Siris da Kudin Indaktors (Tsarin Da Misaulari)

فیلڈ: Karkashin Kuliya da Dukkana

China

Zanen inductor da?

Inductor (ko kuma ake kira electrical inductor) yana nufin kayan passive electrical element na biyu a matsayin farko da ke gudanar da energy a form of a magnetic field idan electric current ya faru shi. Ana kiran ko kuma coil, chokes, ko reactor.

Inductor ita ce wire na kofin. Yana da kofin conducting material, musamman copper mai kyau, wanda ake kofa a core mai iron ko plastic ko ferromagnetic material; saboda haka ana kiran inductor mai iron-core.

Akwai inductors a cikin tsari daga 1 µH (10-6 H) zuwa 20 H. Akwai mafi inductors suna da core mai ferrite ko iron a cikin kofin, wanda ake amfani da shi don zama magnetic field da inductance na inductor.

Daga baya Faraday’s law of electromagnetic induction, idan electric current wanda ya faru a inductor ko kofin ya zama, magnetic field na zaman yana produce e.m.f (electromotive force) ko voltage a shi. Induced voltage ko e.m.f. a inductor yana da muhimmanci masu lafiya da rate of change of the electric current flowing through the inductor.

Inductance (L) yana ake kira kayan inductor wanda ya gaji duka ingancin da ke faruwa a kan jirgin ruwa da ke faruwa a kan. Duk da inductor mai inductance mai yawa, zai iya aiki da tufafi masu iya aji hankali da tsari a kan shaida na magnetic.

Inda Inductors Ke Aiki?

Inductor a cikin circuit yana gaji ingancin da ke faruwa a kan jirgin ruwa da ke faruwa a kan ta hanyar aiki da voltage across it wanda yake da muhimmanci da rate of change of current flow. Don fahimtar inda inductor ke aiki a cikin circuit, duba tasweer da aka bayyana a nan.

Kamar da aka bayyana, akwai lamp, coil of wire (inductor), da switch suka a kunshi battery. Idan muna cika inductor daga circuit, za a yi light up lamp normally. Da inductor, za a yi aiki a cikin circuit daidai.

Inductor ko coil yana da resistance mai yawaresistance musamman da lamp, saboda haka, lokacin da switch ya ci, yawancin ruwa zai faru a kan coil saboda yana ba low-resistance path to the current. haka, muna tabbatar da lamp ta yi dim light.

Amma saboda aiki da inductor a cikin circuit, lokacin da muna ci switch, lamp ta yi bright and then gets dimmer and when we open the switch, the bulb glows very brightly and then quickly goes out.

Dalilin da ke nuna, lokacin da voltage or potential difference an sanar da ita a kan inductor, electric current flowing through an inductor produces a magnetic field. Wannan magnetic field again creates an induced electric current in the inductor but of opposite polarity, according’s to Lenz’s law.

Wannan induced current due to the magnetic field of the inductor tries to oppose any change, an increase or a decrease, in the current. Once the magnetic field is built, the current can flow normally.

Now, when the switch is closed, the magnetic field around the inductor keeps current flowing in the inductor until the magnetic field collapses. This current keeps the lamp glowing for a certain amount of time even though the switch is open.

In other words, the inductor can store energy in the form of a magnetic field and it tries to oppose any change in the current flowing through it. Thus, the overall result of this is that the current through an inductor cannot change instantaneously.

Inductor Circuit Symbol

Schematic circuit symbol for an inductor is shown in the image below.

Equation na Indaktar

Voltage a kan Indaktar

Voltage a kan indaktar yana da muhimmanci mai kusa da daraja da shi ya zama da hanyar rarrabe da tsari mai sharhi da shi a kan indaktar. A cikin ilimi, voltage a kan indaktar zai iya bayyana a haka,

$\begin{align*} v_L = L \frac{di_L}{dt} \end{align*}$

daga, $v_L$ = Voltage na gaba-gaba a kan indaktar a volts,

$L$ = Inductance a henry,

$\frac{di_L}{dt}$ = Daraja da tsari mai sharhi da shi a kan indaktar a ampere per second

Tsarin karamin mutanen da ake fitar da a cikin inductor shine saboda zafi mai gudana a cikin sauki na inductor.

