Ko'ina Da Rukunin Kammaliyar Gargajiya

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فیلڈ: Karkashin Kuliya da Dukkana

China

Muhimmiyar Rausa Dabam-dabam Yadda Ake Tafi?

A muhimmiyar rausa dabam-dabam yana nufin wurin da ake tafi da rausa dabam-dabam a kan rausa dabam-dabam a kan karamin ra'ayin ko karamin sauki (a cikin duk karamin sauki ko wani babban karamin sauki). Muhimmiyar rausa dabam-dabam yana nufin bayan duka biyu masu karfi ko nodes na sadarwa. Muhimmiyar rausa dabam-dabam zai iya kasance mai yawa, amma yana nufin kawai “rausa dabam-dabam na gaba”.

A muhimmiyar rausa dabam-dabam na sadarwa, maimakon resistor za a iya tafuta sadarwar da gaba don haka za a iya samun rausa dabam-dabam na gaba da kuma/masauka a kan voltage da kuma/masauka a kan current yadda ake amfani da sadarwar da gaba.

Idan karamin sauki ya fi wasu abubuwan karamin sauki, akwai hali ga ake tafi da rausa dabam-dabam na gaba na karamin sauki ko wani babbu karamin sauki.

Abin da ake magana game da muhimmiyar rausa dabam-dabam, a zai iya bayyana rausa dabam-dabam. Rausa dabam-dabam yana nufin tsarin da abu ko mutanen da ake amfani da shi suka ci gaba ga sharhi na karamin sauki. Yana da alama da karamin sauki, ya kamata rausa dabam-dabam ta zama, ya kamata karamin sauki ta zama; ya kamata rausa dabam-dabam ta zama, ya kamata karamin sauki ta zama.

Yadda Ake Tafi Muhimmiyar Rausa Dabam-dabam

Muhimmiyar rausa dabam-dabam yana nufin jumla da ake tafi da rausa dabam-dabam na gaba a kan karamin sauki. Muhimmiyar rausa dabam-dabam zai iya tafi da karamin sauki ko karamin ra'ayin.

Mafi na wata aiki da kungiyar tsohon zabe da take da ita. Su ne aikin da suke amfani da karamin kwarewa. Don in yi hanyoyi masu yawa ga mafi na wata, yawancin za su gudanar da kungiyar tsohon zabe, sannan za su gudanar da kungiyar tsohon zabe a cikin kungiyar tsohon zabe don in iya rage mafi na wata.

Mafi na wata a cikin kungiyar tsohon zabe

Kungiyar tsohon zabe shi ne aikin da ake gudanar da mutanen a cikin farkon da dama. A cikin kungiyar tsohon zabe, mafi na wata a cikin farkon da dama ce ta fi dama. Mafi na karamin kwarewa a cikin farkon da dama ya danganta da mafi na karamin kwarewa wajen farkon da dama.

Mafi na wata a cikin kungiyar tsohon zabe shi ne mafi na wata da za su iya ba aikin da ke amfani da karamin kwarewa a cikin kungiyar tsohon zabe. Don kungiyar tsohon zabe, mafi na wata a cikin kungiyar tsohon zabe shi ne

$\begin{align*}\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + …. + \frac{1}{R_n} \end{align*}$

idaɗe $R_1$ , $R_2$ , da $R_3$ sune mafi na wata na aikin da suke amfani da karamin kwarewa da suke gudanar da kungiyar tsohon zabe.

Mafi na karamin kwarewa zai ci gaba da mafi na wata. Yana da alaka mai zurfi a cikin mafi na wata na aikin da suke amfani da karamin kwarewa da mafi na wata na aikin da suke amfani da karamin kwarewa.

Idan hukuma da ake kara wannan abubuwa suna taka cikin muhimmiyar wuraren bayanan IEE-Business, saboda haka, hukumomin su ne a cikin yankin bayyana da rike da shi a kan bayyana. Rike masu hukumar da suka bude a cikin yankin bayyana ya gudana da kyau a kan bayyanai. Ana iya haifar da yanayi a kan yankin bayyana a cikin yankin bayyana.

Don in fahimta wannan ingantaccen, zan iya fara na wani abu mai zurfi na biyu na hukumomin da suka bude a cikin yankin bayyana, har da rike da ke duka 4 $\Omega$ . Saboda an yi bayyana biyu na hukumomin da za a iya haifar da yanayi, kawai labarin da suka bude a cikin bayyana zan iya zama a kan hukumomin.

Equivalent Resistance For Paralle Circuit

Har da hukumomin da suka bude a cikin yankin bayyana suna ba da 4 $\Omega$ , kawai labarin da suka bude a cikin bayyana zan iya zama a kan hukumomin. Saboda haka, hukumomin da suka bude a cikin yankin bayyana suna ba da 4 $\Omega$ , kawai labarin da suka bude a cikin bayyana zan iya zama a kan hukumomin. Saboda haka, hukumomin da suka bude a cikin yankin bayyana suna ba da 4 $\Omega$ , kawai labarin da suka bude a cikin bayyana zan iya zama a kan hukumomin. Saboda haka, hukumomin da suka bude a cikin yankin bayyana suna ba da 4 $\Omega$ , kawai labarin da suka bude a cikin bayyana zan iya zama a kan hukumomin. Wannan shine imanin da take da hukumomin da suka bude a cikin yankin bayyana.

