Tahalluci na LC: Masana da Kashiya, Tashin Aljuburi da Fankin Nemi

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فیلڈ: Karkashin Kuliya da Dukkana

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Zanen da LC Circuit?

LC circuit (ko kuma ake kira LC filter ko kuma LC network) yana nufin karkashin kuliya wanda ya shafi elemento mai kula an indaktar (L) da kapasitar (C) wadanda suka zama daga baya. Ana kiran haka a matsayin karkashin resonance, tank circuit, ko kuma tuned circuit.

Saboda rashin resistanci a cikin formar ideal na karkasha, LC circuit ba ta gudanar da energy ba. Wannan shine mafi kyau daga formar ideal na RC circuits, RL circuits, ko kuma RLC circuits, wadanda suke gudanar da energy saboda presence of a resistor.

Amma a cikin karkashin yawan tattalin arziki, LC circuit za a gudanar da abubuwa ne kafin da ita saboda resistance na biyu a cikin komponentoci da kable da ke sambata.

Miye da LC Circuit Ya Kiranta Tuned Circuit ko Tank Circuit?

Aiki ya haɗa daga kuma zuwa cikin capacitor da kuma inductor. Enerji na haɗa waɗanda da waɗanda hata da capacitor da inductor har zuwa lokacin da resistance na gida na component da wires yana ba haɗar da shi.

Aiki na wannan circuit ce mai kyau da aiki na harmonic oscillator, wanda ya fi dace da pendulum yana haɗa kai tsaye ko miji yana haɗa daga kuma zuwa cikin tank; saboda haka, ana kiranta tuned circuit ko tank circuit.

E circuit zai iya aiki a matsayin electrical resonator da ke taka enerji na haɗa waɗanda da waɗanda hata da frequency wanda ake kira resonant frequency.

Series LC Circuit

A cikin series LC circuit, inductor da capacitor suna haɗa daga kuma zuwa cikin series wanda ake nuna a cikin figure.

Saboda a cikin series circuit current ta fi dace daidai a cikin circuit, domin kuma flow of current ta fi dace da current through both the inductor and the capacitor.

$\begin{align*} i = i_L = i_C \end{align*}$

Yanzu aikin karamin da ke cikin terminali ya zama da sunan karamin da ke cikin kapasita da karamin da ke cikin induktar.

$\begin{align*} V = V_L + V_C \end{align*}$

Resonance in Series LC Circuit

Idan yawan ma'ana na fadada yake, tsari na induktivi reactance yana zama da shi.

$\begin{align*} X_L = \omega L = 2 \pi fL \end{align*}$

kuma tsari na kapasitivi reactance yana ƙarin.

$\begin{align*} X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} \end{align*}$

A nan da shi'arar tushen bayanai, yadda inductive reactance da capacitive reactance suna zama mafi yawan.

A nan an impedance na series LC circuit ta ba da

$\begin{align*} \begin{split} & Z_L_C_(_s_e_r_i_e_s_) = Z_L + Z_C\ &= j \omega L + \frac{1}{j \omega C}\ &= j \omega L + \frac{j}{j^2 \omega C}\ &= j \omega L - \frac{j}{\omega C}\ &= j (\frac{\omega^2 LC - 1}{\omega C}) (where, j^2 = -1)\ \end{split} \end{align*}$

A nan da shi'arar tushen bayanai, yadda inductive reactance da capacitive reactance suna zama mafi yawan.

$\begin{align*} \begin{split} & X_L = X_C\\ & \omega L = \frac{1}{\omega C}\\ & \omega^2 = \frac{1}{LC}\\ & \omega = \omega_0 = \frac{1}{\sqrt {LC}}(where, \omega = angular frequency)\\ & 2 \pi f =\omega_0 = \frac{1}{\sqrt {LC}}\\ & f_0 =\frac{\omega_0}{2\pi} = \frac{1}{2 \pi \sqrt {LC}}\\ \end{split} \end{align*}$

Idan, $\omega_0$ shi ne resonant angular frequency (radians per second).

Sau don haka, resonant angular frequency ta ce $\omega_0 = \frac{1}{\sqrt{LC}}$ , sannan impedance ta zama

(1) $\begin{equation*} Z_L_C(\omega)_(_s_e_r_i_e_s_) = j L (\frac {\omega^2 - \omega_0^2} {\omega}) \end{equation*}$

Saboda haka, a cikin lokacin da $\omega = \omega_0$ total electrical impedance Z ta zama zero, yana nufin XL da XC suna kawo suke. saboda haka, current ta da ita a series LC circuit ya zama maximum ( $I = \frac {V} {Z}$ ).

