# Self-inductance; RL circuit



My notes of the excellent lecture 20 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

We will quantify the ability of a circuit to fight the electric flux that is produced by the circuit itself.

If you run a current through a circuit, then you create some magnetic fields. If the current is changing, then the magnetic fields are changing. There will be an induced EMF in that circuit that fights the change. We express that in terms of self-inductance $$L$$.

The magnetic flux that is produced by the circuit is always proportional to the current $$\shaded{ \phi_B = L\, I } \tag{self-inductance} \label{eq:L}$$

Therefore the induced EMF $$\shaded{ \mathcal E = – \frac{d\phi}{dt} = -L \frac{dI}{dt} } \tag{EMF self-inductance} \label{eq:Lemf}$$

This $$L$$ is only a matter of geometry. It is not a function of the current itself.

## Solenoid

We will calculate the self-inductance of a solenoid with radius $$r$$, length $$l$$ and $$N$$ windings.

Using Ampère’s law we derived

$$B = \frac{\mu_0 I N}{l} \label{eq:B}$$

We assign an open surface to this closed loop. The surface is complicated to imagine, but has a spiral staircase-like surface inside the solenoid. That magnetic field penetrates that surface $$N$$ times, because we have $$N$$ loops. Assuming the magnetic field is constant inside the solenoid, and zero outside, the magnetic flux is \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \phi_B &= \iint_R \vec B\cdot d\vec A, & \left(\vec B\parallelsum\vec A\right) \\ &= B \iint_R dA \\ &= B\, N\, \pi r^2 \end{align*}

Substituting $$B$$ from equation $$\eqref{eq:B}$$ \begin{align*} \phi_B &= B\, N\, \pi r^2 \\ &= \frac{\mu_0 I N}{l} N\, \pi r^2 \\ &= \frac{\mu_0 I N^2 \pi r^2}{l} \end{align*}

This must equal the definition of $$\eqref{eq:L}$$ $$\frac{\mu_0 \bcancel{I} N^2 \pi r^2}{l} = L\,\bcancel{I} \nonumber$$

The self-inductance of the solenoid follows as $$\shaded{ L = \mu_0\, \pi r^2 \frac{N^2}{l}\quad \Big[\rm{H}\Big] } \tag{$$L$$ solenoid} \label{eq:l2}$$

The SI unit for inductance is Henry, named for the American physicist Joseph Henry.

Every circuit has a finite value for the self-inductance, however small that may be. Sometimes it’s so small that we ignore it.

## RL Circuit and DC (step response)

Let’s look at a circuit with a battery, a switch, a self-inductor in series with a resistor.

Note that we use the lowercase $$i$$ for current to indicate that it is not a constant. It is a function of time.

### Battery provides energy

The switch is in position $$a$$ and there is no current. Then, we flip the switch to position $$b$$

1. Before we flip the switch there is no current running. At time $$t_0$$ the current $$i=0$$.
2. As we close the switch the current wants to increase, but the self-inductance fights the change in current.
3. If we wait long enough, the self-inductance loses the fight, and the current reaches a maximum value dictated by Ohm’s law, because the (ideal) self-inductance has no resistance.

We can’t use Kirchhoff’s rule, because there is a changing electric flux. The electric fields inside the conducting wires become non-conservative, and the closed loop integral around the circle is not zero. (see the lecture supplement)

$$\oint_C \vec E\cdot d\vec l = -\frac{d\phi_B}{dt} \nonumber$$

Substituting $$\phi_B$$ from equation $$\eqref{eq:L}$$ $$\phi_B = L\, i$$ $$\oint_C \vec E\cdot d\vec l = -L\frac{di}{dt} \nonumber$$

Going around in the direction of the current, the electric fields are

$$U$$Instead of $$\Delta V$$, we use the European symbol for voltage difference $$U$$.

Applying Faraday’s law \begin{align*} -U + 0 + i R &= -L\frac{di}{dt} \\ \implies U – L\frac{di}{dt} &= i\, R \end{align*}

Since $$di/dt$$ is positive and growing with time, the induced EMF, $$-L\frac{di}{dt}$$, is always opposing the voltage difference of the battery. That is where Lenz’s law is all about. It is not until you wait long enough, and $$di/dt$$ has become zero, that $$U=iR$$.

