Kula da Tatabbatarwa na RC: Sauran Daga Baya Da Kadan Yana Gaba Da Sabon Fage

Electrical4u

فیلڈ: Karkashin Kuliya da Dukkana

China

Zai yin da RC Circuit?

RC circuit (ko kuma ana kira RC filter ko kuma RC network) na nufin cikakken resistor-capacitor. RC circuit ya nufin cikakken cikakken kula wadda ita ce komponento masu cikakken kula mai muhimmanci kamar resistor (R) da capacitor (C), wanda ke fitarwa da mashin voltage ko mashin current.

Saboda ina da resistor a cikakken ideal na cikakken, za a iya gano energy a cikakken RC circuit, kamar RL circuit ko RLC circuit.

Wannan ba shi ne da LC circuit a cikakken ideal, wanda bai gano energy ba saboda ina da resistor. Amma wannan shine kawai a cikakken ideal, a haka a yi aiki, akwai zai gano energy wajen LC circuit saboda ina da resistance ta zero ba tare da komponento da takalma.

Series RC Circuit

A cikin kabilu RC, na iya samun kula mai takammi R a ciki ohms da kula mai kafin kula C a ciki Farads suna fi sanya.

Haka $I$ shine babban halin takammi a cikin kabilu.

$V_R$ shine takammi a kan kula R.

$V_C$ shine takammi a kan kafin kula C.

$V$ shine babban halin takammi na tattalin arziki.

Takarda ya nuna diagram mai hanyar kabilu RC dabbata.

Daga cewa a tsarin zabe mai tsarki karamin hanyar $'I'$ ita ce mafi yawan sama, don haka ita ce mafi yawan sama.

$V_R = IR$ ana sanya da hanyar karamin $'I'$ saboda a tsarin resistor mai tsarki na gaba ta voltage da karamin suna da tsari a gaba.

$V_C=I X_C$ ana amfani da jerin karamin karamin $'I'$ da $90^0$ saboda a kondansar mai tsarki masu hukuma da karamin karamin suna da $90^0$ daga cikinsu ya kuma idan karamin karamin ta gama da jerin karamin karamin da $90^0$ ko karamin karamin ta fi je jerin karamin karamin da $90^0$ .

A haka $V$ yana da zama cikin jamiyar $V_R$ da $V_C$ .

$\begin{align*} \,\, therefore, \,\, V^2 = {V_R}^2 + {V_C}^2 \end{align*}$

$\begin{align*} \begin{split} V = {\sqrt{{V_R}^2 + {V_C}^2}} \ & = {\sqrt{{IR}^2 + {IX_C}^2}} \ & = I {\sqrt{{R}^2 + {X_C}^2}} \ & = IZ \ \end{split} \end{align*}$

Impedansu na R-C series circuit shine

$\begin{align*} Z = {\sqrt{{R}^2 + {X_C}^2}} \end{align*}$

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

A fassara da kudin fassara sun fi siffar.

Kamar yadda aka fice, $V$ ya zama na $I$ da daraja ø inda

$\begin{align*} tan{\phi} = \frac{IX_C}{IR} \end{align*}$

$\begin{align*} {\phi} =tan^-^1 \frac{X_C}{R} \end{align*}$

Don haka, a cikin kabilu R-C na tsakiyar karami $'I'$ yana hada da takardun sashiya $'V'$ tare da zabi

$\begin{align*} {\phi} =tan^-^1 \frac{X_C}{R} \end{align*}$

$\begin{align*} \,\, i.e. \,\ if \,\,V = V_m sin{\omega}t \end{align*}$

$\begin{align*} i = I_m sin({\omega}t + {\phi}) \end{align*}$

$\begin{align*} \,\, where, \,\, I_m = \frac{V_m}{Z} \end{align*}$

Wata na tsawon karamin da tsari a cikin wuraren R-C suna cikakken bayanin fig.

