anonymous
  • anonymous
Let us say that there is an electrical circuit and it also has a inductor of inductance L. The current on the circuit is a function of time, I(t). So now we can be sure that there is always a potential drop across the inductor. And let us assume that dI(t)/dt = a, so potential difference across the inductor is L x a. So, here's my question! Will a Voltmeter connected across the inductor show the same value of potential difference, i.e. L x a ? (Also, Potential difference holds no meaning here as the Electric Field induced is non-conservative!) Help!!
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
thats a very good question and i think the answer.. will totally depend on the orientation of the voltmeter.. U should watch http://www.youtube.com/watch?v=eqjl-qRy71w although its not exactly what you asked.. the concept is the same.. calculating "potential difference" (which really makes no sense when we have induced fields) between 2 points by using a voltmeter.. and two same voltemters give different readings.. based on their orientation.. Great/Legendary Professor Walter lEwin demonstrates this spectacular phenomena.. which for me is trully amazing!!!
anonymous
  • anonymous
I have seen the whole lecture by the way! But this question is still unanswered! :(
anonymous
  • anonymous
It will read L·a

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
why?
anonymous
  • anonymous
Because voltage across the inductor is: Ldi/dt and di/dt=d(at)/dt=a, then V=La
anonymous
  • anonymous
This is not as simple as you feel it is! Voltage Difference golds no meaning when you talk about non conservative electric field. You may easily solve a problem by using Kirrchoff rules here but that may be mathematically correct but is actually wrong. It "might be" correct i.e., L x a may be the answer, but a proper explanation would help!
anonymous
  • anonymous
If we have to hold a discussion about inductors behavior based on your understanding of the relation V=d(Li)/dt as a consequence of Kirchoff´s Laws, then I give up here.
anonymous
  • anonymous
I just want a proper explanation, buddy! Tell me if I went wrong somewhere! Thats it.
anonymous
  • anonymous
The law that relates induced voltage and current in an inductor can be found in Maxwell´s equations (3rd equation in differential form) or in Farady's Law
anonymous
  • anonymous
I know that. Let's not be mathematical, why is it, that's my question!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
It's because in the case of the inductor, the wire connecting the voltmeter to the system does not go INSIDE the coil, but is always outside. Th coil has a symmetry of revolution about its axis, so, whatever the place the voltmeter is, it will always read the same.
anonymous
  • anonymous
yup.. that is a good enough explanation i feel :) :) by mister Vincent!
anonymous
  • anonymous
Yo! Thanks.
anonymous
  • anonymous
@Vincent-Lyon.Fr. Some contradiction in your theory? If there is symmetry outside, there is also inside then the fact that the wire goes inside the coil or not is meaningless.
anonymous
  • anonymous
i think by symmetry what he means is that, all the path that the current can choose to flow through will turn out to be the same, and thus the "potential difference" will have the same value!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
True, if the connecting wire goes inside, the revolution symmetry will give the same reading on the voltmeter, whatever the orientation, BUT it will be different from the one measured outside. The one measured outside is not affected by the changing B-field and induced emf of the coil.

Looking for something else?

Not the answer you are looking for? Search for more explanations.