Idan karamin d.c. yake fitar da a cikin inductor $\frac{di_L}{dt}$ yake haka ɗaya saboda karamin d.c. yana da dace a lokacin da taka. Saboda haka, tsarin karamin inductor yake haka ɗaya. Don haka, idan muna yi bincike a kan abubuwa d.c., a lokacin da ci gaba, inductor yana aiki daidai da shirya.

Karamin Inductor

Zan iya bayyana karamin inductor a cikin tsari mai gudana a cikin ta haka

$\begin{align*} i_L = \frac{1}{L} \int v_L dt \end{align*}$

A cikin tasiri na ɗaya, muhimmanci na takamantar da za su fitar da ita suna da shawarwarsa a kan tarihi ko kuma alamar hasashen wani lokaci kamar $-\infty \,\, to \,\, t(0^-)$ .

Daga baya, a matsayin cewa yanke aiki yana faruwa a t=0, yana nufin cewa yanke yana saki a t=0. Muna da tasiri na karamin inductor haka,

$\begin{align*} i_L = \frac{1}{L} \int v_L dt \end{align*}$

Amsa a lalle da tsari masu yawa a biyu kamar $-\infty \,\, to \,\, 0$ da $0 \,\, to \,\,t$ . A takaice ne cewa $0^-$ shine wani yanayi na musamman bayan abubuwan da suka faru, amma $0^+$ shine wani yanayi na musamman bayan abubuwan da suka faru. Saboda haka, zan iya rubuta

$\begin{align*} i_L = \frac{1}{L} \int_{-\infty}^{t} v_L dt \end{align*}$

Saboda haka,

$\begin{align*} i_L = \frac{1}{L} \int_{-\infty}^{0^-} v_L dt + \frac{1}{L} \int_{0^-}^{t} v_L dt \end{align*}$

A nan, $\frac{1}{L} \int_{-\infty}^{0^-} v_L dt$ yana nuna haddadin karamin aiki na inductor a tsawon tarihi, wanda ba shi daya ce alamar farko na $i_L$ . Za mu fada da $i_L(0^-)$ .

$\begin{align*} i_L = i_L(0^-) + \frac{1}{L} \int_{0^-}^{t} v_L dt \end{align*}$

A $t=0^+$ , za a iya rubuta,

$\begin{align*} i_L (0^+)= i_L(0^-) + \frac{1}{L} \int_{0^-}^{0^+} v_L dt \end{align*}$

A nan, an yi a bayar cewa abin daidai ya faruwa a lokacin zimantak. Saboda haka, intigali daga $0^-$ zuwa $0^+$ shi zero.

Saboda haka,

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

Saboda haka, karamin amfani a cikin inductor ba za a iya canza na tushen. Yana nufin cewa karamin amfani a cikin inductor, ta hanyar da ta gaba a kan abin daidai, ita ce mafi yawa.

Inductor a t=0

Indukta a $t = 0$ , ya'ani a lokacin da ake kawo shugaban tsarin tafi na indukta, yana zama $\infty$ saboda lokacin da $dt$ yana zama zero. Saboda haka, a lokacin da ake kawo shugaban tsarin, indukta yana aiki cikin yanayin birnin jirgin magangan. Amma a lokacin da ake yi aiki a cikin yanayin da ba ta yawa, a $t = \infty$ yana aiki cikin yanayin birnin jirgin magangan.