Abubuwa a wuraren RC Series

Babban abubuwan da ya dace abubuwa shi ne jami'an babban abubuwan da ya dace karami da tsari

$\begin{align*} P = V I \end{align*}$

$\begin{align*} = (V_m sin{\omega}t) [I_m sin({\omega}t + {\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [2sin{\omega}t * sin({\omega}t + {\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos[{\omega}t-({\omega}t+{\phi})] - cos[{\omega}t+({\omega}t+{\phi})]] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos({-\phi}) - cos({2\omega}t+{\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos{\phi} - cos({2\omega}t+{\phi})] \end{align*}$

$\begin{align*} \,\, [where, \,\, cos ({-\phi}) = cos {\phi} \,\, because \,\, cos \,\, curve \,\, is \,\, symmetric] \,\, \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} cos{\phi} - \frac{V_m I_m}{2} cos({2\omega}t+{\phi}) \end{align*}$

Saboda haka, karshe da takwas ta yanzu na biyu abu.

1. Abin da ba ci gaba = $\frac{V_m I_m}{2} cos{\phi}$

2. Abin da ci gaba = $\frac{V_m I_m}{2} cos({2\omega}t+{\phi})$ wanda ya ci gaba a matsayin daga baya zuwa biyu na tarihin takwas.

Misali na karshe da takwas ta yanzu na biyu a cikin waɗannan abin da ci gaba a kan zamaɓi biyu na takwas shine sifili.

Saboda haka, misali na karshe da takwas ta yanzu na biyu a cikin zamaɓi biyu na RC series circuit shine

$\begin{align*} \begin{split} P = \frac{V_m I_m}{2} cos{\phi} \ & = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} cos{\phi} \ & = V I cos{\phi} \ \end{split} \end{align*}$

Idan $V$ da $I$ suna suka cikin RMS values na voltage da karamin kasa a kanwani.

Power Factor a Kanwani RC Series

Gaskiya figure ta wanda yake nuna power da impedance triangles.

$\begin{align*} \begin{split} \,\, (power \,\, factor) \,\, cos{\phi} = \frac{P \,\, (active \,\, power)\,\,} {S \,\, (apparent \,\, power)\,\,} \ & = \frac{R} {Z} \ & = \frac{R} {\sqrt{{R}^2 +{X_C}^2}} \ \end{split} \end{align*}$

Kanwani RC Parallel

A cikin kyakkyawan R-C na iya samu resistance $R$ a ohms da kuma capacitor mai capacitance $C$ a Farads suna yara.

Duk fadada hanyoyi a cikin kyakkyawan R-C na iya sama, saboda haka babban hanyo na iya sama da hanyar hanyoyi a kan tashin da kuma hanyoyi a kan capacitor. Amfani a cikin kyakkyawan R-C na iya sama da amfani a kan tashin da kuma amfani a kan capacitor.

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

$\begin{align*} I = I_R + I_C \end{align*}$

Don ruwa, hanyar zama ta kadan da ke cikin ita a cikin hukummi na Ohm:

$\begin{align*} I_R = \frac {V_i_n} {R} \end{align*}$

Hanyar zama da tasirin kadan da ke cikin kapasita shi ne:

$\begin{align*} I_C = C \frac {dV_i_n} {dt} \end{align*}$

A yi amfani da Hukummi na Kirchhoff (KCL) zuwa circuit R-C ta kadan

$\begin{align*} I_R + I_C = 0 \end{align*}$

$\begin{align*} \frac{v} {R} +C \frac {dV} {dt} = 0 \end{align*}$

Tambayar da ya shafi shi ita ce tushen kimiyyar birnin R-C.

Funshin Nemiya na R-C na Tashada:

$\begin{align*} H(s) = \frac {V_o_u_t} {I_i_n} = \frac {R}{1+RCs} \end{align*}$

Tambayar R-C

Kapasitor C yana aiki a $\frac {1} {sC}$ a cikin yankin daɗi da sorsa tsari na $\frac {vC(0^-)} {s}$ a gaba da shi inda $vC (0^-)$ shine tsari mai zuwa na kapasitor.

Matsayin: Matsayin C ya kula da $Z_C$ ya zama

$\begin{align*} Z_C = \frac {1} {sC} \end{align*}$

$\begin{align*} \,\, Where, \,\, s = j{\omega} \end{align*}$

$\,\,1.\,\, j$ tana nufin babban alamomin $j^2 = -1$

$\,\,2.\,\, \omega$ tana nufin tsarin dandano (radians per second)

$\begin{align*} Z_C = \frac{1}{j\omega C} = \frac{j}{j2\omega C} = -\frac{j}{\omega C} \end{align*}$

Na fari: Na fari yana da dama cikin kabilan R-C na gida.

