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Voltage resonance and Thomson’s formula in alternating current. Examples

current

Voltage resonance and Thomson’s formula are terms related to alternating current. Voltage resonance is only possible in the case of a series connection of a thermogenic resistor, induced thread, and capacitor. Thomson’s formula gives a connection between the circular frequencies of the natural oscillations of a simple oscillator circuit without losses and the parameters of the L-C circuit. Voltage resonance and Thomson’s formula occurs at the RLC series connection under the condition that the reactive voltages are equal. In other words, when the reactive resistances are equal. The apparent resistance of the circuit is the smallest, the current is the largest, and it is in phase with the voltage.

If the above three elements are connected to the AC voltage in series, then alternating current will flow through the circuit i={{I}_{m}}\sin \omega t Then the voltage at the ends of each of these three elements will be:

{{u}_{R}}={{U}_{Rm}}\sin \omega t
{{u}_{L}}={{U}_{Lm}}\sin \left( \omega t+\frac{\pi }{2} \right)
{{u}_{C}}={{U}_{cm}}\sin \left( \omega t-\frac{\pi }{2} \right)

And their effective values will be:
{{U}_{R}}=RI
{{U}_{L}}={{X}_{L}}I
{{U}_{c}}={{X}_{cL}}I=~\frac{I}{\omega C}

RLC circuit

By applying the Second Kirchhoff’s law for this circuit, we have:
{{\underline{u}}}={{\underline{u}}_{R}}+{{\underline{u}}_{L}}+{{\underline{u}}_{C}}
That is, for effective values:
{{\underline{u}}}={{\underline{U}}_{R}}+{{\underline{U}}_{L}}+{{\underline{U}}_{C}}

The corresponding phasor diagram
The corresponding phasor diagram

The intensity of the total voltage phasor is calculated using the Pythagorean theorem:
U=\sqrt{U_{R}^{2}+{{({{U}_{L}}-{{U}_{C}})}^{2}}}

the Pythagorean theorem

In the phase diagram for this circuit, the current phasor is shifted backward by the voltage angle phi if the inductive character of the circuit predominates. If the reactive character of the circuit predominates, the current phasor is moved backward by an angle \frac{\pi }{2} from the voltage phasor. The common voltage has two components: active {{U}_{R}}=I\cdot R and reactive \mathop{U}_{X}=XI=\left( \mathop{X}_{L}-\mathop{X}_{C} \right)I

Where, based on Ohm’s law:
{{U}_{R}}=I\cdot R
{{U}_{L}}=I\cdot {{X}_{L}}
{{U}_{C}}=I\cdot {{X}_{C}}
On the other hand, the intensity of the total voltage phasor is: U=I\cdot Z
where Z is the total resistance in the AC circuit (impedance):

Z=\sqrt{\mathop{R}^{2}+\mathop{X}^{2}}=\sqrt{{{R}^{2}}+{{\left( L\omega -\frac{1}{C\omega } \right)}^{2}}}

Where X=L\omega -\frac{1}{C\omega } represents the reactive resistance of the circuit. If it is \omega L<\frac{1}{\omega C} then then the current will precede the voltage. But if it is \omega L>\frac{1}{\omega C} \\ then the current will lag behind the voltage. Connected elements, in this example, according to Ohm’s law, have the effective value of current:

I=\frac{U}{\sqrt{{{R}^{2}}+{{\left( L\omega -\frac{1}{C\omega } \right)}^{2}}}}
phase difference
\cos \varphi =\frac{R}{Z}=\frac{R}{\sqrt{\mathop{R}^{2}+\mathop{\left( \frac{\omega L-1}{\omega C} \right)}^{2}}}
alternating voltage
u=\mathop{U}_{m}\sin \left( \omega t+\varphi  \right)