Saboda haka, series LC circuit, idan an haɗa shi a series da load, za a yi waɗanda band-pass filter wanda ya fi zero impedance a resonant frequency.

A cikin kisan da yanki na biyu a fadin daɗi, $f < f_0$ , $X_C >> X_L$ . Saboda haka, cikin daɗi ya fi mai yara.
A cikin kisan da yanki na biyu a fadin daɗi, $f>f_0$ , $X_L >> X_C$ . Saboda haka, cikin daɗi ya fi mai tashar.
A cikin kisan da yanki na biyu a fadin daɗi, $f = f_0$ , $X_L = X_C$ . Amfani ya fi mai tsawo da zama mai yawa daɗi. A wannan yanayi, cikin daɗi zai iya aiki a matsayin cikin daɗi mai karɓar daɗi.

Cikin Daɗi LC Tabbata

A cikin daɗi LC tabbata, inductor da capacitor suna ƙunshi a matsayin tabbata wanda aka nuna a cikakken baka.

Parallel LC Circuit — Cikin Daɗi LC Tabbata

A cikin kabilu daɗi na gida, tsari a kan wani abubuwa suna da shi a kan biyu. Saboda haka, tsari a kan abubuwan da suka dace ita ce da tsari a kan induktori da kuma tsari a kan kapasiti.

$\begin{align*} V = V_L = V_C \end{align*}$

Tana da yawa, adadin kayan aiki da ke cikin kabilu daɗi na gida ita ce da adadin kayan aiki da ke cikin induktori da kuma adadin kayan aiki da ke cikin kapasiti.

$\begin{align*} i = i_L + i_C \end{align*}$

Resonance a cikin Kabilu Daɗi na Gida LC

A lokacin da resonance, inda tsarin kayan aiki na induktori ( $X_L$ ) ita ce da tsarin kayan aiki na kapasiti ( $X_C$ ), adadin kayan aiki na zama daɗi da maɗa. Saboda haka, suka canza waɗannan zuwa adadin kayan aiki mai nuna a cikin kabilu. A wannan lokaci, adadin kayan aiki na daɗi na gida.

Tsarin kayan aiki na resonance tana da shi a cikin

$\begin{align*} f_0 = \frac {\omega_0} {2 \pi} = \frac {1} {2 \pi \sqrt{LC}} \end{align*}$

A nan za Impedance na Parallel LC circuit ta shafi da

$\begin{align*} \begin{split} Z_L_C_(_P_a_r_a_l_l_e_l_) = \frac {Z_L Z_C} {Z_L + Z_C}\ &= \frac {j \omega L \frac{1}{j \omega C}} {j \omega L + \frac{1}{j \omega C}}\ &= \frac{\frac{L}{C}} { \frac{- \omega^2 LC + 1}{j \omega C}}\ &= \frac {j \omega L} {1 - \omega^2 LC} \ \end{split} \end{align*}$

A nan za angular resonant frequency ta shafi da $\omega_0 = \frac{1}{\sqrt{LC}}$ , kuma ya zama

(2) $\begin{equation*} Z_L_C(\omega)_(_p_a_r_a_l_l_e_l_) = - j (\frac {1}{C}) (\frac {\omega}{\omega^2 - \omega_0^2}) \end{equation*}$

Saboda haka a yanayin juna na yadda ake kula da $\omega = \omega_0$ zan iya zama tsari mai karfi a kan Z kuma karamin tashar LC na doka za a gaba da adadin da ke daidai ( $I = \frac {V} {Z}$ ).

Saboda haka a lokacin da an sa LC na doka da wani abu da ake kula ta a kan gabas, zan iya yi aiki a matsayin band-stop filter da zan iya haɗa da tsari mai karfi a kan frequency na juna. Idan an sa LC na doka da wani abu da ake kula ta a kan gabas, zan iya yi aiki a matsayin band-pass filter.

A frequency da take daɗi da frequency na juna, ya'ni f<f0, XL >> XC. Saboda haka abu na iya zama inductive.
A frequency da take fiye da frequency na juna, ya'ni f>f0, XC >> XL. Saboda haka abu na iya zama capacitive.
A frequency na juna, ya'ni f = f0, XL = XC, adadin da ke daidai ita ce mafi kyau kuma tsarin da ke karfi ita ce mafi kyau. A wannan halin, abun da zan iya yi aiki a matsayin rejector circuit.