#### Solve using the antiderivative

Details

\require{physics} \begin{align*} L\dv{i}{t} + Ri &= U \\ \implies \dv{i}{t} &= \frac{U – Ri}{L} \\ \implies \frac{L}{U – Ri}\dd i &= \dd t \\ \implies \int \frac{L}{U – Ri}\,\dd i &= \int \dd t + C\\ \end{align*}

Substitute $$v\equiv U-Ri$$ $$\implies$$ $$\dv{v}{i}=-R$$ $$\implies$$ $$\dd i = -\frac{\dd v}{R}$$ \newcommand{dv}{\frac{\dd #1}{\dd #2}} \begin{align*} \int \frac{L}{\underline{v}} \left(\underline{-\frac{\dd v}{R}}\right) &= \int \dd t + C \\ \implies -\frac{L}{R} \int \frac{1}{v}\,\dd v &= \int \dd t + C \\ \implies -\frac{L}{R} \ln v &= t + C \\ \implies \ln v &= -\frac{R}{L} t + C_2 & \text{(subst. back)} \\ \implies \ln U-Ri &= -\frac{R}{L} t + C_2 \\ \implies U-Ri &= C_3\,\rm{e}^{-\frac{R}{L} t} \\ \implies Ri &= U – C_3\,\rm{e}^{-\frac{R}{L} t} \\ \implies i &= \frac{U}{R} – C_4\,\rm{e}^{-\frac{R}{L} t}, & \left(i(0)=0\right) \\ &= \frac{U}{R} \left( 1 – \rm{e}^{-\frac{R}{L} t} \right) \end{align*}

#### Solve using Laplace transform

Details

Use a Laplace transform to solve this differential equation \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align*} L\frac{di}{dt} + Ri &= \gamma(t)\,U & \left(\laplace\right) \\ \implies \mathfrak{L} \left\{ L\frac{d\,i(t)}{dt} + R\,i(t) \right\} &= \mathfrak{L} \left\{ \gamma(t)\,U \right\} \\ \implies L\left(sI(s) – i(0^+)\right) + R\,I(s) &= \frac{U}{s} & \left(i(0^+)=0 \right) \\ \implies I(s) &= \frac{U}{s(Ls + R)} = \frac{U}{L}\frac{1}{s\left(s + \frac{R}{L}\right)} \\ &= K\frac{1}{s\left(s – p\right)}, \quad K=\frac{U}{L},\quad p=-\frac{R}{L} \\ \end{align*}

Split up this complicated fraction into forms that are in the Laplace Transform table. According to Heaviside, this can be expressed as partial fractions. [swarthmore, MIT-cu] $$I(s) = K\frac{1}{s\left(s – p\right)} \equiv\frac{c_0}{s}+\frac{c_1}{s-p} \label{eq:heavi}$$

Multiply with respectively $$s$$ and $$(s-p)$$ \left\{ \begin{align*} K\frac{\cancel{s}}{\cancel{s}\left(s – p\right)} &\equiv \frac{\cancel{s}\,c_0}{\cancel{s}}+\frac{s\,c_1}{s-p} \\ K\frac{\cancel{s-p}}{s\cancel{(s-p)}} &\equiv \frac{c_0(s-p)}{s}+\frac{c_1\cancel{(s-p)}}{\cancel{s-p}}\\ \end{align*} \right.

Given that these equations are true for any value of $$s$$, choose a convenient value that help us find $$c_0$$, and substitute $$K$$ and $$p$$ back \left\{ \begin{align*} c_0 &= \left.\frac{K-sc_1}{s-p}\right|_{s=0} = -\frac{K}{p} = \frac{U}{R} \\ c_1 &= \left.\frac{K – (s-p)c_0}{s}\right|_{s=p} = -\frac{U}{R} \\ \end{align*} \right.