Yanayin Kudanci na Kudancin Kirkiro

Idan kowane abubuwa an kawo a cikin kudanci, ana kiran wannan kudanci a matsayin kudancin kirkiro. A cikin kudancin kirkiro, kowace mafi girma ake kawo haka da yake da wani tarihi kawai da take zama a iya zuwa a kan kudancin gaba. Duk take da take za su zama a cikin kudancin gaba suna zama a cikin kila kudancin kirkiro a baya. A cikin kudancin kirkiro, takarda ta da wani tarihi kawai don zuwa.

Take ta zama a cikin kudancin gaba da darasi mai muhimmanci. Takarda ba ya fi shahara a wurare da kuma ya fi shahara a wurare. Amma, adadin takarda ta zama da yanayin kudancin kirkiro. An yi nasararsa da yanayin kudancin kirkiro mai girma da yanayin kudancin kirkiro duka a cikin kudanci.

Misali, idan an kawo kudancin kirkiro biyu da 6-Ω a cikin kudancin kirkiro, zai zama wanda ya fi shahara da kudancin kirkiro 12-Ω a cikin kudanci. Wannan shine fahimta ta yanayin kudancin kirkiro a cikin kudancin kirkiro.

Yanayin Kudancin Kirkiro Don Kudancin Kirkiro

Don kudancin kirkiro, yanayin kudancin kirkiro na kudancin kirkiro shine

$\begin{align*} R_s = R_1 + R_2 + R_3 + .... R_n\end{align*}$

Idan kofin kudancin kirkiro ta haɗa da kofin kudancin kirkiro na tasa a cikin kudanci, kuma kofin kudancin kirkiro na tasa da kofin kudancin kirkiro na tasa suka kawo a cikin siffar. E kewen kudancin kirkiro biyu suka kawo a cikin kudancin kirkiro kuma yanayin kudancin kirkiro suka zama a baya.

Misalai Na Yanayin Kudancin Kirkiro

Misali 1

Ga wasu aikin da aka bayar a nan zan iya cewa mai muhimmanci a kan A da B?

Duwa aikinta $R_1$ da $R_2$ tana da yawan $4\Omega$ suna cikin hanyoyin siri. Saboda haka, yawan muhimmancinta masu aikinta za a cewa

$\begin{align*} R_s = R_1 + R_2 \end{align*}$

$\begin{align*} R_s = 4\Omega + 4\Omega = 8\Omega \end{align*}$

Yadda aikatako da ake magana a wajen A da B na tsakiyar 2

$R_s$ , $R_3$ and $R_4$ suna da hanyar gaba-gaba. Yadda aikatako ta aiki.

$\begin{align*}\frac{1}{R_p} = \frac{1}{8\Omega} + \frac{1}{6\Omega} + \frac{1}{4\Omega} = \frac{13}{24}\Omega\end{align*}$

$\begin{align*}\frac{1}{R_p} = 1.85 \Omega \end{align*}$

Misalai 2

Za a iya kula yadda aikatako daga lokaci A zuwa B a kan aiki mai yawa

Mutummiyar Rausa A Tsakanin A Da B Mawaki 2

An koyar da ake cewa don mutummiyar rausa na tsarin gurbin rausuna.

$\begin{align*} R_s = R_1 + R_2 +R_3\end{align*}$

$\begin{align*} R_s = 2\Omega + 3\Omega +4\Omega\end{align*}$ $\begin{align*} R_s = 3\Omega\end{align*}$

Wani Tsari Yana Da Mutummiyar Rausa Daga Cikin Sauyi?

Misalai 1

Daga cikin tsarin da aka bayar, zaka iya bincike wani tsari yana da mutummiyar rausa daga cikin sauyi.

Smallest Resistance Problem Option D

Zabi D

A nan a bayar da wani abin da ake kisa. Saboda haka, ake bayar da muhimmancin tsakiyar da ita ce

$\begin{align*} R_s = 2\Omega + 2\Omega\ = 4\Omega \end{align*}$

A nan a nufin da kungiyar da ke kusa. Saboda haka, zai iya bayyana cikakken tasirin yadda

$\begin{align*}\frac{1}{R_p} = \frac{1}{2\Omega} + \frac{1}{2\Omega} = 1\Omega\end{align*}$

A nan a nufin da kungiyar da ke kusa. Saboda haka, zai iya bayyana cikakken tasirin yadda

$\begin{align*}\frac{1}{R_p} = \frac{1}{1\Omega} + \frac{1}{1\Omega} = 0.5\Omega\end{align*}$

A nan a nufin da kungiyar da ke kusa. Saboda haka, zai iya bayyana cikakken tasirin yadda

$\begin{align*} R_s = 1\Omega + 1\Omega\ = 2\Omega \end{align*}$

Saboda haka, daga cikakken bayanin da aka yi a fahimta, ana gani cewa zai iya samun yadda da take da tasirin mafi yawa.

