Nonstandard RL circuit

The Maxwell-Faraday relationship describes the connection between the electromotive potential around a closed loop and the change in magnetic flux across the enclosed surface.

$\displaystyle -\frac{\partial \phi_B}{dt} = \oint E\cdot dl = \sum_i V_i$ .

The changing flux through the ring creates an electromotive potential which is opposed by a voltage in the ring. This voltage has contributions from the resistive ( $V_R = IR$ ) and inductive ( $V_L = L\frac{\partial I}{\partial t}$ ) characteristics of the ring. The Maxwell-Faraday relationship for the ring is given by

$\begin{aligned} \displaystyle -\frac{\partial\phi_B}{\partial t} &= \sum_i V_i\\ &= V_R + V_L \\ &= IR + L\frac{\partial I}{\partial t} \end{aligned}$

As the flux is a periodic signal with frequency $\omega$ , and there is only resistive and inductive behavior in the ring, the current will be periodic with the same frequency.

We can represent any set of sinusoidal signals with the same frequency as the real components of arrows spinning around the complex plane. The arrows are represented by a complex amplitude describing the signal amplitude and a complex exponential responsible for the oscillation with an offset angle $\theta$ describing the position of the arrow at time zero: $\mathbf{A}e^{i(\omega t+\theta)}$ .

We're told that the flux is given by $\phi_0\cos\omega t$ so we know that the amplitude is real and that the phase offset is zero, i.e. $\phi_B \rightarrow \Re \left[\phi_0e^{i\omega t}\right]$ (where $\Re$ indicates that we take the real component).

The current is given by $I \rightarrow \Re \left[\mathbf{I}e^{i(\omega t+\theta_I)}\right]$ which allows for a phase offset in the current.

Making these substitutions in the Maxwell-Faraday relationship, we find

$\displaystyle i\omega\phi_0 = \mathbf{I}e^{i\theta_I } R + i\mathbf{I}L\omega e^{i\theta_I}$

$\displaystyle \mathbf{I} = e^{-i\theta_i}\frac{\omega\phi_0}{R+i\omega L}$

The magnitude of the amplitude for the current is then $\displaystyle \mathbf{I} = \frac{\omega\phi_0}{\sqrt{R^2+\omega^2 L^2}}$ which is $\frac{1}{\sqrt{5}}$ given the values in the problem description.

The rate of heat dissipation through the resistor is then $P = I^2R$ which is the square of a sinusoid with amplitude $\frac15 10 = 2$ . The time averaged value of the square of a sinusoid is half its amplitude, so we have $\langle P\rangle_T = 1$ W .

When the thin ring is rotating with the constant angular speed in the external $B$ field, it experiences two $emf$ , one resulting from the changing magnetic flux created by the external field, $\frac{\mathrm d \Phi(t)}{\mathrm dt}$ , and the other from changing magnetic flux created by self-inductance, $L\frac{\mathrm di}{\mathrm dt}$ .

Method 1:

Hence we get the equation, $\frac{\mathrm d \Phi}{\mathrm dt} - L \frac{\mathrm di}{\mathrm dt} - Ri=0$ . The sign in front of $L\frac{\mathrm di}{\mathrm dt}$ is negative because the self-induced $emf$ is always opposing the external field, which creates the current, which causes the self-induced $B$ field.

This equation is a first-order differential equation. To make it looks better, we arrange it to be: $\frac{\mathrm di}{\mathrm dt} + \frac{R}{L}i=\frac{\mathrm d \Phi}{\mathrm dt} = \frac{\omega \Phi_0}{L}\sin \omega t$

To solve the differential equation, we multiply both sides by the integrating factor, $e^{\frac{R}{L}t}$ .

$\frac{\mathrm di}{\mathrm dt}(e^{\frac{R}{L}t}) + \frac{R}{L}i = \frac{\omega \Phi_0}{L} e^{\frac{R}{L}t }\sin \omega t$

$\implies \frac{\mathrm d}{\mathrm dt}(i e^{\frac{R}{L}t})= \frac{\omega \Phi_0}{L} e^{\frac{R}{L}t } \sin \omega t$

$\implies i e^{\frac{R}{L}t} = \int \frac{\omega \Phi_0}{L} e^{\frac{R}{L}t } \sin \omega t \mathrm dt$

$\implies i e^{\frac{R}{L}t} = \frac{\omega \Phi_0}{L (\frac{R^2}{L^2} + \omega^2)}(\frac{R}{L} e^{\frac{R}{L}t} \sin \omega t - \omega e^{\frac{R}{L}t} \cos \omega t) + c$ , where $c$ is a arbitrary constant.