Transfer equation $$\eqref{eq:heavi}$$ back to the time domain, and fill in the values for $$c_0$$, $$c_1$$ and $$p$$ \begin{align*} i(t) &= \mathcal{L}^{-1} \left\{ \frac{c_0}{s} \right\} + \mathcal{L}^{-1} \left\{ \frac{c_1}{s-p} \right\}, & t\geq 0 \\ &= c_0 + c_1\,\rm{e}^{pt} \\ &= \frac{U}{R} – \frac{U}{R}\rm{e}^{-\frac{R}{L}t} \end{align*}

#### Solution

The current in the RL circuit as a result of the step-function $$\shaded{ i(t) = i_{max}\left( 1 – e^{-\frac{R}{L}t} \right) }, \quad i_{max} = \frac{U}{R} \tag{$$LR$$ battery} \label{eq:lron}$$

Plotting $$i(t)$$

### Magnetic field supplies energy

All of a sudden, we’re making the voltage difference zero by flipping the switch to position $$a$$. The current is still running, and at a new time $$t_0$$, the voltage difference, $$U$$, is zero.

The self-inductance is going to fight the change of the current going down. You expect that the current will not die off right away.

The time behavior can be calculated by going back to our differential equation and making $$U$$ zero. \begin{align*} L\frac{di}{dt} + Ri &= 0 \end{align*}

#### Solve using the antiderivative

Details

\newcommand{dv}{\frac{\dd #1}{\dd #2}} \begin{align*} L\dv{i}{t} + Ri &= 0 \\ \implies \dv{i}{t} &= -\frac{Ri}{L} \\ \implies \frac{1}{i}\dd i &= -\frac{R}{L}\dd t \\ \implies \int \frac{1}{i}\dd i &= -\frac{R}{L} \int \dd t + C\\ \implies \ln i &= -\frac{R}{L}t + C \\ \implies i &= C\,\rm{e}^{-\frac{R}{L}t}, & \left( i(\infty)=\tfrac{U}{R}\right) \\ &= \frac{U}{R}\,\rm{e}^{-\frac{R}{L}t} \end{align*}

#### Solve using Laplace transform

Details

Use a Laplace transform to solve this differential equation \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align*} L\frac{di}{dt} + Ri &= 0 & \left(\laplace\right) \\ \implies \mathfrak{L} \left\{ L\frac{d\,i(t)}{dt} + R\,i(t) \right\} &= \mathfrak{L} \left\{ 0 \right\} \\ \implies L\left(sI(s) – i(0^+)\right) + R\,I(s) &= 0 & \left( i(0^+)=\frac{U}{R} \right) \\ \implies L\left(sI(s) – \frac{U}{R}\right) + R\,I(s) &= 0 \\ \implies sLI(s) + R\,I(s) &= \frac{UL}{R} \\ \implies I(s) &= \frac{UL}{R} \frac{1}{sL + R} = \frac{U}{R}\frac{1}{s + \frac{R}{L}} \\ &= K\frac{1}{s – p}, \quad K=\frac{U}{R},\quad p=-\frac{R}{L} \\ \end{align*}

This is already in a form that we can find in the Laplace Transform table. Transfer this equation back to the time domain, and fill in the value for $$p$$ \begin{align*} i(t) &= \mathcal{L}^{-1} \left\{ K\frac{1}{s-p} \right\}, & t\geq 0 \\ &= K\,\rm{e}^{pt} \\ &= \frac{U}{R}\rm{e}^{-\frac{R}{L}t} \end{align*}

#### Solution

The current in the RL circuit as a result of the step-function $$\shaded{ i(t) = i_{max}\,e^{-\frac{R}{L}t} }, \quad i_{max} = \frac{U}{R} \tag{$$LR$$ magnetic field} \label{eq:lroff}$$

Plotting $$i(t)$$

Note

• For $$t=0$$, we find $$i(t)=U/R$$
• For $${t\to\infty}$$, we find $$i(t)\to 0$$
• For $$t=\frac{L}{R}$$ seconds, then $$i(t)\approx 0.37\, i_{max}$$

This is all the consequence of the circuit is capable of fighting the change its own magnetic flux that it is creating.