Equations na LC Circuit

Equations na karamin da shi da voltage

A lokacin da farkon:

$\begin{align*} I(0) = I_0 sin\phi \end{align*}$

$\begin{align*} V(0) = -\omega_0 L I_0 sin\phi \end{align*}$

A cikin zama:

$\begin{align*} I(t) = I_0 sin (\omega_0 t + \phi) \end{align*}$

$\begin{align*} V(t) =\sqrt {\frac{L}{C}} I_0 sin (\omega_0 t + \phi) \end{align*}$

Tarhima na LC da tushen kalkulusa

$\begin{align*} \frac {d^2 i(t)}{dt^2} + \frac{1}{LC} i(t) = 0 \end{align*}$

$\begin{align*} S^2 i(t) + \frac{1}{LC} i(t) = 0 \end{align*}$

$\begin{align*} S^2 + \omega_0^2 = 0 \,\, (where, \omega = \omega_0 = \frac{1}{\sqrt{LC}}) \end{align*}$

Na'urar da Series LC circuit

$\begin{align*} Z_L_C(\omega)_(_s_e_r_i_e_s_) = j L (\frac {\omega^2 - \omega_0^2} {\omega}) \end{align*}$

Inganci na Parallel LC circuit

$\begin{align*} Z_L_C(\omega)_(_p_a_r_a_l_l_e_l_) = - j (\frac {1}{C}) (\frac {\omega}{\omega^2 - \omega_0^2}) \end{align*}$

Lokaci na Setting

LC circuit zai iya aiki a matsayin electrical resonator kuma energy zai tafi daga electric field zuwa magnetic field a lokacin da ake kira resonant frequency. Saboda haka, saboda waɗannan oscillatory system zai samu steady-state condition a lokacin da yake, wanda ake kira lokaci na setting.

Lokacin da ya kamata amsa ta zama steady da shi a cikin firfirarren ta da suka fi sani da +- 2% daga balon bayanensa ake kira lokaci na setting.

Jerin LC Circuit

Za ka fara $I(t)$ ita ce jerin da ya faru a cikin circuit. Voltage drop across the inductor yana nufin a kan jerin $V = L \frac{dI(t)} {dt}$ kuma voltage drop across the capacitor shine $V = \frac {Q}{C}$ , inda Q shine charge stored on the positive plate of the capacitor.

Daga kalmomin Kirchhoff na hukuma, yadda adadin gudummawa ta shiga da shiga na birnin tafkin masu wasan kisa mai karfi ya zama zero.

(3) $\begin{equation*} L \frac {dI(t)}{dt} + \frac {Q}{C} = V \end{equation*}$

Idan an saka ta hanyar L da kuma an yanayi da t, za a samun

$\begin{align*} \frac{d^2 I(t)}{dt^2} + \frac{d}{dt} \frac{Q}{LC} = \frac{dV}{dt} \end{align*}$

$\begin{align*} \frac{d^2 I(t)}{dt^2} + \frac{1}{LC} \frac{d}{dt} (It) = 0 (where, Q = It) \end{align*}$

$\begin{align*} \frac{d^2 I(t)}{dt^2} + \frac{1}{LC} I(t) = 0 \end{align*}$ (4) $\begin{equation*} \frac{d^2 I(t)}{dt^2} = - \frac{1}{LC} I(t) \end{equation*}$

A nan da yawan daɗi a cikin harmonika na nufin:

(5) $\begin{equation*} I (t) = I_0 sin (\omega t + \phi) ( I = I_m sin \omega t ) \end{equation*}$

Idan $I_0 > 0$ da $\phi$ suke kowane.

Za mu yi lalacewar (5) zuwa (4) muna sami,

$\begin{align*} \frac{d^2}{dt^2}I_0 sin(\omega t+\phi) = - \frac{1}{LC}I_0 sin(\omega t+\phi) \end{align*}$

$\begin{align*} \frac{d}{dt} [\frac{d}{dt}I_0 sin(\omega t+\phi)] = - \frac{1}{LC}I_0 sin(\omega t+\phi) \end{align*}$

$\begin{align*} \frac{d}{dt} [\omega I_0 cos(\omega t+\phi)] = - \frac{1}{LC}I_0 sin(\omega t+\phi) [\frac{d}{dx} sinax = acosax] \end{align*}$

$\begin{align*} -\omega^2 I_0 sin(\omega t+\phi) = - \frac{1}{LC}I_0 sin(\omega t+\phi) [\frac{d}{dx} cos ax = -asinax] \end{align*}$

$\begin{align*} - \omega^2 = - \frac{1}{LC} \end{align*}$

(6) $\begin{equation*} \omega = \frac{1}{\sqrt{LC}} \end{equation*}$

Saboda haka daga likita ta tsakiya, za a iya cewa LC circuit shi ne wata circuit mai kawo-kawo da yake kawo-kawo a matsayin frequency masu sunan resonant frequency.