$\begin{align*} I(s) = \frac{V_i_n(s)}{R+\frac{1}{Cs}} = {\frac{Cs}{1+RCs}}V_i_n(s) \end{align*}$

Sahara: Ta haka da amfani da sahara divider rule, sahara wanda yake kan kapasita yana cewa:

$\begin{align*} \begin{split} V_C(s) = \frac {\frac{1}{Cs}}{{R+\frac{1}{Cs}}} V_i_n(s) \ & = \frac {\frac{1}{Cs}}{{\frac{1+RCs}{Cs}}} V_i_n(s) \ & = \frac{1}{1+RCs}V_i_n(s) \ \end{split} \end{align*}$

kuma sahara wanda yake kan resistor yana cewa:

$\begin{align*} \begin{split} V_R(s) = \frac{R}{R+\frac{1}{Cs}} V_i_n(s) \ & = \frac{R}{\frac{1+RCs}{Cs}} V_i_n(s) \ &= \frac{RCs}{1+RCs}V_i_n(s) \ \end{split} \end{align*}$

Na fari a kabilan RC

Na fari yana da dama cikin kabilan R-C na gida.

$\begin{align*} I(s) = \frac{V_i_n(s)}{R+\frac{1}{Cs}} = {\frac{Cs}{1+RCs}}V_i_n(s) \end{align*}$

Na iya mulkin da RC Circuit

Na iya mulkin da mulkin daga fassara na gaba zuwa fassara a cikin kofin kapasita ita ce

$\begin{align*} H_C(s) = \frac{V_C(s)}{V_i_n(s)} = \frac{1}{1+RCs} \end{align*}$

Duk da haka, na iya mulkin da fassara na gaba zuwa fassara a cikin kofin tashin takalmi ita ce

$\begin{align*} H_R(s) = \frac{V_R(s)}{V_i_n(s)} = \frac{RCs}{1+RCs} \end{align*}$

Tarhin Mulkin da RC Circuit

Idan zaka samu wani abu a cikin tarhinta, kamar samun saki, fassara da karamin tashin takalmi suna canzawa zuwa alamomin da suka faru. Idan wannan samun yana nuna halin abubuwa ta hanyar tsawon, tarhin ya zama shi ne da aka fi sani da tsawo.

Jami'ar tasirin wata tafiya ce ta hanyar tasirin zama da tasirin dabi'a. Zan iya kara waɗannan tasiri da sauka su ne da amfani da siffar addinin superposition.

Tasirin zama shi ne a cikin da aka gira maimakon tasirin bayan aiki amma da rarrabe (kamfanin koɗa a kan) an yi lura zuwa zero.

Tasirin dabi'a shi ne a cikin da aka kusa maimakon tasirin bayan aiki amma da wata tafiya ta ƙunshi rarrabe (voltage na farko a kan capacitors da current a kan inductors). Tasirin dabi'a yana da sunan kuma da zero input response saboda maimakon tasirin bayan aiki an kusa.

Saboda haka, jami'ar tasir = tasirin zama + tasirin dabi'a

Muhiimmanci ya Daukan Tsari?

A nan inductor, ba za a iya canza current a kan inda ɗaya. Yana nufin cewa current a kan inductor a lokacin $t=0^-$ yana ɗauka a lokacin da aka yi ƙungiyar a lokacin $t=0^+$ . Na iya cewa,

$\begin{align*} i (0^-) = I_0 = 0 = i (0^+) \end{align*}$

A cikin karamin konsa, ba a iya canza tasirin konsa tare da kyau. Yana nufin cewa tasirin konsa a lokacin $t=0^-$ zai ci gaba a cikin hanyar sama a lokacin $t=0^+$ . Ya'ni,

$\begin{align*} V_C (0^-) = V_0 = V = V_C (0^+) \end{align*}$

Bayanin Gajarta na R-C Circuit mai Kafa

Za a bayar cewa konsa ta shafi daidai da suka bace da karamin kafa (K) ta shiga a lokacin da yake da wani lokaci mai yawa da ita ce a lokacin $t=0$ .

