# Tag Info

## Hot answers tagged inductance

42

Suppose an inductor is connected to a source and then the source is disconnected. The inductor will have energy stored in the form of magnetic field. But there is no way/path to discharge this energy? Short answer: It will find a way/path to discharge this energy. Longer answer: Let's have this simple electric circuit consisting of a battery (...

31

It depends. You cannot disconnect an ideal inductor from an ideal voltage source with an ideal switch. These ideal things will break your calculations and you will get an infinite voltage on disconnect. A real inductor has its coil resistance, a capacitance between coils and an insulation between coils that has some great, but pretty much nonlinear ...

20

The energy is not used up. It goes to the magnetic field, and when the magnetic field is at its strongest value there is no energy left in the electric field of the capacitor. But then the magnetic field starts decreasing as the capacitor charges back up because the current starts decreasing. And when the capacitor is fully charged there is no current and no ...

17

Capacitors and inductors are images of one another under the self-inverse mapping that transforms a linear electrical network to its dual network. The network duality transformation maps the network's graph to its topological dual graph,then all the impedances (either as lone-frequency complex scalars or as Laplace transfer functions) in the dual graph ...

13

There is no analogy of "voltage", "current", "charge" or "flux" to electromagnetism for the weak force, at least none that would be helpful. The reason for this is that all of these are classical concepts, while the notion of the weak force is completely quantum. Taking the classical limit just makes it vanish because the classical force law of forces with ...

12

It depends as to whether you are considering the LC(R) circuit in isolation or with a alternating voltage supply as the driver. The series of diagrams illustrate what might happen if the capacitor is initially charged and then connected to an inductor (with resistance). So what is happening is essential an exchange of energy associated with a magnetic ...

11

But there is no way/path to discharge this energy? What will happen to the stored energy, current and voltage of the inductor in this case? In that case it makes its own circuit with its own path to ground. Often, that is through dielectric breakdown at the switch itself, but the details are highly unpredictable and depend very sharply on environmental ...

9

In the limit of long times, the currents are steady, so the magnetic fields they create are steady so there is no induced EMF. This situation is usually tagged "steady state". That said, there will be a period of time when you have just switched a circuit on or off during which things have not settled down and then there will in general be effects not seen ...

9

In this answer, I'm going to use $Q$ instead of $q$ for the capacitor charge. There is only one certain rule for finding Lagrangians: The Lagrangian is chosen such as to get the correct equations of motion. Never forget that. In the case of a circuit problem, the most sure way to know you got the right Lagrangian is to see if it gives you the right ...

9

My book says the current through the inductor would pick up to 0.2A, while the current through the capacitor drops to 0A. This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit. Why? In DC steady state (the solution as $t\rightarrow\infty$), all the ...

9

Usually this extra energy creates a spark due to the high back emf produced. But it is not always possible for a coil to create sparks. It is clear If we try out the experiment. So what happens to the magnetic energy if no sparks are generated? firstly, The sudden switching off would create a potential. difference between the ends of the coil. This means ...

8

When we have a DC voltage source with a switch in series with RL and the switch is closed at t=0 then it is said that current is zero initially, but the voltage across inductor is same as that of applied voltage( according to kirchhoff voltage law) so there should be current( according to v=L(di/dt) )but it contradicts the initial statement so how do I ...

7

There does not need to be a magnetic field in the inductor for there to be "back emf" (I would prefer "induced emf"). The induced emf is the consequence of a changing magnetic field and not of a magnetic field itself and hence there can be a changing magnetic field even at zero magnetic field (something like a positive acceleration downwards for a ball ...