LC Circuit Voltage

Daga baya a likita (3), voltage mai inda a inductor ita ce minus voltage a capacitor.

$\begin{align*} V = -L \frac {dI(t)}{dt} \end{align*}$

Amsa harsunan jama'a mai karfi (5), muna samun

$\begin{align*} \begin{split} V(t) = - L \frac{d}{dt} [I_0 cos (\omega t + \phi)] \ &= - L I_0 \frac{d}{dt} [cos (\omega t + \phi)] \ &= - L I_0 [-\omega sin (\omega t + \phi)] \ &= \omega L I_0 [sin (\omega t + \phi)] \ &= \frac{1}{\sqrt{LC}} L I_0 [sin (\omega t + \phi)] (where,\omega = \frac{1}{\sqrt{LC}}) \\ V(t) = \sqrt\frac{L}{C} I_0 [sin (\omega t + \phi)] \ \end{split} \end{align*}$

Wasu haka, tsarin kashi yana ci gaba a lokacin da amfani ya shiga zero da kuma karon. Amfani na kashi yana da muhimmanci wadda ke da amfani na kashi yana ba da $\sqrt\frac{L}{C}$ .

Tsarin Amfani na LC Circuit

Tsarin amfani daga fadin kashi zuwa kashi ta kan capacitor shine

$\begin{align*} \begin{split} H_C(s) = \frac{V_C(s)} {V_i_n(s)}\ &= \frac{Z_C}{Z_C + Z_L}\ &= \frac{\frac{1}{j \omega C}} {j \omega L + \frac{1}{j \omega C}}\ &= \frac {\frac{1}{j \omega C}} {\frac{j^2 \omega^2 LC + 1}{j \omega C}}\ &= \frac{1} {-\omega^2 LC + 1}\\ H_C(s) = \frac{1}{1 - \omega^2 LC} (where, j^2 = -1)\ \end{split} \end{align*}$

Yanzu, tushen da mutanen tsarin karamin kirkiro zuwa karamin kirkiro a kan kapasita

$\begin{align*} \begin{split} H_L(s) = \frac{V_L(s)} {V_i_n(s)}\ &= \frac{Z_L}{Z_C + Z_L}\ &= \frac{j \omega L} {j \omega L + \frac{1}{j \omega C}}\ &= \frac{j \omega L} {\frac{j^2 \omega^2 LC + 1}{j \omega C}}\ &= \frac{j^2 \omega^2 LC} {-\omega^2 LC + 1}\\ H_L(s)= -\frac{\omega^2 LC}{1 - \omega^2 LC}\ \end{split} \end{align*}$

Tsari Mai Bazuwa na Kirkiro LC

Tambayi da kapasita ta shahara a gaba da yawa da karamin sakamakon da ya zama a gaba da lokacin da karamin sakamakon K ta shahara don lokacin da dama.

A t=0– karamin sakamakon K ta shahara

Wannan shi muryar kawo na tsohuwar haka, saboda haka za a iya rubuta

$\begin{align*} I_L(0^-) = 0 = I_L(0^+) \end{align*}$

$\begin{align*} V_C(0^-) = 0 = V_C(0^+) \end{align*}$

Saboda hanyar zuwa ta inductor da fasahohin a cikin capacitor ba su zama lura-luran.

Daga cikin duka t>=0+ switch K ya ci

A nan an sanya mai fadada fasahohi a cikin circuit. Saboda haka amfani da KVL a cikin circuit, muna samu

$\begin{align*} \begin{split} - V_L(t) - V_C(t) + V_S = 0 \\ V_L(t) + V_C(t) = V_S \\ L \frac{di(t)}{dt} + \frac{1}{C} \int i(t) dt = V_S \\ \end{split} \end{align*}$

A nan fasahohin a cikin capacitor an nufin da ita a cikin hanyar zuwa.