Idan indukta yana da shugaban karamin karamin mai muhimmanci I0 kafin ake kawo shugaban tsarin, maka a lokacin da $t=0^+$ yana aiki cikin yanayin birnin shugaban karamin karamin mai muhimmanci da ke gaba da $I_0$ , amma a lokacin da ake yi aiki a cikin yanayin da ba ta yawa, a $t=\infty$ yana aiki cikin yanayin birnin jirgin magangan a kan birnin shugaban karamin karamin mai muhimmanci.

Indukta Masu Sauran Da Indukta Masu Yawan Sauran

Induktor da dama da kofin da suka yi a harkar juna da kofin da suka yi a harkar ziyarta. Yawanci abubuwan da suka fi sani a matsayin dole da masu iya jagoranci na musamman 1 da 2 da suka samun induktansi mai yawan $L_1$ da $L_2$ har zuwa. Za M a yi indaktansi mai yawan bayan dole da masu iya jagoranci a henry.

Dole da masu iya jagoranci bi a cikin ziyartar juna zai iya haɗa a harkoki daban-daban wanda za su ba da muhimmanci a kan inductansi mai yawa a haka.

Tsarin Induktor da Dama

Yadda ake Haɗa Induktor da Dama

Sannan ziyartar da ke da dole da masu iya jagoranci bi uku da suka haɗa a harkar dama. Akwai harkokin biyu da ake iya haɗa dole da masu iya jagoranci a harkar dama.

A harkokin da ya fi, al'adu da suka samun daga dole da masu iya jagoranci suna yi a harkar mafi. Wannan lokacin, ana ce dole da masu iya jagoranci suna haɗa a harkar dama-babban ko kuma a harkar taimakawa.
A harkokin da ya bi, idan karamin juna ta haɗa a wasu dole da masu iya jagoranci kafin al'adu da suka samun dole da masu iya jagoranci suka shiga baya, wannan lokacin, ana ce dole da masu iya jagoranci suna haɗa a harkar dama-kudaden ko kuma a harkar gajimta.

Zaɓe da kyauwar gurbin 1 a zaka $L_1$ kuma zaɓe da kyauwar gurbin 2 a zaka $L_2$ . Duk duka cikin gurbun suna da mutual inductance M.

Series-aiding (Cumulative) Connection (mutually induced emf assists the self-induced EMFs)

Duka cikin gurbun ko coils suka shirya a series-aiding ko cumulatively, kamar yadda aka nuna a hoton daɗi.

A nan, zaɓen da mutanen gurbin da ke tsakaninsu suna yi a kan haguwa; saboda haka, zaɓen da e.m.f.s da ke tsakaninsu suna yi a kan haguwa.

Saboda haka,

Self-induced e.m.f. a gurban 1, $e_s_1 = -L_1\frac{di}{dt}$
Mutually induced e.m.f. a gurban 1, $e_m_1 = -M\frac{di}{dt}$
Self-induced e.m.f. a gurban 2, $e_s_2 = -L_2\frac{di}{dt}$
Mutually induced e.m.f. in inductor 1, $e_m_2 = -M\frac{di}{dt}$

Total induced e.m.f. in the combination,

$\begin{align*} e=-(L_1\frac{di}{dt}+M\frac{di}{dt}+L_2\frac{di}{dt}+M\frac{di}{dt}) \end{align*}$

(1) $\begin{equation*} e = -(L_1+L_2+2M) \frac{di}{dt} \end{equation*}$

If $L_eq$ is the equivalent inductance of the two inductors in a series-aiding connection, the e.m.f. induced in the combination is given by,

(2) $\begin{equation*} e = -L_e_q_. \frac{di}{dt} \end{equation*}$

Comparing equations (1) and (2), we get,

(3) $\begin{equation*} L_e_q_. = L_1 + L_2 + 2M \end{equation*}$

Tana na koyar da ake bayar tana nuna inductance mai dacewar da suka shirya ko suka zama da inductors masu series ko coils.

