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Mai zama wani Current Divider?

Current divider tana nufin wani circuit mai tsari na lafiya wanda yake bayar da aiki a cikin output current wanda yake fiye da sararin input current. Wannan ana samun ita ne tun daga hanyar rarrabe duk wata kamar mafi girma da suka rarraba a cikin parallel, current a cikin har daidai na biyu ya kunshi zaka iya gudanar da hakan da ake amfani da total energy an yi a cikin circuit ya shiga minimum.

Duk da haka, a parallel circuit, supply current tana kusa da duk wata kamar mafi girma. Ana kiran wannan da “current divider rule” ko kuma “current divider law”.

Parallel circuit tana kiran current divider inda terminal da duka muhimmanci suka rarraba a cikin haka da suke share da abubuwa biyu na nodes. Wannan ta ba da hanyoyi da zaka iya gudanar da current a fuskantar path da branches.

Saboda haka current a cikin duk branches na parallel circuit tana bambanta amma voltage tana daidai a cikin duk path da suka rarraba. i.e. $V_R_1 = V_R_2 = V_R_3$ …. etc. Saboda haka, babu kyau da aka bukatar don in tabbatar da voltage na biyu a cikin resistor wanda yake ba da hakan da branch currents za su iya samun da KCL (Kirchhoff’s Current Law) da ohm’s law.

Kuma, a cikin kyauka da sauran, zai kawo wasu kadan daga cikin wasu kadan.

Tsarin Kyautar Wasu

An bayyana tsari na biyu wajen kyautar wasu a cikin kyauka da sauran:

$\begin{align*} I_X = I_T [\frac {R_T}{R_X}] \end{align*}$

Idan,

$I_X$ = Kyau a kan kyautar da sauransu a cikin kyauka = $\frac{V}{R_X}$

$I_T$ = Kyau mai zurfi a cikin kyauka = $\frac{V}{R_T}$

$R_T$ = Kyakkyawan hambata na kungiyar daɗi

$V$ = Fassara a cikin kungiyar daɗi = $I_T R_T$ = $I_X R_X$ (saboda fassara ce ta duka muhimman kungiyar daɗi)

A nan hanyar hambatan daɗi, tushen bayan gudummawa mai tsayi shine

$\begin{align*} I_X = I_T [\frac {Z_T}{Z_X}] \end{align*}$

A nan hanyar tashirya, tushen bayan gudummawa mai tsayi shine

$\begin{align*} I_X = I_T [\frac {Y_X}{Y_T}] \,\,\,\, (as \,\, Z = \frac{1}{Y}) \end{align*}$

Na'urar da Tsari na RC Circuit na Karamin HaguRC Parallel Circuit

Idan kana iya amfani da na'urar da tsari a wannan circuit, za ta bayarwar da tsari a cikin resistor shine,

$\begin{align*} I_R = I_T [\frac {Z_C}{R+Z_C}] \end{align*}$

Amsa, $Z_C$ = Zanar ta capacitor = $\frac{1}{j\omega C}$

Saboda haka muke so,

$\begin{align*} \begin{split*} & I_R = I_T [\frac {\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}]\\ = I_T [\frac {\frac{1}{j\omega C}}{\frac{j\omega CR+1}{j\omega C}}]\\ \end{split*} \end{align*}$

$\begin{align*} I_R = I_T [\frac{1}{1+j\omega RC}] \end{align*}$

Bayani amsa Current Divider Rule

Fara cikin kabilu na tsawo da ke R1 da R2 da aka gudanar da V volts.

Resistive Current Divider Circuit

Kayan Aiki na Tsakiyar Kirkiya

Sallama cewa tsari mai gudanar da kirkiyoyin kayan aiki yana nuna IT. Tsari mai gudanar da IT ya zama biyu, I1 da I2 idanci, inda I1 shine tsari mai haɗa a kan kirkiya R1 da I2 shine tsari mai haɗa a kan kirkiya R2.

Saboda haka, tsari mai gudanar yana nuna

(1)

$\begin{equation*} I_T = I_1+I_2 \end{equation*}$

(2)

$\begin{equation*} I_1 = I_T-I_2 \end{equation*}$

(3)

$\begin{equation*} I_2= I_T-I_1 \end{equation*}$

Daga nan, idan kuma mafaridu biyu suna hankali a kan, mafaridar da take daɗe Req ya shafi ne

$\begin{align*} R_e_q = R_1 // R_2 \end{align*}$

(4)

$\begin{equation*} R_e_q = \frac {R_1 * R_2}{R_1 + R_2} \end{equation*}$

Daga nan a cikin tarihin Ohm iya bayyana $I=\frac{V}{R}$ , alama mai tsarki a kan mafaridar R1 ya shafi ne

$\begin{align*} I_1 = \frac{V}{R_1} \end{align*}$

$\begin{equation*} V = I_1 R_1 \end{equation*}$

Duk da haka, tsari na gudanar da yake cikin kufin R2 ita ce

$\begin{align*} I_2 = \frac{V}{R_2} \end{align*}$

(6)