Force Response Of Driven Series R C Circuit

A $t=0^-$ yanayin K ta kusa

Wannan shi gaskiya na bayanai, saboda haka za a iya rubuta,

(1)

$\begin{equation*} V_C (0^-) = V_0 = V = V_C (0^+) \end{equation*}$

Saboda babu zama da ya fi yawa kan fassara a cikin konsolanta.

Kafin tana da $t\geq0$ yanayin K ta magance.

Daga nan ana iya girma siffar da ke canza a cikin kaya. Saboda haka waɗanda ake amfani da KVL a cikin kaya, muna samun,

$\begin{align*} -R i(t) - V_c(t) + V_s =0 \end{align*}$

(2)

$\begin{equation*} R i(t) + V_c(t) = V_s \end{equation*}$

A haka, i(t) shi wata da yake karkashin capacitor, kuma zai iya bayyana a cikin sakamakon da ke kan capacitor

$\begin{align*} i (t) = i_c (t) = C \frac {dV_c(t)}{dt} \end{align*}$

Za suka fitar da wannan a cikin tushen (2), muna samu

$\begin{align*} RC \frac {dV_c(t)}{dt} + V_c (t) = V_s \end{align*}$

$\begin{align*} RC \frac {dV_c(t)}{dt} = V_s - V_c (t) \end{align*}$

Daga bincike mai girma, muna samu

$\begin{align*} \frac{dV_c(t)} {[V_s - V_c (t)]} = \frac {1} {RC} dt \end{align*}$

A kammala duka hafin

$\begin{align*} \int \frac {dV_c(t)} {[V_s - V_c (t)]} = \int \frac {1} {RC} dt \end{align*}$

(3)

$\begin{equation*} -ln [V_s - V_c (t)] = \frac {t} {RC} + K^' \end{equation*}$

Idan $K^'$ shi daidai na musamman

Don samu $K'$ : Tana da yadda ake fara i.e. tana da yadda ake koyar (1) zuwa (3), zan iya samun

$\begin{align*} -ln [V_s - 0] = \frac {0} {RC} + K^' \end{align*}$

(4)

$\begin{equation*} {K^'} = -ln [V_s] \end{equation*}$

Tana da yadda ake koyar K’ zuwa (3) zan iya samun

$\begin{align*} -ln [V_s - V_c (t)] = \frac {t} {RC} - ln[V_s] \end{align*}$

$\begin{align*} -ln [V_s - V_c (t)] + ln[V_s] = \frac {t} {RC} \end{align*}$

$\begin{align*} ln [V_s - V_c (t)] - ln[V_s] = -\frac {t} {RC} ([ln[a] - ln[b] = ln \frac{a}{b}]) \end{align*}$

$\begin{align*} ln \frac {V_s - V_c (t)}{V_s} = -\frac {t} {RC} \end{align*}$

A nan antilog, muke so,

$\begin{align*} \frac {V_s - V_c (t)}{V_s} = e^ {-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_s - V_c (t) = V_s e^ {-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s - V_s e^ {-\frac {t} {RC}} \end{align*}$

(5)

$\begin{equation*} V_c (t) = V_s (1 - e^ {-\frac {t} {RC}}) V \end{equation*}$

Tushen bayanin ya nuna halin kima ta tarihi na farko na kawar da R-C.

Tushen bayanin ya shafi tushen daidai iya cewa $V_S$

da tushen tsakanin iya cewa $V_s * e^ {-\frac {t} {RC}}$

Tushen Bayani na Tukunomi na Kawar da R-C Da Baki

Tushen bayanin da baki shi ne karkashin kawar da R-C da yake.