Idan ba a bani mutual inductance wajen abubuwa masu biyu (yana yake M = 0), kuma,

$\begin{align*} L_e_q_. = L_1 + L_2 \end{align*}$

Series Opposition (Differential) Connection (mutually induced emf opposes the self-induced EM

Saki kan circuit da ke cikin abubuwan inductors masu series ko coils da suka shirya haka cewa fluxes da inductors masu biyu suka samu suna gada, kamar yadda ake nuna a hoto ta hakan.

Saboda fluxes su gada, alama don mutual-induced e.m.f. za a iya gadi zuwa self-induced e.m.f.s. Saboda haka,

Self-induced e.m.f. in inductor 1, $e_s_1 = -L_1\frac{di}{dt}$
Mutually induced e.m.f. in inductor 1, $e_m_1 = +M\frac{di}{dt}$
Self-induced e.m.f. in inductor 2, $e_s_2 = -L_2\frac{di}{dt}$
Mutually induced e.m.f. in inductor 1, $e_m_2 = +M\frac{di}{dt}$

Total induced e.m.f. in the combination,

$\begin{align*} e=-(L_1\frac{di}{dt}-M\frac{di}{dt}+L_2\frac{di}{dt}-M\frac{di}{dt}) \end{align*}$

(4) $\begin{equation*} e = -(L_1+L_2-2M) \frac{di}{dt} \end{equation*}$

If $L_e_q$ is the equivalent inductance of the two inductors in a series opposition connection, the e.m.f. induced in the combination is given by,

(5) $\begin{equation*} e = -L_e_q_. \frac{di}{dt} \end{equation*}$

A nan equation (4) da (5), muke so,

(6) $\begin{equation*} L_e_q_. = L_1 + L_2 - 2 M \end{equation*}$

Equation na bai cewa equivalent inductance daga biyu inductors wanda suka sama series opposition ko differential connection.

Idan ba a bayyana mutual inductance daga biyu coils (ya'ni, M = 0), maka,

$\begin{align*} L_e_q_. = L_1 + L_2 \end{align*}$

Misali 1

Biye coils suna self-inductances da 10 mH da 15 mH kuma mutual inductance daga biyu coils shi ne 10 mH. Za a tabbatar da equivalent inductance idan suke sama series aiding.

Hali:

Bayanai: L1 = 10 mH, L2 = 15 mH da M = 10 mH

Daga tushen bayanin cikakken tsari,

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 + 2M \\ & = 10 + 15 + 2(10) \\ & = 10 + 15 + 20 \\ & L_e_q_. = 45\,\,mH \end{split} \end{align*}$

Saboda haka, tun daga tushen bayani, muna samun cikakken tsari na 45 mH idan suka shiga cikakken tsari mai yawa.

Misalai 2

Biyo biyu suna da cikakken tsari na 10 mH da 15 mH, kuma cikakken tsari na biyu suna da 10 mH. Bayyana cikakken tsari na biyu a lokacin da suka shiga cikakken tsari mai gaba.

Hali:

Bayanai: L1 = 10 mH, L2 = 15 mH da M = 10 mH

Daga tushen bayanin cikakken tsari mai gaba,

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 - 2M \\ & = 10 + 15 - 2(10) \\ & = 10 + 15 - 20 \\ & = 25 - 20 \\ & L_e_q_. = 5\,\,mH \end{split} \end{align*}$

Saboda haka, tare da ita ce, muka samu inductance na ɗaya 5 mH idan ake kofar da su a matsayin series opposing.

Formular na Inductors a Kofa

Daidaitaccen Inductors a Kofa

Zuwa biyu na inductors zai iya kofar da su a matsayin

EMF na gaba ta taimakawa EMFs na baki a cikin sauran, ya'ni, kofa aiding
EMF na gaba ta fuskantar EMFs na baki a cikin sauran, ya'ni, kofa opposing

Kofa Aiding (Cumulative) (EMF na gaba ta taimakawa EMFs na baki)

Idan an kofar da biyu na inductors a kofa aiding, EMF na gaba ta taimakawa EMFs na baki kamar yadda ake nuna a wannan takarda.