7

Notice, the magnetic magnetic field $B$ at the center of a coil carrying current $i$, with radius $r$ & having $n$ no. of turns $$B=\frac{\mu_0}{2}\frac{ni}{r}$$ hence, magnetic flux $\phi$ linked to the coil is given as $$\Phi=BA=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2=\frac{\mu_0 \pi nir}{2}$$ Now, setting the value of $\phi$, we get $$L=\frac{n\Phi}{i}=... 7 however I can't think of the fundamental reason as to why it is completely impossible to charge a capacitor with anything but 50% of the battery energy whereas an inductor could theoretically store 100%. Essentially, to charge a capacitor with finite current from a voltage source requires dissipation. Why? Consider the ideal capacitor equation (in ... 7 Using imaginary numbers for current in reactive components just happens to make the maths a lot simpler. In AC circuits there is typically some phase difference between the voltage and the current. Manipulating these quantities without the use of complex numbers, but instead just keeping track of the phase differences (such as the power factor), is a right ... 7 An important point which is somewhat addressed by others but perhaps not clearly enough is (quoting Scotty) "Y' canna break the laws of Physics". You can make everything ideal - semiconductor wire, perfect instantaneously acting switch, infinite insulation - and the basic "rules" governing an inductor still apply. The fact that current flow cannot ... 6 I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn. 1. Why self-inductance is not considered when solving Faraday's law problems Self inductance should be ... 6 It's a reasonable question, and the answer is: one can't prove it, without introducing induction. Consider a conducting loop with no current. Then someone starts creating a current in it, using, for example, a battery. The question is: why should we perform work to create this current, if we know that magnetic force$$ \mathbf {F} = \frac{q}{c}[\mathbf{v},\...

6

Does that mean when I apply a voltage, the current will be infinite large? No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid. When those assumptions don't hold, we have to see what the 'infinity' (division by zero) ...

6

Transformers generate oscillating magnetic fields at the mains frequency and the fields produce an oscillating force on: anything nearby that's ferromagnetic (like the core) anything nearby that is carrying a current (like the windings) The sound you hear is because various bits of the transformers are moving in response to the oscillating fields and this ...

6

If the emf due to the solenoid is assumed to oppose the applied voltage and have equal magnitude (in volts), there is zero electromotive intensity in the wire. Since current is assumed to be present, this means the current flows even while total electromotive force vanishes. This is possible for wire made of perfect conductor (superconductor). In practice, ...

6

If you have a single tube, the current will flow on it directly without making the $N$ loops. It will result a different direction, i.e. different magnetic field, its magnetic field will be much weaker. Having the loops, the magnetic fields created by the induvidual loops is added. Actually, you have "the same current" using $N$ times to produce the field. ...

6

Yes, Kirchhoff's voltage law (KVL): Sum of voltage drops across all elements connected via perfect conducting wire in series in to a closed circuit is zero. is valid for lumped element RLC circuits, so also for inductors (for currents that do not change too fast, so voltage can be measured in practice). In practical circuits designed not to radiate, ...

6

Since you want to avoid differential equations, I will instead consider the so-called Phasor domain, which is actually nothing but the Fourier transform of the original signals. In phasor domain, we will be basically considering complex values: Complex voltages, complex resistances (which are denoted by $Z$ and called impedance): This is simply the ...

6

Therefore, current will increase, and voltage will also increase across the entire circuit. It's not clear exactly what you mean by "across the entire circuit", but think about your model of a battery. For a simple analysis that is usually used for circuits like this, the battery is considered as a constant voltage source. Therefore, by Kirchoff's Voltage ...

6

Your reasoning is correct. If you store energy in a system, mass-energy equivalence provides the conversion for calculating the matter equivalent which, if you had a weighing device sensitive enough, you could (in principle) measure. For things like charged capacitors and stretched springs and pressurized gas tanks and cans filled with gasoline, the mass ...

6

First of all, let us consider only a solenoid (without iron core) of length ${l}$ and area of cross-section ${A}$. The magnetic moment of the solenoid is given by:- $${M}_{solenoid}=({n}{l}){I}{A}$$ Where,n=number of turns per unit length of solenoid and ${I}$ = current flowing in the solenoid Now, the net magnetic field is:- $${B}={\mu_0}{n}{I}$$ Now if an ...

5

When a current passes through a closed loop, a magnetic flux through the cross-sectional area of that loop arises due to that current. This magnetic flux is related to the current by the relation: $$\phi=Li$$ $L$ is called the self-inductance of the loop and depends upon the configuration of that loop. $L$ remains constant as long as the configuration of ...

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