Natural Response Of Source Free Series R C Circuit

Duka $t>=0^+$ na kama K

A cikin hakan na iya bayar da KVL, muna samun,

$\begin{align*} -R i(t) - V_c(t) = 0 \end{align*}$

(6)

$\begin{equation*} R i(t) = - V_c(t) \end{equation*}$

$\begin{align*} \,\, Now \,\, i(t) = i_c (t) = C \frac {dV_c(t)} {dt} \end{align*}$

Idan an yi amfani da wannan balonin karamin kamar (6), muna samun,

$\begin{align*} R C \frac {dV_c(t)} {dt} = - V_c (t) \end{align*}$

A nan kwa muhimmanci, muna samun

$\begin{align*} \frac {dV_c(t)} {V_c(t)} = - \frac {1} {R C} dt \end{align*}$

Duk da yin kudaden duk fadada

$\begin{align*} \int \frac {dV_c(t)} {V_c(t)} = \int - \frac {1} {R C} dt \end{align*}$

(7)

$\begin{equation*} ln [{V_c(t)}] = - \frac {1} {R C} + K^' \end{equation*}$

Idan da $K^'$ yana cikin abu mai kawo

Don samun $K^'$ : Tum da shaida na farko, yana nufin tsububinsa (1) a tsububin (7), muke so,

$\begin{align*} ln [V_0] = - \frac {0} {RC} + K^' \end{align*}$

(8)

$\begin{equation*} {K^'} = ln [V_0] \end{equation*}$

Tum da darajar $K^'$ a tsububin (7) muke so,

$\begin{align*} ln [V_c (t)] = - \frac {t} {RC} + ln[V_0] \end{align*}$

$\begin{align*} ln [V_c (t)] - ln[V_0] = -\frac {t} {RC} \end{align*}$

$\begin{align*} ln \frac {V_c (t)} {V_0} = -\frac {t} {RC} \end{align*}$

A cikin antilog, zan iya,

$\begin{align*} \frac {V_c (t)} {V_0} = e^{-\frac {t} {RC}} \end{align*}$

(9)

$\begin{equation*} V_c (t)} = V_0 e^{-\frac {t} {RC}} \end{equation*}$

Tambayar da ya bayyana tasirin mai zurfi na kungiyar RC.

Sau, tasirin da duka = tasirin mai zurfi + tasirin mai zurfi

$\begin{align*} V_c (t) = V_s (1 - e^{-\frac {t} {RC}})+ V_0 e^{-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s - V_s e^{-\frac {t} {RC}}+ V_0 e^{-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s + (V_0 - V_s) e^{-\frac {t} {RC}} \end{align*}$

Idan, $V_S$ shi ne tasirin mai zurfi.

$V_0$ shi ne tasirin mai gaba a konsolansa.

Wasu da RC Circuit

Wasu da R-C circuit yana nufin lokacin da zama da tsari a kan capacitor ya samu dukkan hukumomi na musamman.

Wasu kafuwa ce lokaci da take da tsarin hukumomi ta zama 0.632 dukkan hukumomi na musamman ko lokacin da take da tsarin current ta rage 0.368 dukkan hukumomi na musamman.

Wasu da R-C circuit shine darasi da resistance da capacitance.

$\begin{align*} \tau = R C \end{align*}$

Tsanoni shi shine seconds.

Tsarin Aiki na RC Circuit

Daga Fadada Impedance: Tushen daɗi da ake amfani da ita a cikin tsarin aiki na frequency response system shine

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

A haka, amfani da siffar tashin kudin kan gyaran da ya bayar

(10)

$\begin{equation*} V_o_u_t = V_i_n \frac {Z_c} {Z_c + R} \end{equation*}$

Idan, $Z_C$ = Tashin kudin kafin kapasita

$\begin{align*} Z_c = \frac {1} {j\omega C} \end{align*}$

Amfani da wannan a tsarin (10), muna samun,

$\begin{align*} V_o_u_t = V_i_n \frac {\frac{1}{j\omega C}}{{\frac{1}{j\omega C} + R}} \end{align*}$

$\begin{align*} \frac {V_o_u_t} {V_i_n} =\frac {\frac{1}{j\omega C}}{\frac{1+j\omega RC}{j\omega C}} \end{align*}$

$\begin{align*} \frac {V_o_u_t} {V_i_n} = \frac {1} {1+j\omega R C} \end{align*}$

$\begin{align*} H (\omega) = \frac {V_o_u_t} {V_i_n} = \frac {1} {1+j\omega R C} \end{align*}$

Tsunanin da ya koyar ta hanyar R-C na tsari a wurin mafi yawan bayani.