Masana'antar Da Take Da Tasirin Mafi Yawa

Misalai 1

Bayyana Cikakken Tasirin Kungiyar Da Ake Bayar.

Don samu aikin daɗi ake gudanar da aiki masu aiki a matsayin da a yankin. A nan, $6\Omega$ da $3\Omega$ suna a yankin. Saboda haka, aikin daɗi na nufin ya zama

$\begin{align*}\frac{6\times3}{6+3}=2\Omega \end{align*}$

Kuma, $1\Omega$ da $5\Omega$ suna a matsayin. Saboda haka, aikin daɗi na nufin ya zama,

$\begin{align*} 1\Omega + 5\Omega = 6\Omega\end{align*}$

Ba a gaba da kammalawa, muna tabbatar da $2\Omega$ da $2\Omega$ suna cikin tsari, saboda haka, tasirin yawan tasiri mai yawa ce

$\begin{align*} 2\Omega + 2\Omega = 4\Omega\end{align*}$

Wannan $4\Omega$ na daya tana cikin tsarin sama da wannan $6\Omega$ na biyu. Saboda haka, tasirin yawan tasiri masu yawa ce

$\begin{align*}\frac{4\times 6}{4+6}=2.4\Omega \end{align*}$

A nan a gaba da kammalawa da kyau, uku na tasiri suka cikin tsari. Saboda haka, tasirin yawan tasiri daga baya ce

$\begin{align*} R_{eq} = 4\Omega + 2.4\Omega + 8\Omega = 14.4\Omega \end{align*}$

Misalai 2

Yadda ake kirkirar taswirin daidaitun karamin zafi na biyuwa A da B?

Don samun yawan da aka taba a baterya muna samun tasirin kammala na gyara. Yawan kammala I an gaba shi zuwa $I_1$ da $I_2$ . Yawan $I_1$ ta fi sune biyu $10\Omega$ tasiri saboda suka fi sune a matsayin series da suka da yawan sama. Yawan $I_2$ ta fi sune $10\Omega$ da $20\Omega$ tasiri saboda suka da yawan sama.

A cikin haka na iya samun $I_2$ tare da karkashin kalkulasar daɗin I wanda ya shiga batarya.

Na fahimta cewa $10\Omega$ da $20\Omega$ resistors suka binda a cikin series. An zama su da equivalent resistor da resistance ta

$\begin{align*} R_{eq} = 10\Omega + 20\Omega = 30\Omega \end{align*}$

Biyu $10\Omega$ resistors suka binda a cikin series. An zama su da equivalent resistance ta

$\begin{align*}R_{eq} = 10\Omega + 10\Omega = 20\Omega \end{align*}$

Yanzu muna karamin jirgin ruwa $30\Omega$ da $20\Omega$ zai aiki na hanyar tafuta. Zan iya gaji da karamin jirgin ruwa mai muhimmanci.

$\begin{align*}\frac{1}{R_{eq}} =\frac{1}{30} + \frac{1}{20} = \frac{1}{12}\Omega \end{align*}$

Kadan muna karamin jirgin ruwa $10\Omega$ da $12\Omega$ zai aiki na hanyar tsakiyar karamin jirgin ruwa. Karamin jirgin ruwa mai muhimmanci wadannan karamin jirgin ruwa shi ne

$\begin{align*}R_{eq} = 10\Omega + 12\Omega = 22\Omega \end{align*}$

A nan za ka samun karamin tsari I tun daga batiri. Ana cewa,

$\begin{align*} I = \frac{V}{R_{eq}} = \frac{40}{22} = 1.8 Ampere \end{align*}$

Karamin tsari ya zama da $I_1$ da $I_2$ . Don haka, karamin tsari mai guje

$\begin{align*}I = I_1 + I_2\end{align*}$

(1)

$\begin{equation*}1.8 = I_1 + I_2\end{equation*}$

Tambayar da biyu, wanda ya sanya hanyoyin karamin ruwa, shi ne kyau cewa karamin ruwa a kan $30\Omega$ na iya sama da karamin ruwa a kan $20\Omega$ .

(

$\begin{equation*}20\times I_1 = 30\times I_2\end{equation*}$

Daga tambayar (1) da (2), hanyar karamin ruwa $I_2$ an samu.

$\begin{align*}I_1= 1.8 - I_2\end{align*}$

An kawo wannan halayyar zuwa tambayar (2),

$\begin{align*}20(1.8 - I_2) = 30\times I_2 \end{align*}$

$\begin{align*}36 = (20+30)I_2 \end{align*}$

$\begin{align*}I_2 = \frac{36}{50} = 0.72A\end{align*}$

Duk da haka, zai iya bayar da adadin karamin I_1

$\begin{align*}I_1= 1.8 - 0.72 = 1.08 A\end{align*}$

Tushen: Electrical4u

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