no losses due to household devices for example.
 
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  • #2
The results seem good,as long as the initial equation (the Ohm's law) is correct,regarding the signs...
The tension varies sinusoidally (damped,c'est vrai) and it passes through zero many times,and there's no surprise...

For the second part,i didn't get your reasoning...What are u talking about...?

Daniel.
 
  • #3
I conclude from my calculations that the potential drop is 0V, so the power dissipated is I.V=0 W and there are no losses of energy.
 
  • #4
Yes,but that's happening at a moment of time,not an interval...

Daniel.
 
  • #5
what's so characteristic about that single moment of time?
 
  • #6
Well,nothing really,just the fact that at that moment of time,the potential difference between the ends of the circuit is zero...Didn't u ever see graphs of intensity or potential difference in alternative current?? :rolleyes:

Daniel.
 
  • #7
bulbanos said:
I've been thinking while studying physics and came up with the following problem:

A current I, starting at a point A, flows through a resistor with R=25 kΩ, a capacitor with C=4milliF, and an inductor with L=7 Henry to finish in point B. At a time when Q=8 milliC and the current I=0.2 milliAmps and is changing at a rate dI/dt=-1.0 A/s, what is the potential difference VB-VA (inVolts)?

Solution: Moving in the direction of the current flow we have
VB-VA=-IR-Q/C-L dI/dt
IR =5 V
Q/C = 2 V
L dI/dt = -7 V

Thus, VB-VA = -5V -2V +7V = 0V (?)

how is this possible? There must be some loss due to the resistor, otherwise we could be able to make perfect circuits with no losses due to household devices for example.

This correct and you need to look at it like this : the potential difference is zero initially, because given your specifications, the self induced emf and the decrease of i is just big enough to cancel any potential differences in this chain. But ofcourse, you need to realize that in the very "beginning" there was a potential difference in A and B because otherwise the current cannot flow. When the current has passed the induced emf of the coil, that's when the potential difference evolves to zero. If you were to look at this chain a little bit later, you shall see that the influence of the inductor is just that of an ordinary conducting wire. One can prove that by solving the differential equation for i. this equation comes from the loop-rule which is a mere manifestation of conservation of energy. But this result has nothing to do with the fact that the solution for i passes through zero many times (which it does ofcourse). However this solution is valid for a certain time-interval, NOT just one moment. Once the inductor starts acting like a wire, this solution is useless. You can calculate this time-periode based upon the formula for the decay of the current

regards
marlon

ps : why are you not in my lessons victor ??
 

1. What is potential difference in series circuits?

Potential difference in series circuits refers to the difference in electric potential between two points in a series circuit. It is measured in volts (V) and represents the amount of energy required to move an electric charge from one point to another.

2. How is potential difference calculated in a series circuit?

In a series circuit, potential difference is calculated by adding up the individual potential differences of each component. This can be done by using Ohm's law, which states that potential difference (V) equals current (I) multiplied by resistance (R).

3. What is the purpose of potential difference in series circuits?

The purpose of potential difference in series circuits is to provide a driving force for the flow of electric current. It is necessary for the functioning of electrical devices and allows for the transfer of energy from a power source to the components in the circuit.

4. How does potential difference affect the behavior of components in a series circuit?

Potential difference affects the behavior of components in a series circuit by determining the rate of flow of electric current. Components with a higher resistance will experience a greater potential difference and thus a smaller current, while components with a lower resistance will experience a smaller potential difference and a larger current.

5. Can potential difference be negative in a series circuit?

Yes, potential difference can be negative in a series circuit. This occurs when the direction of current flow is opposite to the direction of the electric field, resulting in a decrease in potential energy. Negative potential difference can also occur when the potential difference across a load is greater than the potential difference of the power source.

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