With the battery still connected, the battery was providing the heat that was dissipated in the resistor. Now, with no battery, there is still current. While the current is dying, there is still heat produced in that resistor. That energy must come from the magnetic field in the solenoid.

The idea that energy comes out as heat, that was there earlier in the form of a magnetic field, allows us to evaluate what we call the magnetic energy field density.

### RL Circuit and AC (frequency response)

This time, let’s look at a circuit with an AC power source, and a self-inductor in series with a resistor.

The AC power source has angular frequency $$\omega$$ $$U = U_0 \sin\omega t$$

Applying Faraday’s law is the same as before, except that $$U = U_0 \cos\omega t$$ $$L\frac{di}{dt} + i\, R = U_0 \sin\omega t$$

### Solve using Fourier Series

Details

Solve using Fourier Series $$L \frac{di}{dt} = Ri = U_0\sin\omega t \label{eq:dv}$$

Let $$i(t)$$ \begin{align} i(t) &= \sum_{n=\infty}^\infty c_n\, e^{\rm{j}nt} \label{eq:i} \\ \implies \frac{di}{dt} &= \sum_{n=\infty}^\infty \rm{j}n\,c_n\, e^{\rm{j}nt} \nonumber \end{align}

Substitute $$i(t)$$ and $$\frac{di}{dt}$$ in the differential equation $$\eqref{eq:dv}$$, and express the sine term using Euler’s formula \begin{align*} L \sum_{n=\infty}^\infty jn\,c_n\, e^{jnt} + R\sum_{n=\infty}^\infty c_n\, e^{jnt} &= U_0\left( \frac{\mathrm{e}^{j\omega t} – \mathrm{e}^{-j\omega t}}{2j} \right) \\ \implies \sum_{n=\infty}^\infty c_n\, e^{\rm{j}nt} \left( R + \rm{j}nL \right) &= -j\,\tfrac{U_0}{2}\,\mathrm{e}^{j\omega t} + j\,\tfrac{U_0}{2}\,\mathrm{e}^{-\rm{j}\omega t} & \text{(only $$n=\omega$$ matches)} \\ \end{align*}

The only values for $$n$$ that fit are $$\omega$$ and $$-\omega$$ \left\{ \begin{align*} c_\omega\, e^{j\omega t} ( R + j\omega L) &= -j\tfrac{U_0}{2}\,\mathrm{e}^{j\omega t} \\ c_{-\omega}\, e^{j\omega t} ( R + j\omega L) &= j\tfrac{U_0}{2}\,\mathrm{e}^{-j\omega t} \\ \end{align*} \right.

Now, we need to find the coefficients $$c_\omega$$ and $$c_{-\omega}$$. Let $$c_\omega=a_\omega + j\,b_\omega$$ \begin{align*} (a_\omega + j\,b_\omega)( R + j\omega L) &= j\tfrac{U_0}{2} \\ \implies (a_\omega R – b_\omega\omega L) + j(a_\omega\omega L + b_\omega R) &= j\tfrac{U_0}{2} \end{align*}

The real and imaginary parts should match \begin{align} a_\omega R – b_\omega\omega L &= 0 \implies a_\omega = b_\omega\omega \frac{L}{R} \label{eq:re}\\ a_\omega\omega L + b_\omega R &= \frac{U_0}{2} \label{eq:im} \end{align}

To find $$b_\omega$$, substitute $$a_\omega$$ from equation $$\eqref{eq:re}$$ in $$\eqref{eq:im}$$ \begin{align} b_\omega\omega \tfrac{L}{R} \omega L + b_\omega R &= \tfrac{U_0}{2} \nonumber \\ \implies b_\omega(\omega^2 \tfrac{L}{R} L + R) &= \tfrac{U_0}{2} \nonumber \\ \implies b_\omega &= \tfrac{U_0R}{2(\omega^2 L^2 + R^2)} \label{eq:bw} \\ \end{align}