Tushen da ake bayar a nan sunan integro-differential equation. A yi wannan tushen da ake bayar a kan t, muna samu

$\begin{align*} L \frac{d^2i(t)}{dt^2} + \frac{i(t)}{C} = 0 \end{align*}$

(7) $\begin{equation*} \frac{d^2i(t)}{dt^2} + \frac{1}{LC} i(t) = 0 \end{equation*}$

Tambayar (7) ya nuna taurari na biyu mai kusa da LC.

Saka $\frac{d^2}{dt^2}$ da s2, za a samun,

(8) $\begin{equation*} S^2i(t) + \frac{1}{LC} i(t) = 0 \end{equation*}$

Daga baya roots na tambayar na da ke

$\begin{align*} S_1,_2 = \frac {\sqrt{\frac{4}{LC}}} {{2}} = \frac {\frac{2}{\sqrt{LC}}} {2} = \frac{1}{\sqrt{LC}} \end{align*}$

A nan $\frac{1}{\sqrt{LC}}$ shi ne tsohon yawan yanka.

Yawan Yanka na Tafin LC

Daga fadada Impedance: Tushen masu yawan yanka ta da

$\begin{align*} H(\omega) = \frac{Y(\omega)}{X(\omega)} = \frac{V_o_u_t}{V_i_n} \end{align*}$

Idan kana neman cikakken voltage a kan tsakiyar capacitor, amfani da siffar Potential Divider don tafin da aka bayar

(9) $\begin{equation*} V_o_u_t = V_i_n \frac {Z_C}{Z_C + Z_L} \end{equation*}$

Daga cewa, $Z_C =$ Impedance of the capacitor $= \frac{1}{j \omega C}$

$Z_L =$ Impedance of the inductor $= {j \omega L}$

Sauke ita a (9), muke sami

$\begin{align*} \begin{split} \frac{V_o_u_t}{V_i_n}\ &= \frac{Z_C}{Z_C + Z_L}\ &= \frac{\frac{1}{j \omega C}} {j \omega L + \frac{1}{j \omega C}}\ &= \frac {\frac{1}{j \omega C}} {\frac{j^2 \omega^2 LC + 1}{j \omega C}}\ &= \frac{1} {-\omega^2 LC + 1} (where, j^2 = -1)\\ \end{split} \end{align*}$

(10) $\begin{equation*} H(\omega) = \frac{V_o_u_t}{V_i_n} = \frac{1}{1 - \omega^2 LC} \end{equation*}$

Tambayi cewa tsari na duka yana faruwa a kan inductor, yi amfani da siffar daɗi na bayanin wannan kafa

(11) $\begin{equation*} V_o_u_t = V_i_n \frac {Z_L}{Z_C + Z_L} \end{equation*}$

Bayyana ma'ana na $Z_C$ da $Z_L$ a tushen bayanin ya fi, zaka iya samun

$\begin{align*} \begin{split} \frac{V_o_u_t}{V_i_n}\ &= \frac{Z_L}{Z_C + Z_L}\ &= \frac{j \omega L} {j \omega L + \frac{1}{j \omega C}}\ &= \frac{j \omega L} {\frac{j^2 \omega^2 LC + 1}{j \omega C}}\ &= \frac{j^2 \omega^2 LC} {-\omega^2 LC + 1}\ \end{split} \end{align*}$

(12) $\begin{equation*} H(\omega) = \frac{V_o_u_t}{V_i_n} = -\frac{\omega^2 LC}{1 - \omega^2 LC} \end{equation*}$

Tambayar (10) da (12) yana nuna tashin karamin L-C na daga wani hankali zuwa wani hankali a wurin mafi inganci.

LC Circuit Differential Equation

$\begin{align*} L \frac{di(t)}{dt} + \frac{1}{C} \int i(t) dt = V \end{align*}$

Tambayar da ya bace ita ce tambayar intigro-differensiyal. A kan wannan, tsari a cikin konsoladura an bayyana a kan tasirin karamin.

Daga baya, a yi differensiyar da tambayar da ya bace ita daga baya da karkasha t, muna samun,

$\begin{align*} L \frac{d^2i(t)}{dt^2} + \frac{i(t)}{C} = 0 \end{align*}$

(13) $\begin{equation*} \frac{d^2i(t)}{dt^2} + \frac{1}{LC} i(t) = 0 \end{equation*}$

Tambayar da ya bayyana tushen tafiya na LC.