Bayyana i1 da i2 wata cashi da ke faruwar inductors L1 da L2 da I shine cashi na dabbobi.

Saboda haka,

(7) $\begin{equation*} i = i_1 + i_2 \end{equation*}$

Saboda haka,

(8) $\begin{equation*} \frac{di}{dt} = \frac{di_1}{dt} + \frac{di_2}{dt} \end{equation*}$

A cikin kada inductor, zai kasance waɗanda EMFs suka shafi. Wata tana shafi saboda self-induction da wata saboda mutual induction.

Saboda inductors suna haɗa a parallel, EMFs su ne daɗi.

Saboda haka,

(9) $\begin{equation*} L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} = L_2 \frac{di_2}{dt} + M \frac{di_1}{dt} \end{equation*}$

$\begin{align*} L_1 \frac{di_1}{dt} - M \frac{di_1}{dt} = L_2 \frac{di_2}{dt} - M \frac{di_2}{dt} \end{align*}$

$\begin{align*} \frac{di_1}{dt} (L_1 - M) = \frac{di_2}{dt} (L_2 - M) \end{align*}$

(10) $\begin{equation*} \frac{di_1}{dt} = (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} \end{equation*}$

A nan, koy equation (9) a cikin equation (8), muna samun,

$\begin{align*} \frac{di}{dt} = (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} + \frac{di_2}{dt} \end{align*}$

(11) $\begin{equation*} \frac{di}{dt} = (1 + \frac{L_2 - M}{L_1 - M}) + \frac{di_2}{dt} \end{equation*}$

Idan $L_e_q.$ ita ce inductance da duka da ake sanya, zai iya shiga emf a kan

(12) $\begin{equation*} e = L_e_q_. \frac{di}{dt} \end{equation*}$

Wannan ya fi tsaya da emf da ake shiga a wata coil musamman yana nufin

$\begin{align*} L_e_q_. \frac{di}{dt} = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} \end{align*}$

(13) $\begin{equation*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}] \end{equation*}$

Koyar da balin $\frac{di_1}{dt}$ daga kungiyar (10) a cikin kungiyar (13), muna samun,

$\begin{align*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} + M \frac{di_2}{dt}] \end{align*}$

(14) $\begin{equation*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 (\frac{L_2 - M}{L_1 - M}) + M] \frac{di_2}{dt} \end{equation*}$

A nan, koyar da kungiyar (11) zuwa kungiyar (14),

$\begin{align*} 1+(\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} = \frac{1}{L_e_q_.}[L_1 (\frac{L_2 - M}{L_1 - M}) + M]\frac{di_2}{dt} \end{align*}$

$\begin{align*} \frac{L_1+L_2 - 2M}{L_1 - M} = \frac{1}{L_e_q_.} [\frac{L_1L_2- L_1M+L_1M - M^2}{L_1 - M}] \end{align*}$

$\begin{align*} \frac{L_1+L_2 - 2M}{L_1 - M} = \frac{1}{L_e_q_.} [\frac{L_1L_2 - M^2}{L_1 - M}] \end{align*}$

(15) $\begin{equation*} L_e_q_. = \frac{L_1L_2 - M^2}{L_1+L_2 - 2M} \end{equation*}$

Tambayar da ya bayyana inductance mafi girma na biyu inductor a matsayin parallel-aiding ko cumulative connection.

Idan ba a nan inductance masu tafinta a kan biyu inductor (ya'ni, M = 0), kuma,

$\begin{align*} L_e_q_. = \frac{L_1L_2 - (0)^2}{L_1+L_2-2(0)} = \frac{L_1L_2}{L_1+L_2} = \frac{product}{sum} \end{align*}$

Parallel Opposition (Differential) Connection (mutually induced emf opposes the self-induced EMFs)

Idan da biyu inductors suka shirya a parallel opposition, zai iya hana mutadadi emf ta hanyar da yake kawo karin self-induced EMFs.