To find $$a_\omega$$, substitute $$b_\omega$$ from equation $$\eqref{eq:bw}$$ in $$\eqref{eq:re}$$ \begin{align*} a_\omega &= \frac{U_0R}{2(\omega^2 L^2 + R^2)} \omega\frac{L}{R} \\ &= \frac{U_0 L \omega}{2(\omega^2 L^2 + R^2)} \\ \end{align*}

Putting the real an imaginary parts back together, $$c_\omega=a_\omega + j\,b_\omega$$ \begin{align*} c_\omega &= \frac{U_0\omega L}{2(\omega^2 L^2 + R^2)} + j\left(\frac{U_0R}{2(\omega^2 L^2 + R^2)}\right) \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega + jR) \\ \end{align*}

Since $$i(t)$$ is real, $$c_{-\omega} = c_\omega^* = a_\omega – j b_\omega$$ \begin{align*} c_{-\omega} &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega – jR) \end{align*}

Subst. $$c_\omega$$ and $$c_{-\omega}$$ back in equation $$\eqref{eq:i}$$ \begin{align} i(t) &= c_\omega\, \rm{e}^{\rm{j}\omega t} + c_{-\omega}, \rm{e}^{-\rm{j}\omega t} \nonumber \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega + jR)\, \rm{e}^{\rm{j}\omega t} + \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega – jR), \rm{e}^{-\rm{j}\omega t} \nonumber \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}\, \left( (L\omega + jR)\,\rm{e}^{\rm{j}\omega t} + (L\omega – jR)\,\rm{e}^{\rm{j}\omega t} \right) \nonumber \\ &= \frac{U_0}{\omega^2 L^2 + R^2}\, \left( \omega L \frac{\rm{e}^{\rm{j}\omega t} + \rm{e}^{-\rm{j}\omega t}}{2} – R \frac{\rm{e}^{\rm{j}\omega t} – \rm{e}^{-\rm{j}\omega t}}{2} \right) \nonumber \\ &= \frac{U_0}{\omega^2 L^2 + R^2}\, \left( \underline{\omega L \cos\omega t – R\sin\omega t} \right) \label{eq:i2} \end{align}

Examine the term: $$\omega L \cos\omega t – R\sin\omega t$$, using the Sinusoidal identity \begin{align} \omega L \cos\omega t – R\sin\omega t &= A\sin(\omega t – \phi) \label{eq:i3} \\ &= A\sin(\omega t) \cos(-\phi) + A\cos(\omega t) \sin(-\phi) \nonumber \end{align}

Sine and cosine terms should match \left\{ \begin{align*} \omega L \cancel{\cos(\omega t)} &= A\cancel{cos(\omega t)} \sin(-\phi) \\ – R\cancel{\sin(\omega t)} &= A\cancel{\sin(\omega t)} \cos(-\phi) \end{align*} \right.

Angle of $$\phi$$, and modules $$A$$ \left\{ \begin{align*} \tan\phi &= -\tfrac{\sin(-\phi)}{\cos(-\phi)} = \tfrac{\omega L}{R} \implies \underline{\phi = \atan\tfrac{\omega L}{R}} \\ A^2\left( \sin(-\phi)^2+\cos(-\phi)\right) &= (\omega L)^2 + (-R)^2 \implies \underline{A = \sqrt{\omega^2 L^2 + R^2}} \end{align*} \right.

Substituting $$A$$ back in equation $$\eqref{eq:i3}$$ \begin{align*} \omega L \cos\omega t – R\sin\omega t &= A\sin(\omega t – \phi) \\ &= \sqrt{\omega^2 L^2 + R^2}\,\sin(\omega t – \phi) \end{align*}

Substituting this term back in $$\eqref{eq:i2}$$ \begin{align*} i(t) &= \frac{U_0}{\omega^2 L^2 + R^2}\,\sqrt{\omega^2 L^2 + R^2}\,\sin(\omega t – \phi) \\ &= \frac{U_0}{\sqrt{\omega^2 L^2 + R^2}}\,\sin(\omega t – \phi),\quad \phi = \atan\tfrac{\omega L}{R} \end{align*}