Therefore, the current is in phase with the voltage if it is L\omega =\frac{1}{C\omega } and the reactive resistance is then
X=L\omega -\frac{1}{C\omega }=0 so the apparent resistance of the circuit is Z=\sqrt{{{R}^{2}}+{{\left( \frac{\omega L-1}{C\omega } \right)}^{2}}}=\sqrt{\mathop{R}^{2}}=R

Then the impedance has the smallest value, Z = R. The current is then highest and is in phase with the voltage. Then, a voltage resonance occurred in the circuit.
The apparent resistance then has the lowest possible value since Z> R is still. The current then get the highest value and is in phase with the voltage \cos \varphi =\frac{R}{Z}=1 and depends only on the resistance R. The phasor diagram took the form as in the previous two figures. The voltages at the ends of the capacitor and the inductive thread are of equal intensity but in the opposite direction: there is a voltage resonance in the circuit. Formula X=L\omega -\frac{1}{C\omega }=0 is a condition for the appearance of voltage resonance in the circuit that can be achieved in three ways:

  1. ω and L are constant and C has a value \mathop{L}_{0}=\frac{1}{\mathop{\omega }^{2}C} \\
  2. ω and C are constant and L has value \mathop{C}_{0}=\frac{1}{\mathop{\omega }^{2}L} \\
  3. L and C are constant and ω has value \mathop{\omega }_{0}=\frac{1}{\sqrt{LC}}

    Respectively {{f}_{0}}=\frac{1}{2\pi \sqrt{LC}} It is the resonant frequency of oscillations, whose period is T=\frac{1}{\mathop{f}_{0}}=2\pi \sqrt{LC} This is Thomson’s formula. Voltage resonance and Thomson’s formula have a great application in low current techniques. The voltage resonance and Thomson’s formula provide a connection between the circular frequencies of the natural oscillations of a simple oscillator circuit without losses and the parameters of the circuit LC.

Examples

Problem no. 1

A series RLC circuit has the following data:
L= 5H
C= 200mF
a) Calculate the voltage resonance and Thomson’s formula.
b) Calculate Lo and Co, which occur at a frequency of 50 Hz

a) \mathop{f}_{0}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{5H\cdot 200\cdot \mathop{10}^{-3}F}}\left( Hz \right)
b) \mathop{L}_{0}=\frac{1}{\mathop{\mathop{\omega }_{0}}^{2}C}=\frac{1}{\mathop{\left( 2\pi 50 \right)}^{2}\cdot 200\cdot \mathop{10}^{-3}}\left( H \right) \\
\mathop{C}_{0}=\frac{1}{\mathop{\mathop{\omega }_{0}}^{2}L}=\frac{1}{\mathop{\left( 2\pi 50 \right)}^{2}\cdot 5}\left( mF \right) \\

Problem no. 2

Problem no. 2. A series RLC circuit has the following data:
R= 55Ω
L= 4H
U= 60 V
Voltage resonance and Thomson’s formula performed in the circuit frequency of 100Hz. Calculate Co, Io.

\mathop{C}_{0}=\frac{1}{\mathop{\mathop{\omega }_{0}}^{2}L}=\frac{1}{\mathop{\left( 2\pi \mathop{f}_{0} \right)}^{2}\cdot L}=\frac{1}{\mathop{\left( 2\pi 100 \right)}^{2}\cdot 4}\left( mF \right) \\
\mathop{I}_{0}=\frac{U}{R}=\frac{60V}{55\Omega }

Summary | Voltage resonance and Thomson’s formula

During the day, individual parts in the alternating current network are constantly switched, on and off, so the total inductance and capacitance change. Therefore, the inductive and capacitive resistance may be equalized, so a voltage resonance occurs, which can cause disturbances in the network (an increase of current and voltage in certain parts of the system). We conclude, based on Voltage resonance and Thomson’s formula, that the occurrence of voltage resonance is undesirable in the technique of strong current (energy). However, in the low current (electronics) technique, voltage resonance is very often used to amplify signals. Follow more on Instagram.

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