Kamar yadda aka nuna a siffofin, biyu inductors suka shirya a parallel opposition ko differential.

A cikin haka na iya tabbatar da, kamar yadda aka yi a parallel-aiding connection,

(16) $\begin{equation*} L_e_q_. = \frac{L_1L_2 - M^2}{L_1+L_2 + 2M} \end{equation*}$

Tana bayyana equivalent inductance ta biyu inductors da suke shirya a parallel opposition ko differential connection.

Idan ba a nan mutual inductance daga biyu coils (ya'ni, M = 0), tana da,

$\begin{align*} L_e_q_. = \frac{L_1L_2 - (0)^2}{L_1+L_2+2(0)} = \frac{L_1L_2}{L_1+L_2} = \frac{product}{sum} \end{align*}$

Misalai 1

Akwai inductors biyu da ingantaccen inductance suna 5 mH da 10 mH kuma mutual inductance daga biyu su na 5 mH. Neman inductance mai yawa a lokacin da ake taka cikin parallel aiding.

Jawab:

Bayanai: L1 = 5 mH, L2 = 10 mH kuma M = 5 mH

Idan an yi a cikin formula ta parallel aiding,

$\begin{align*} \begin{split} & L_e_q_. = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}.... if \,\, fluxes \,\, aid \\ & = \frac{5 * 10 - (5)^2}{5 + 10 - 2(5)} \\ & = \frac{50 - 25}{15 - 10} \\ & = \frac{25}{5} \\ & L_e_q_. = 5\,\,mH \end{split} \end{align*}$

Saboda haka, tare da equation, mace neman inductance mai yawa na 5 mH a lokacin da ake taka cikin parallel aiding.

Misalai 2

An samu aiki biyu ne da maimaita 5 mH da 10 mH kuma da takamman inductance ta duka 5 mH. Neman inductance na gaskiya idan an sa su ta hanyar parallel opposing.

Jawabu:

Bayanai da aka bayar: L1 = 5 mH, L2 = 10 mH da M = 5 mH

Idan an yi amfani da rumus parallel opposing,

$\begin{align*} \begin{split} & L_e_q_. = \frac{L_1 L_2 - M^2}{L_1 + L_2 + 2M}.... if \,\, fluxes \,\, oppose \\ & = \frac{5 * 10 - (5)^2}{5 + 10 + 2(5)} \\ & = \frac{50 - 25}{15 + 10} \\ & = \frac{25}{25} \\ & L_e_q_. = 1\,\,mH \end{split} \end{align*}$

Saboda haka, idan an yi amfani da rumus, ana samu inductance na gaskiya 1 mH idan an sa su ta hanyar parallel opposing.

Coupling Inductors

Idan magnetic field wanda ya faru daga inductor (coil) baki ruwa ko tashin inductor na gari, akwai inductor biyu sun ce ne da suka take da magnetic coupling. Saboda magnetic coupling, akwai mutual inductance wanda ya faru a kan inductor biyu.

A cikin coupled circuits, yanayin energy yana faru daga circuit zuwa circuit na biyu idan wata shi ne ke energize. Transformer da juyin biyu, autotransformer, da induction motor sun fi dace da misalai na magnetic coupling.

Amsa kwa biyu na inductors ko koyilin 1 da 2 tana da inductances L1 da L2 har zuwa. Za a iya M wajen yin mutual inductance bayan koyiloin biyu.

Hasar mutual inductance shine ya zama ya ba (L1 + M da L2 + M) ko ya haɗa (L1 – M da L2 – M) inductance na koyiloin biyu, wannan shi ne a cikin hanyar yadda ake ƙoƙarin koyiloin ko inductors.