### Solve using Laplace transform

Details

Let’s approach this as a LR-filter, where output $$y(t)$$ depends on input $$u(t)$$. The transfer function $$H(s) = \frac{Y(s)}{U(s)} = \frac{R}{sL + R} = \frac{R}{L} \frac{1}{s+\frac{R}{L}} \nonumber$$

The denominator is a first-order polynomial. The root is called the system’s pole $$\shaded{ H(s) = K \frac{1}{1-p} },\quad K=\frac{R}{L},\quad p=-\frac{R}{L} \nonumber$$

The frequency response $$y_{ss}(t)$$ is defined as the steady state response to a sinusoidal input signal $$u(t)=\sin(\omega t)γ(t)$$. In Evaluating Transfer Functions, we have proven that

$$y_{ss}(t)=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t) \nonumber$$

Based on Euler’s formula, we can express $$H(s)$$ in polar coordinates \left\{ \begin{align} H(s) &= |H(s)|\ e^{j\angle{H(s)}}\nonumber \\ |H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|} =K \frac{1}{\left|s-p\right|}\nonumber \\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) =-\angle(s-p)\nonumber \end{align} \right.

The transfer function with pole $$p$$, evaluated for $$s=j\omega$$ can be visualized with a vector from the pole to $$j\omega$$.

The length of the vector corresponds to $$|(H(j\omega)|$$, and minus the angle with the real axis corresponds to phase shift $$\angle H(j\omega)$$. \left\{ \begin{align} |H(j\omega)| &=K \frac{1}{\left|j\omega-p\right|}= K \frac{1}{\sqrt{\omega^2+p^2}}\nonumber \\ \angle{H(j\omega)}&=-\angle(j\omega-p)=\mathrm{atan2}(\omega,-p)\nonumber\\ &=-\arctan\frac{\omega}{-p},\ p\lt0\land p\in\mathbb{R}\nonumber \end{align} \right. \label{eq:polar1}

Substitute $$K=\frac{R}{L}$$ and $$p=-\frac{R}{L}$$ \left\{ \begin{align*} |H(j\omega)| &=\frac{R}{L} \frac{1}{\left|j\omega-p\right|} = \frac{R}{\sqrt{(\omega L)^2+R^2}} \\ \angle{H(j\omega)} &= -\angle\left(j\omega+\frac{R}{L}\right) = -\arctan\left(\frac{\omega L}{R}\right) \end{align*} \right.

The output signal $$y_{ss}$$ \begin{align*} y_{ss}(t) &= \frac{U_0 R}{\sqrt{(\omega L)^2+R^2}}\,\sin(\omega t – \phi)\,\gamma(t)\nonumber \\ \phi &= \arctan\left(\frac{\omega L}{R}\right) \end{align*}

#### Solution

The steady state current $$i_{ss}=y_{ss}/R$$ for a sinusoidal input signal $$U_0\sin(\omega t)γ(t)$$ $$\shaded{ i_{ss}(t) = \frac{U_0}{\sqrt{(\omega L)^2+R^2}}\,\sin(\omega t – \phi) },\quad \tan\phi = \frac{\omega L}{R} \tag{Current AC} \label{eq:currentac}$$

The term in front of the cosine, is the maximum possible current. The cosine term is just oscillating between $$+1$$ and $$-1$$.

• The $$\omega L$$ has the role of resistance. Its dimension is $$\Omega$$. If $$\omega$$ is very high, then the resistance becomes very high, so the current becomes very low. If $$\omega$$ is high, then the changes $$di/dt$$ are very high. The induced EMF is going to be high, so the current will be low.
• Also, if $$L$$ is very high, then the system is capable of fighting back very hard, so it puts up a large resistance.
• If $$\omega$$ is very low, say zero, you simply get $$i(t)=U_0/R$$

The current comes later than the driving voltage, because the self-inductance is fighting the change in current. It is not so surprising that there’s a delay. Looking at the phase shift $$\tan\phi=\omega L/R$$

• If the self-inductance is very large, the system has a strong ability to fight back. It can delay that current by a large amount.
• Same when $$\omega$$ is high.

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