$\begin{equation*} V = I_2 R_2 \end{equation*}$

a kunna tasirin (5) da (6) zan iya samun

$\begin{align*} V = I_1 R_1 = I_2 R_2 \end{align*}$

$\begin{align*} I_1 = I_2 \frac{R_2}{R_1} \end{align*}$

A wannan yadda kananan I1 da ake koyar da (1) muna tana cika,

$\begin{align*} \begin{split*} & I_T = I_2\frac{R_2}{R_1}+I_2\\ = I_2 [\frac{R_2}{R_1}+1]\\ = I_2 [\frac{R_2+R_1}{R_1}] \end{split*} \end{align*}$

(7)

$\begin{equation*} I_2 = I_T [\frac{R_1}{R_1+R_2}]\end{equation*}$

Tana iya koyar da wannan tambayar na I2 da ake koyar da (2), muna tana cika

$\begin{align*} \begin{split*} & I_1 = I_T - I_T [\frac{R_1}{R_1+R_2}]\\ = I_T [1-\frac{R_1}{R_1+R_2}]\\ = I_T [\frac{R_1+R_2-R_1}{R_1+R_2}] \end{split*} \end{align*}$

(8)

$\begin{equation*} I_1 = I_T [\frac{R_2}{R_1+R_2}] \end{equation*}$

Kuma haka, daga (7) da (8) zan iya cewa, karamin tsakiyar a gida yana cikin jirgin da ya shafi shine haske na jirgin da ya shafi ba zuwa ita ce, kafin tare da adadin karamin tsakiyar a gida a cikin jirgin.

A gaba daya,

$\,\,Branch\,\,Current\,\,=\,\,Total\,\,Current*(\frac{resistance\,\,of\,\,opposite\,\,branch}{sum\,\,of\,\,the\,\,resistance\,\,of \,\,the\,\,two\,\,branch})$

Misalai na Current Divider

Current Divider for 2 Resistors in Parallel With Current Source

Misali 1: Yana da duwatsu 20Ω da 40Ω suna daɗi a gida ta hanyar current source 20 A. Tabbatar da adadin karamin tsakiyar a cikin duk wata duwatsu a gida.

Abubuwa aika: R1 = 20Ω, R2 = 40Ω da kuma IT = 20 A

Sauƙaƙin maganar da ta goza resistors R1 yana wasu

$\begin{align*} \begin{split} & I_1 = I_T [\frac{R_2}{R_1+R_2}] = 20[\frac{40}{20+40}] = 20[\frac{40}{60}] = 20[0.67] =13.33 A \end{split} \end{align*}$

(9)

$\begin{equation*} I_1 = 13.33 A \end{equation*}$

Sauƙaƙin maganar da ta goza resistors R2 yana wasu

$\begin{align*} \begin{split} & I_2 = I_T [\frac{R_1}{R_1+R_2}] = 20[\frac{20}{20+40}] = 20[\frac{20}{60}] = 20[0.33] =6.67 A \end{split} \end{align*}$

(10)

$\begin{equation*} I_2 = 6.67 A \end{equation*}$

A nan, zama ake gara (9) da (10) za a iya samun,

$\begin{align*} I_1 + I_2 = 13.33 + 6.67 = 20 A = I_T \end{align*}$

Saboda haka, a cikin Kirchhoff’s Current Rule, kowane saki na fasahar jirgin sama da fasahar yadda. Saboda haka, za a iya tabbatar da cewa fasahar yadda (IT) ya ciyarwa a kan tarihin fasahar saki na fasahar.

Current Divider for 2 Resistors in Parallel With Voltage Source

Misali 2: Yana da shawarwari 10Ω da 20Ω suka haɗa a kan parallel ta hanyar voltage source na 50 V. Neman fasahar yadda da kuma fasahar saki na shawarwari a kan tasiri.

When You Can Use the Current Divider Rule

Za a iya amfani da current divider rule a kan yanayi:

Amfani da current divider rule idan akwai duwatsu ko kadan suna haɗa a kan voltage source ko current source.

Za a iya amfani da dabin da ke tsakanin sauri don samun saurin kowane jerin da ke cikin shago kamar yadda aka fi sani sauri na sakonnin shagon da kuma alama ta hanyar daidaitawa.
Lokacin da wasu resistors suna da makamanin cire sabon shagon maka, sauran za a yi raguwar sauri a kowanne shagon (IT)). Idan wasu resistors suna da darajar daidai, sai dai sauran zaiyi raguwa akan duk wadannan shaguna.
Lokacin da uku ko masu resistors suna da makamanin cire sabon shagon maka, alama ta hanyar daidaitawa (Req.) zai yi amfani shi don raguwar sauri na gabaɗayan shagon akan sauran sauƙi a kowane shagon a cikin shagon makamanin cire sabon maka.

Kayyade: Electrical4u

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