Gina $\frac{d^2}{dt^2}$ da s2, muna samun,

(14) $\begin{equation*} S^2i(t) + \frac{1}{LC} i(t) = 0 \end{equation*}$

Daga baya, $\omega_0 = \frac{1}{\sqrt{LC}}$ saboda haka, $\omega_0^2 = \frac{1}{LC}$ , kara wannan a cikin tambayar da ya bayyana muna samun,

$\begin{align*} S^2i(t) + \omega_0^2 i(t) = 0 \end{align*}$

$\begin{align*} S^2 + \omega_0^2 = 0 \end{align*}$

Na Kirkiya LC na Karkashin Da Taushe

A cikin kirkiya LC, inductor da capacitor suna suka taka rawar da suke kai tsaye, ya'ni inductor yake kara energy a maganin kayan (B), ta hanyar current da ke kan, da kapacitor yake kara energy a electric field (E) a nan gida da sauran platoci, ta hanyar voltage da ke kan.

Sannan za a sani cewa, ya kawo charge q a cikin kapacitor, kuma energy duka cikin kirkiya an yi a cikin electric field da ke kan kapacitor. Energy da ke kan kapacitor shine

$\begin{align*} \begin{split} E_C =\frac{1}{2} CV^2 \ &= \frac{1}{2} C \frac{q^2}{C^2} \ &= \frac{1}{2} \frac{q^2}{C^2} (V = \frac{q}{C}) \ \end{split} \end{align*}$

Idan kana sanya inductor zuwa shiga capacitor da ya karkasha, zai iya haƙa hanyar voltage wajen karkashin capacitor, wanda yake iya haƙa hanyar inductor, wanda ke yi magnetic field a gaba daya inductor, capacitor yana faruwar da kafin karkasha da voltage wajen capacitor yana ƙara zuwa zero idan kafin an amfani da kafin karkasha ( $I = \frac{q}{t}$ ).

Daga baya, capacitor ta ƙara faruwa da kafin karkasha, kuma duka energy ta shiga a magnetic field na inductor. A wannan lokacin, current ta taka muhimmanci da energy wajen inductor ta ƙara bayyana ( $E_L = \frac{1}{2} LI^2)$ .

Saboda rashin resistor, ba na iya haƙa hanyar circuit. Saboda haka, maximum energy wajen capacitor ta ƙara sama da maximum energy wajen inductor.

A wannan lokacin, energy wajen magnetic field a gaba daya inductor ta ƙara haƙa hanyar voltage wajen coil kamar faraday’s law of electromagnetic induction ( $e = N \frac{d\phi}{dt}$ ). Wannan voltage ta ƙara haƙa hanyar capacitor da capacitor yana faruwar da kafin karkasha da voltage na farko.

Wannan prosesi na karkashin da kafin karkasha zai ƙara faruwa, da current ta ƙara haƙa hanyar inductor kamar yadda ake samu a gaba daya.

Saboda haka, karkashin da kafin kadan na LC circuit zai iya zama a cikin tsarin da yake kawo-karfi da energy ya zama ta ci gaba da kafin kadan bayan kapasita da indukta har zuwa lokacin da rukunin da ke cikin systemi ya shiga mummunan kawo-karfi.

Tambayar ta nuna tsarin karkashin da kafin kadan voltage da current waveform.

Karkashin da Kafin Kadan LC Circuit Waveform — Karkashin da Kafin Kadan Voltage da Current Waveform

Matattaccen LC Circuit

Matattaccen LC Circuits sun hada:

Matattaccen LC circuit suna cutar da manyan wurare daga cikin abubuwan elektronika, musamman wurare radiokin kamar transmitter, radio receivers, da TV receivers, amplifiers, oscillators, filters, tuners, da frequency mixers.
Akwai kuma amfani da LC circuits wajen kawo signals a matsayin nau'i daɗi ko kada signal daga sabon signal a nau'i daɗi.
Mafi kyau na matattaccen LC circuit shine zama a cikin tsarin da yake kawo-karfi da damping mafi kyauta, saboda haka rukunin da ke cikin systemi ana yi shi da kalmomi.
Series resonance circuit tana ba da voltage magnification.
Parallel resonance circuit tana ba da current magnification.

Me Damping?

Damping shine kudurwar da amplitude na kawo-karfi ko wave motion yana ƙasa a lokacin lokaci. Resonance shine ƙasar da amplitude a lokacin da damping yana ƙasa.

Bayanin: Fara mulkin asalin, babban bayanai za a tabbatar, idama da hakkin haɗa, zaka iya gani don hare.

Ba da kyau kuma kara mai rubutu!

Makarantarƙi:

Kawarwari Yawan Kukuna

Tsatofin RLC

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