$\implies i= \frac{\omega \Phi_0}{L(\frac{R^2}{L^2} + \omega^2)}(\frac{R}{L}\sin \omega t - \omega \cos \omega t )+ ce^{- \frac{R}{L}t}$

Now we sub all the values in and get:

$i= 0.2 \sin 100t - 0.4 \cos 100t + ce^{-50t}$

The decaying exponential on the right is transient, so it will vanish as time passed. Hence $i= 0.2 \sin 100t -0.4 \cos 100t = \frac{1}{\sqrt{5}}\sin (100t - \tan ^{-1}(\frac{0.4}{0.2}))$

The power dissipated = $i^2 R=\frac{10}{5} \sin^2 (100t - \tan^{-1}2)$

The average value of sine squared is $\frac{1}{2}$ , so average power dissipated = $1 W$ .

Method 2:

We can get the steady-state answer by using impedance, which can greatly simplify the working.

As we have analysed, the system resembles a circuit with R and L connected in series and an alternating $emf$ of $\omega \Phi_0 \sin \omega t$ . The impedance of inductor is $j\omega L$ , where $j$ is the imaginary number $\sqrt{-1}$ .

The previous differential equation is thus simplified to $(R+j\omega L)i=Im(\omega \Phi_0 e^{j\omega t})$ , where $Im(x)$ means the imaginary part of x.

It can be proven that $i=Im(\frac{e^{j\omega t}}{R+j\omega L})$

Sub all the values in, we get: $i=Im(\frac{e^{j(100)t}}{1+2j}) =Im( \frac{1}{5}(2j-1)(e^{j(100)t})= 0.2 \sin 100t - 0.4 \cos 100t$

And finally , we get the same answer: average power dissipated = $1 W$ .

Let us begin by writing Kirchhoff's law for the ring $\mathcal{E}_{total}=\sum_{k}\mathcal{E}_{k}= IR.$ From Faraday's Law we know that there is an induced emf in the ring given by $\mathcal{E}_{total}=-\frac{d\Phi}{dt}- L \frac{dI}{dt}.$ Here, the term $-L \frac{dI}{dt}$ accounts for the magnetic flux created by the current in the ring, i.e., self- induction. Thus, Kirchhoff's law assumes the form $L \frac{dI}{dt}+I R= \omega \Phi_{0}\sin(\omega t).$ Note that the system is equivalent to a an RL circuit connected to an AC power supply. Therefore, we look for a solution of the form $I(t)=I_{m}\sin(\omega t-\phi).$ The values of $I_{m}$ and $\phi$ are chosen so that $I(t)$ is a solution of the equation $L \frac{dI}{dt}+I R= \omega \Phi_{0}\sin(\omega t).$ Making use of the trigonometric identity $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$ we find that $\tan(\phi)=\frac{\omega L}{R} \quad \textrm{and} \quad I_{m}=\frac{\omega \Phi_{0}}{\sqrt{R^{2}+\omega^{2}L^{2}}}.$ Now we can compute the average power dissipated in the system. If $T=\frac{2\pi}{\omega}$ we can simply write $\bar{P}=\frac{1}{T}\int_{0}^{T} I(t)^{2} R dt = I_{m}^{2}R \frac{1}{T}\underbrace{\int_{0}^{T}\sin^{2}(\omega t-\phi)dt}_{=\frac{1}{2}}= \frac{1}{2} I_{m}^{2}R = \frac{1}{2}\frac{\Phi_{0}^{2}\omega^{2}R}{R^{2}+\omega^{2}L^{2}}=1~\mbox{W}.$ In order to compensate for the dissipated (ohmic losses), the external forces must deliver $1~\mbox{W}$ of power. It is also possible to arrive at the same solution by looking at the torque $\vec{\tau}=\vec{\mu} \times \vec{B}$ produced by the external magnetic field. Here, $\mu= I A$ is the magnetic moment of the ring. This torque can be rewritten as $\tau(t)=I(t) \Phi(t)\sin(\omega t).$ It is important to note that this torque tends to slow down the ring. The corresponding mechanical power is $P_{\textrm{mechanical}}(t)= \omega \tau(t).$ Computing $\bar{P}=\frac{1}{T}\int_{0}^{T} P_{\textrm{mechanical}}(t)$ we obtain the same expression as we did before, that is, $\bar{P}=\frac{1}{2}\frac{\Phi_{0}^{2}\omega^{2}R}{R^{2}+\omega^{2}L^{2}}.$

Take the ring to be a L R circuit with AC source.

The maximum EMF= $phi *omega$ =10V

We get inductive reactance =20 ohm so I_o)\ =\(\frac{1}{\sqrt(5)} )\ and power factor= \(\frac{1}{\sqrt(5)}

So $P_avg= \frac{V_0*I_0*power factor}{2}$ =1