Focus here: uncoupled inductors!
Based on $L = {{ \Psi(t)}\over{i}}$ and Kirchhoff's mesh law ($i=\rm const$) the series circuit of inductions can be interpreted as a single current $i$ which generates multiple linked fluxes $\Psi$. Since the current must stay constant in the series circuit, the following applies for the equivalent inductor of a series connection of single ones:
\begin{align*} L_{\rm eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}
A similar result can be derived from the induced voltage $u_{ind}= L {{{\rm d}i}\over{{\rm d}t}}$, when taking the situation of a series circuit (i.e. $i_1 = i_2 = i_1 = ... = i_{\rm eq}$ and $u_{\rm eq}= u_1 + u_2 + ...$):
\begin{align*} & u_{\rm eq} & = &u_1 & + &u_2 &+ ... \\ & L_{\rm eq} {{{\rm d}i_{\rm eq} }\over{{\rm d}t}} & = &L_{1} {{{\rm d}i_{1} }\over{{\rm d}t}} & + &L_{2} {{di_{2} }\over{dt}} &+ ... \\ & L_{\rm eq} {{{\rm d}i }\over{{\rm d}t}} & = &L_{1} {{{\rm d}i }\over{{\rm d}t}} & + &L_{2} {{di }\over{dt}} &+ ... \\ & L_{\rm eq} & = &L_{1} & + &L_{2} &+ ... \\ \end{align*}
For parallel circuits, one can also start with the principles based on Kirchhoff's mesh law:
\begin{align*} u_{\rm eq}= u_1 = u_2 = ... \\ \end{align*}
and Kirchhoff's nodal law:
\begin{align*} i_{\rm eq}= i_1 + i_2 + ... \\ \end{align*}
Here, the formula for the induced voltage has to be rearranged:
\begin{align*} u_{\rm ind} &= L {{{\rm d}i}\over{{\rm d}t}} \quad \quad \quad \quad \bigg| \int(){\rm d}t \\ \int u_{\rm ind} {\rm d}t &= L \cdot i \\ i &= {{1}\over{L}} \cdot \int u_{\rm ind} {\rm d}t \\ \end{align*}
By this, we get:
\begin{align*} i_{\rm eq} &=& i_1 &+& i_2 &+& ... \\ {{1}\over{L_{\rm eq}}} \cdot \int u_{\rm eq} {\rm d}t &=& {{1}\over{L_1}} \cdot \int u_{1} {\rm d}t &+& {{1}\over{L_2}} \cdot \int u_{2} {\rm d}t &+& ... \\ {{1}\over{L_{\rm eq}}} \cdot \int u {\rm d}t &=& {{1}\over{L_1}} \cdot \int u {\rm d}t &+& {{1}\over{L_2}} \cdot \int u {\rm d}t &+& ... \\ {{1}\over{L_{\rm eq}}} &=& {{1}\over{L_1}} &+& {{1}\over{L_2}} &+& ... \\ \end{align*}
For AC circuits (i.e. with sinusoidal signals) the impedance $Z$ based on the real part $R$ and imaginary part $X$ has to be considered. To do so, one has to solve:
\begin{align*} \underline{Z} = {{\underline{u}}\over{\underline{i}}} = {{1}\over{\underline{i}}} \cdot \underline{u} \end{align*}
With the induction $u_{\rm ind}= L {{{\rm d}i}\over{{\rm d}t}}$ we get:
\begin{align*} \underline{Z} &= {{1} \over {\underline{i}}} \cdot L { { {\rm d}\underline{i} } \over {{\rm d}t} } \\ \end{align*}
Without limiting the generality, one can assume the current $i$ to be: $i = I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}$.
Once inserted, the formula gets:
\begin{align*} \underline{Z} &= {{1} \over {I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L {{ {\rm d}} \over {{\rm d}t} } \left( I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right) \\ &= {{1} \over {I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L I \cdot \sqrt{2} \cdot {{ {\rm d}} \over {{\rm d}t} } \left( {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right) \\ &= {{1} \over {\qquad\quad\; {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L \qquad\; \cdot {{ {\rm d}} \over {{\rm d}t} } \left( {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right) \\ &= {{1} \over {\qquad\quad\; {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L \qquad\; \cdot j\omega \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \\ \\ \underline{Z} &= L \cdot {\rm j}\omega \\ \end{align*}
Charging and discharging an $RL$ circuit is comparable to the RC-circuit in chapter DC Circuit Transients from last semester. Details are not covered here; see OpenStax
Similar to last semester's approach, we now focus on circuits with inductors $L$. For preparation, please recap the chapter Circuits under different Frequencies from last semester.
