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RLC Series Circuit

Important Formula

Formula: Impedance (RLC Series Circuit)

$$|Z| ~=~ \sqrt{ R^2 ~+~ \left( \class{green}{\omega} \, \class{brown}{L} ~-~ \frac{1}{\class{green}{\omega} \, \class{purple}{C}} \right)^2 }$$ $$|Z| ~=~ \sqrt{ R^2 ~+~ \left( \class{green}{\omega} \, \class{brown}{L} ~-~ \frac{1}{\class{green}{\omega} \, \class{purple}{C}} \right)^2 }$$ $$R ~=~ \sqrt{ |Z|^2 ~-~ \left( \omega \, L ~-~ \frac{1}{\omega \, C} \right)^2 }$$

Rearrange formula

What do the formula symbols mean?

Impedance

$$ |Z| $$ Unit $$ \mathrm{\Omega} $$

Impedance is the magnitude of the complex resistance $ Z $. This formula gives the total impedance of a resistor-coil-capacitor series circuit (RLC series circuit). To obtain this formula, the Pythagorean theorem was applied to a phasor diagram on which the three quantities are plotted:

Ohmic impedance: $Z_{\text R } ~=~ R $.
Inductive impedance: $Z_{\text L } ~=~ \text{i} \, \omega \, L $.
Capacitive impedance: $Z_{\text C } ~=~ -\text{i}\frac{1}{\omega \, C} $.

Inductance

$$ L $$ Unit $$ \mathrm{H} = \frac{ \mathrm{Vs} }{ \mathrm{A} } $$

The inductance of the coil connected to the RLC circuit.

Capacitance

$$ C $$ Unit $$ \mathrm{F} = \frac{ \mathrm{C} }{ \mathrm{V} } $$

Capacitance of the capacitor connected to the RLC circuit.

Electrical Resistance

$$ R $$ Unit $$ \mathrm{\Omega} = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A}^2 \, \mathrm{s}^3 } $$

Ohmic resistance of a resistor connected to the RLC series circuit.

Angular frequency

$$ \class{green}{\omega} $$ Unit $$ \frac{\mathrm{rad}}{\mathrm s} $$

The angular frequency $ \omega = 2\pi \, f $ indicates the frequency $f$ at which the LRC circuit oscillates. The frequency $f$ is specified, for example, by the frequency of the applied AC voltage.

Table of contents

Important Formula
Exercises with Solutions

An RLC series circuit consists of a resistor $ R $, an inductor $ L $ and a capacitor $ C $, which are arranged in series in an electrical circuit. These components form a series of consecutive elements through which the current can flow.

RLC series circuit of resistor (R), coil (L) and capacitor (C)

$$ \begin{align} |Z| ~=~ \sqrt{ R^2 ~+~ \left( \class{green}{\omega} \, \class{brown}{L} ~-~ \frac{1}{\class{green}{\omega} \, \class{purple}{C}} \right)^2 } \end{align} $$

$$ \begin{align} \tan(\varphi) ~=~ \frac{ \class{green}{\omega} \, \class{brown}{L} ~-~ \frac{1}{\class{green}{\omega} \, \class{purple}{C}} }{ R } \end{align} $$

$$ \begin{align} I_0 ~=~ \frac{ U_0 }{ \sqrt{ R^2 ~+~ \left( \class{green}{\omega} \, \class{brown}{L} ~-~ \frac{1}{\class{green}{\omega} \, \class{purple}{C}} \right)^2 } } \end{align} $$

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Electric Spark Suppression with Capacitor (Snubber)

When the current flowing through the coil is switched off by a switch, a high voltage is momentarily generated at the switch, leading to an electric spark.

The coil has an inductance $L = 4.2 \, \text{H}$ and a current flows through it $I = 1 \, \text{A}$. To eliminate the arc at the switch, a capacitor is connected in parallel to the coil, which can withstand a maximum of $500 \, \text{V}$.

What capacity $C$ must the capacitor have for this?

Solution to Exercise #1

The induction voltage at the coil is given by: \[ U ~=~ L \, \frac{dI}{dt} \]

In this case, you can write: \[ U ~=~ L \, \frac{I}{\Delta t} \]

Capacitance of a capacitor is: \[ C ~=~ \frac{Q}{U} \] where charge is given by $ Q ~=~ I \, t $: \[ C ~=~ \frac{I\Delta t}{U} \]

Rearrange for time: \[ \Delta t ~=~ \frac{L\,I}{U} \]

Substitute into equation for $ C $: \[ C ~=~ \frac{I^2 L}{U^2} \]

Thus, the capacitance is: \[ C ~=~ 1.68 \,\cdot\, 10^{-5} \, \text{F} ~=~ 16.8 \, \mu\text{F} \]

Alternatively, the conservation of energy $\frac{1}{2}\,C\,U^2 = \frac{1}{2}\,L\,I^2$ can be utilized.