Inductors in alternating circuits

[Bilbo B]

Placing an inductor in the first place in an ac circuit will start opposing the current but after a long time it behaves as it were never there and current flows constantly, problem is what happens to energy while opposing? Does it gets stored in the inductor?
After when you open the circuit how the current's value changes, the inductor like again start to oppose it.

 

[sophiecentaur]

The energy is stored in the magnetic field E = LI²/2 where L is the Inductance. The 'half x squared" form of the formula is the same as for Kinetic Energy and the Energy stored in a Capacitor. See this link.
As you say, Changes are 'opposed' due to Lenz's Law.

 

[vanhees71]

You can describe the conductor effectively with its self-inductance $L$. A pretty complicated analysis using Maxwell's equations shows that the electromotive force is given by $L \dot{I}$, where $I=I(t)$ is the current running through it. Concerning the correct signs in the Kirchhoff circuital Laws you must refer to the right-hand rule in Faraday's Law.

Now also a real coil has a finite resistance. You can describe the real coil as the series of an ideal coil and a resistance connected to an AC voltage source $U(t)$. Then Kirchhoff's Law tells you that
$$R I + L \dot{I}=U(t).$$
Let's take
$$U(t)=U_0 \exp(\mathrm{i} \omega t),$$
understanding that the physical quantities are given by the real parts, i.e., the physical voltage source is
$$\mathrm{Re} U(t)=U_0 \cos(\omega t), \quad U_0 \in \mathbb{R}.$$
The above equation is a linear ordinary differential equation, and you get the general solution by first solving the corresponding homogeneous equation (i.e., the equation with $U(t)=0$) and then adding one special solution of the inhomogeneous equation.

Let's start with the homogeneous equation:
$$R I + L \dot{I}=0.$$
To solve it just rewrite it to
$$\frac{1}{I} \dot{I} = \frac{\mathrm{d}}{\mathrm{d} t} \ln \left (\frac{I}{I_0} \right)=-\frac{R}{L}.$$
Integrating and solving for $I(t)$ gives the general solution of the homogeneous ODE
$$I_{\text{hom}}(t) = I_0 \exp \left (-\frac{R}{L} t \right).$$

The inhomogeneous equation is also easily solved by thinking about the physics. The homogeneous solution is exponentially damped. Thus any "transient state" decays after some time and the current should just be an AC with the frequency imposed by the source, i.e., we make the Ansatz
$$I(t)=A \exp(\mathrm{i} \omega t).$$
Plugging this into the inhomogeneous ODE gives
$$(R + \mathrm{i} \omega L) A \exp(\mathrm{i} \omega t)=U_0 \exp(\mathrm{i} \omega t),$$
and thus
$$A=\frac{U_0}{R+\mathrm{i} \omega L} = \frac{U_0 (R-\mathrm{i} \omega L)}{R^2+\omega^2 L^2}.$$
The solution thus is
$$I_{\text{inh}}(t)=\mathrm{Re} \left [\frac{U_0 (R-\mathrm{i} \omega L)}{R^2+\omega^2 L^2} \exp(\mathrm{i} \omega t) \right]=\frac{U_0}{R^2+\omega^2 L^2} [R \cos(\omega t) + \omega L \sin(\omega t)].$$
In the long-time limit the current just oscillates with the same frequency as the voltage source, but its phase is somewhat behind the voltage. To see this one can rewrite the factor $A$ as
$$A=\frac{U_0}{\sqrt{R^2+\omega^2 L^2}} \exp(\mathrm{i} \varphi),$$
and the phase shift is given by
$$\varphi=-\arccos \left (\frac{R}{\sqrt{R^2+\omega^2 L^2}} \right) \in [-\pi,0].$$
Thus you have
$$I_{\text{inh}}(t)=\frac{U_0}{\sqrt{R^2+\omega^2 L^2}} \mathrm{Re} \exp(\mathrm{i} \omega t + \mathrm{i} \varphi) = \frac{U_0}{\sqrt{R^2+\omega^2 L^2}} \cos(\omega t+\varphi).$$
Since $\varphi<0$ the phase of the current lags behind the phase of the voltage source.

Concerning the energy: part of the energy within the circuit is stored in the magnetic field in the coil (inductor). This energy is given by
$$E_{\text{ind}}=\frac{L}{2} I^2.$$
To see this, consider the real coil connected to a DC voltage source, i.e., $U(t)=U_0=\text{const}$. Then the inhomogeneous equation tells you that the right ansatz is $I=A=\text{const}$ and this leads to
$$R A=U_0 \; \Rightarrow \; A=\frac{U_0}{R}.$$
The full solution then is
$$I(t)=I_0 \exp \left (-\frac{R}{L} t \right) + \frac{U_0}{R}.$$
Now assume there's no current flowing at $t=0$, i.e., we switch on the circuit at $t=0$. Then $I(0)=0$ and $I_0=-U_0/R$, i.e.,
$$I(t) = \frac{U_0}{R} \left [1-\exp \left (-\frac{R}{L} t \right) \right].$$
After a long time you just have a DC $I_{\text{infty}}=U_0/R$, i.e., the ideal coil in our circuit is just a short circuit. This makes sense, because if there's no change of the magnetic field anymore there's no induced emf.

Now we consider the case that we switch off the circuit again by just taking off the battery at $t=0$ and short-circuit the coil+resistor-series. Then we have the solution with $U_0=0$, i.e.,
$$I(t)=I_0 \exp \left (-\frac{R}{L} t \right),$$
With $I_0=U_0/R$. Now the voltage drop at the resistor at any time is
$$U_R(t)=R I(t) =U_0 \exp \left (-\frac{R}{L} t \right).$$
The total energy dissipated as heat thus is
$$E_{\text{diss}} = \int_0^{\infty} \mathrm{d} t U_R(t) I(t) = I_0^2 R \int_0^{\infty} \mathrm{d} t \exp \left (-2\frac{R}{L} t \right)=\frac{L}{2} I_0^2.$$
This total amount of energy must have been stored in the coil at $t=0$, i.e., it's the energy of the magnetic field in the coil at $t=0$:
$$E_{\text{coil}}(t=0)=\frac{I}{2} I_0^2,$$
and $I_0$ is the current running through the coil at $t=0$. This is still true within the quasistatic approximation of Maxwell's equations, employed here for standard AC circuit theory. It neglects the radiation of electromagnetic waves and the corresponding energy dissipation of the full solution of Maxwell's equations.

 

[Bilbo B]

thanks vanhees71 for helping me out!

 

[Herman Trivilino]

Placing an inductor in the first place in an ac circuit will start opposing the current

No, it opposes the change in the current.

 

[Delta2]

Summary:: If we place an inductor in an a.c circuits how their behavior affects the current flowing in it.What happens with the energy? On closing or opening the switch, the value of current, to what extent changes drastically.

Placing an inductor in the first place in an ac circuit will start opposing the current but after a long time it behaves as it were never there and current flows constantly, problem is what happens to energy while opposing? Does it gets stored in the inductor?
After when you open the circuit how the current's value changes, the inductor like again start to oppose it.

In DC circuits energy is stored in the inductor for some time and after that time (theoretically infinite time but practically time equal to 5x constants of time of the circuit) the inductor is like it doesn't exist and the circuit is affected only by the ohmic resistance, if any, of the inductor.
In AC circuits, during half cycle ,energy is stored in the inductor. During the other half cycle ,energy is removed form the inductor and is given back to the voltage source or to other elements of the circuit.