As seen last semester, the circuits with complex impedances can be interpreted as four-terminal networks. There, we will again look at „output versus input“, i.e: $A_V = {{U_O}\over{U_I}} \rightarrow \underline{A}_V = {{\underline{U}_O}\over{\underline{U}_I}} $.
In this chapter, we look at a combination where all three components resistor $R$, capacitor $C$, and inductance $L$ are used.
If a resistor $R$, a capacitor $C$, and an inductance $L$ are connected in series, the result is a series resonant circuit. In this case, it is not clearly defined, what the output voltage is. Consequently, it must be considered how the voltages behave across all the individual components in the following. The total voltage (= input voltage $U_I$) results to:
\begin{align*} \underline{U}_I = \underline{U}_R + \underline{U}_L + \underline{U}_C \end{align*}
Since the current in the circuit must be constant, the total impedance can be determined here in a simple way:
\begin{align*} \underline{U}_I &= R \cdot \underline{I} + {\rm j} \omega L \cdot \underline{I} + \frac {1}{j\omega C } \cdot \underline{I} \\ \underline{U}_I &= \left( R + {\rm j} \omega L - {\rm j} \cdot \frac {1}{ \omega C } \right) \cdot \underline{I} \\ \underline{Z}_{\rm eq} &= R + {\rm j} \omega L - {\rm j} \cdot \frac {1}{ \omega C } \end{align*}
By this, the magnitude of the (input) voltage $U_I$, the (input or total) impedance $Z$, and the phase result to:
\begin{align*} U_I &= \sqrt{U_R^2 + (U_Z )^2} = \sqrt{U_R^2 + (U_L - U_C)^2} \end{align*}
\begin{align*} Z &= \sqrt{R^2 + X^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \end{align*}
\begin{align*} \varphi_u = \varphi_Z &= \arctan \frac{\omega L - \frac{1}{\omega C}}{R} \end{align*}
There are now 3 different situations to distinguish:
Again, there seems to be a singular frequency, namely when $U_L = U_C$ or $Z_L = Z_C$ holds:
\begin{align*} \frac{1}{\omega_0 C} & = \omega_0 L \\ \omega_0 & = \frac{1}{ \sqrt{LC}} \\ 2\pi f_0 & = \frac{1}{ \sqrt{LC}} \rightarrow \boxed{ f_0 = \frac{1}{2\pi \sqrt{LC}} } \end{align*}
The frequency $f_0$ is called resonance frequency.
$\quad$ | $f \rightarrow 0$ | $\quad$ | $f = f_0$ | $\quad$ | $f \rightarrow \infty$ | |
---|---|---|---|---|---|---|
voltage $U_R$ at the resistor | $\boldsymbol{\small{0}}$ | $\boldsymbol{\LARGE{U}}$ since the impedances just cancel out | $ \boldsymbol{\small{0}}$ | |||
voltage $U_L$ at the inductor | $\boldsymbol{\small{0}}$ because $\omega L$ becomes very small | $\boldsymbol{\omega_0 L \cdot I = \omega_0 L \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$ | $\boldsymbol{\LARGE{U}}$ since $\omega L$ becomes very large |
|||
voltage $U_C$ at the capacitor | $\boldsymbol{\LARGE{U}}$ because $\frac{1}{\omega C}$ becomes very large | $\boldsymbol{\frac{1}{\omega_0 C} \cdot I = \frac{1}{\omega_0 C} \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$ | $\boldsymbol{\small{0}}$ because $\frac{1}{\omega C}$ becomes very small |
The calculation in the table shows that in the resonance case, the voltage across the capacitor or inductor deviates from the input voltage by a factor $\color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}}$. This quantity is called quality or Q-factor $Q_{\rm S}$:
\begin{align*} \boxed{ Q_{\rm S} = \left.\frac{U_C}{U} \right\vert_{\omega = \omega_0} = \left.\frac{U_L}{U} \right\vert_{\omega = \omega_0} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}} } \end{align*}
The quality can be greater than, less than, or equal to 1. The quality $Q_{\rm S}$ does not have a unit and should not be confused with the charge $Q$.