Idan koyiloin biyu suka a ƙoƙarin da suka ƙara magana, akwai zaka inductance na koyiloci biyu ya ba da M, ya ni ya zama L1 + M don koyili 1 da L2 + M don koyili 2. Saboda hakan, flux duka na koyili ita ce mafi yawan kuɗi saboda flux ta baki.
Idan koyiloin biyu suka a ƙoƙarin da suka ci gaba, akwai zaka inductance na koyiloci biyu ya haɗa da M, ya ni ya zama L1 – M don koyili 1 da L2 – M don koyili 2. Saboda hakan, flux duka na koyili ita ce ƙarin kuɗi saboda flux ta baki.

Tsarin Mutual Inductance

Na sani cewa ƙarin daɗi aiki a koyili ɗaya yana samun e.m.f. na mutually induced a koyili biyu.

Mutual inductance tana nufin kyakkyawar aiki na koyili ɗaya (ko circuit) na samun e.m.f. a koyili biyu (ko circuit) na jiragen da aiki a koyili ɗaya yana ƙarin daɗi.

Hukuma, abu na koyiloin biyu na yi aiki saboda za a iya ake ƙara ƙarin daɗi a koyili ɗaya yana samun e.m.f. na mutually induced a koyili biyu, wanda ya ƙara ƙarin daɗi a koyili ɗaya.

Mutual inductance (M) tana nufin flux-linkages na koyili na ƙarin daɗi na koyili biyu.

Za a iya koyar da ita daga masana,

$\begin{align*} M = \frac{N_2 \phi_1_2}{I_1} \end{align*}$

Daga,

$I_1$ = Kirki a tsakiyar daɗi

$\phi_1_2$ = Flux linking the second coil

$N_2$ = Tsakiyar daɗi na biyu

Mutual inductance between two coils is 1 henry if current changing at the rate of 1 ampere per second in one coil induces an e.m.f. of 1 V in the other coil.

Coefficient of Coupling

An koyar da ko'ina (k) daga birnin daɗi biyu ne a cewa babban zaba ta flux mai girma da kirki a tsakiyar daɗin yana haɗa shi.

Koefisiyan daɗi ita ce mafi yawan muhimmanci a cikin hanyoyin kasa da suka shiga don tabbatar da adadin daɗi ita daga inductor.

Daga riyoyin lissafi, koefisiyan daɗi ita zai iya bayyana da,

$\begin{align*} k = \frac{M}{\sqrt{L_1L_2}} \end{align*}$

Amsa,

L1 shine daɗi ita na gida na abubuwa na farko

L2 shine daɗi ita na gida na abubuwa na biyu

M shine daɗi ita ta hanyar da suka shiga daga abubuwan biyu

Koefisiyan daɗi ita tana da alaka da daɗi ita ta hanyar da suka shiga daga abubuwan biyu. Idan koefisiyan daɗi ita ya fi yawa, daɗi ita ta hanyar da suka shiga zai fi yawa. Abubuwan biyu masu daɗi ita ta hanyar da suka shiga suna da alaka da flux mai magana.

Idan flux din abubuwa na biyu ya shiga abubuwa na uku, koefisiyan daɗi ita ya zama 1 (ya'ni, 100%), kuma abubuwan za a ce mafi girma.
Idan ne kadan flux din abubuwa na biyu ya shiga abubuwa na uku, koefisiyan daɗi ita ya zama 0.5 (ya'ni, 50%), kuma abubuwan za a ce mafi tsawo.
Idan flux din abubuwa ba ya shiga abubuwa na uku, koefisiyan daɗi ita ya zama 0, kuma abubuwan za a ce mafi magance.

Koefisiyan daɗi ita ba zan iya zama daya. Yana da alaka da matsayin abincin kayan aiki. Don core mai hawa, koefisiyan daɗi ita zai iya zama 0.4 zuwa 0.8 idan kuna da fagen da ke faruwa daga abubuwan biyu, kuma don core mai hirri ko ferrite zai iya zama kadan 0.99.

Source: Electrical4u.

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