Why Does a Series Circuit Show Zero Potential Difference Despite Components?

[bulbanos]

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.

 

[dextercioby]

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.

 

[bulbanos]

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.

 

[dextercioby]

Yes,but that's happening at a moment of time,not an interval...

Daniel.

 

[bulbanos]

what's so characteristic about that single moment of time?

 

[dextercioby]

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??

Daniel.

 

[marlon]

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 ??