The reciprocal of the $Q$ is called attenuation $d_{\rm S}$. This is specified when using the circuit as a non-overshooting filter.
\begin{align*} \boxed{ d_{\rm S} = \frac{1}{Q_{\rm S}} = R \sqrt{\frac{C}{L}} } \end{align*}
Simulation in Falstad. Note: The simulation gives a highly simplified picture. The response of the microcontroller is shown reduced to a triangular signal, since the slope of the voltages cannot be represented. A real simulation requires a powerful SPICE program in which the conduction theory can be represented.
Further details can be found here (practice), here (layout), also Layout or Layout.
Simulation in Time Domain
Simulation in Frequency Domain
A $R$-$L$-$C$ series circuit uses a capacity of $C=100 ~\rm µF$. A voltage source with $U_I$ feeds the circuit at $f_1 = 50~\rm Hz$.
A given $R$-$L$-$C$ series circuit is fed with a frequency, $20~\%$ larger than the resonance frequency keeping the amplitude of the input voltage constant. In this situation, the circuit shows a current $30~\%$ lower than the maximum current value.
Calculate the Quality $Q = {{1}\over{R}}\sqrt{{{L}\over{C}}}$.
However, it is possible and there are multiple ways to solve it.
What we know
But first, add some more info, which is always true from resonant circuits at the resonant frequency:
From the task, the following is also known.
Solution 2: The fast path
We start with $Z = \sqrt{R^2 + (X_L + X_C)^2}$ for the cases: (1) at the resonant frequency $f_0$ and (2) at the given frequency $f = 1.2 \cdot f_0 $
\begin{align*} (1): && Z_0 &= R \\ (2): && Z &= \sqrt{R^2 + (X_L + X_C)^2} \\ \end{align*}
In formula $(2)$ the impedance $X_L$ and $X_C$ are:
With $X_{C0} = X_{L0}$ we get for $(1)$:
\begin{align*} Z &= \sqrt{R^2 + \left(1.2\cdot X_{L0} - {{1}\over{1.2}} X_{L0} \right)^2} \\ &= \sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} \\ \end{align*}
Since we know that $Z = {{1}\over{0.7}} \cdot R$ and $Z_0 = R$, we can start by dividing $(2)$ by $(1)$:
\begin{align*} {{(2)}\over{(1)}} : && {{Z}\over{Z_0}} &= {{\sqrt{R^2 + (X_L + X_C)^2} }\over{R}} & &| \text{put in the info from before}\\ && {{1}\over{0.7}} &= {{\sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R}} & &| (...)^2 \\ && {{1}\over{0.7^2}} &= {{ {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R^2}} & &| \cdot R^2 \\ && {{1}\over{0.7^2}} \cdot R^2 &= {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} & &| -R^2 \\ && \left({{1}\over{0.7^2}} -1\right) \cdot R^2 &= { X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} & &| : R^2 \quad | : \left(1.2 - {{1}\over{1.2}} \right)^2 \\ && {{X_{L0}^2}\over{R^2}} &={ { {{1}\over{0.7^2}} -1 } \over { \left(1.2 - {{1}\over{1.2}} \right)^2 } } & &| \sqrt{...} \\ && {{X_{L0}}\over{R}} &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over { 1.2 - {{1}\over{1.2}} } } & &| \text{with } X_{L0} = \omega_0 \cdot L = {{1}\over{\sqrt{LC}}} \cdot L = \sqrt{ {L} \over {C} }\\ && {{1}\over{R}}\cdot \sqrt{ {L} \over {C} } &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over { 1.2 - {{1}\over{1.2}} } } \end{align*}
\begin{align*} Q &= {{ \sqrt{{{1}\over{0.7^2}} - 1} }\over{ 1.2 - {{1}\over{1.2}} }} \\ &= 2.782... \\ \rightarrow Q &= 2.78 \end{align*}