Na'ura Tashin R-C

Na'ura Tashin R-C na Gargajiya

Zama na tashen kapasita tana nuna

(11)

$\begin{equation*} V_c(t) = V - V e^{-\frac {t} {R C}} V \end{equation*}$

A halin da zai iya bayarwa da zan yi a cikin konsansadur

$\begin{align*} i(t) = i_c(t) = C \frac {dV_c(t)}{dt} = C \frac {d}{dt} [V - V e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [0 - V (\frac{-t}{RC})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [- V (\frac{-1}{R})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = \frac{V}{R}e^ {\frac{-t}{RC}} \end{align*}$

(12)

$\begin{equation*} i(t) = \frac{V}{R}e^ {\frac{-t}{\tau}} A \end{equation*}$

Yadda ake kawo wurin RC Discharging Circuit

Wurin da ke cikin capacitor yana nuna

(13)

$\begin{equation*} V_c(t) = V_0 e^{-\frac {t} {R C}} V \end{equation*}$

Sau da haka, jerin da ke cikin capacitor yana nuna

$\begin{align*} i(t) = i_c(t) = C \frac {dV_c(t)}{dt} = C \frac {d}{dt} [V_0 e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [V_0 (\frac{-t}{RC})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [V_0 (\frac{-1}{R})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = -\frac{V_0}{R}e^ {\frac{-t}{RC}} \end{align*}$

(14)

$\begin{equation*} i(t) = -\frac{V_0}{R}e^ {\frac{-t}{\tau}} A \end{equation*}$

Na gaji da kare da kare RC Circuit

Na gaji RC Circuit

A cikin wannan rubutu, yana nuna R-C circuit mai kyau da ke kula da capacitor (C), da shi ne a tarihi na resistor (R) wanda ya fara da sursu na DC voltage via switch na mechanical (K). A lokacin da kapo, capacitor ba ta samu charge bane. Idan an kara switch K, zai iya kula da charge a kan capacitor tun daga resistor har zuwa lokacin da voltage a kan capacitor yake sama da supply voltage source. Charge a kan plates na capacitor yana nufin Q = CV.

$\begin{align*} V_c(t) = V (1 - e^{-\frac {t} {R C}}) V \end{align*}$

Daga wannan equation, ana iya gano cewa voltage na capacitor yana ci abincin tsawon exponential.

Idan,

$V_C$ yana nufin voltage a kan capacitor
$V$ yana nufin supply voltage.

RC yana nufin time constant na RC charging circuit. i.e. $\tau = R C$

A cikin tushen (11) da (12), zaka iya kawo hanyoyi daban-daban na lokacin t, za a samun fassara mafi yawan tsari, ya'ni

$\begin{align*} t = \tau \,\, then \,\, V_c(t) = V - V * e^-^1 = (0.632) V \,\, (where, e = 2.718) \,\, \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, V_c(t) = V - V * e^-^2 = (0.8646) V \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, V_c(t) = V - V * e^-^4 = (0.9816) V \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, V_c(t) = V - V * e^-^6 = (0.9975) V \end{align*}$

da amfani na tsari

$\begin{align*} t = \tau \,\, then \,\, i(t) = \frac {V}{R} * e^-^1 = \frac {V}{R}(0.368) A \,\, (where, e = 2.718) \,\, \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^2 = \frac {V}{R}(0.1353) A \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^4 = \frac {V}{R} (0.0183) A \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^6 = \frac {V}{R}(0.0024) A \end{align*}$

Yadda karamin tsari a cikin kofin karami $V_C(t)$ da karamin da ya fi kan kofin karami $i(t)$ a cikin lokaci an samu a cikin wannan shahararren.

Saboda haka a cikin tashar R-C da ya ci gaba idan karamin tsari a cikin kofin karami yake faru da tsari, karamin da ya fi kan kofin karami yake rage da tsari da dama. Idan karamin tsari a cikin kofin karami yake ci gaba da darajinsa da ya fi, karamin da ya fi kan kofin karami yake rage da zero.

Tashar R-C Da Ya Rage

Idan kofin karami da ya ci gaba ya rage da battery, energy da aka rage a cikin kofin karami a lokacin da ya ci gaba zai duba da ita a matsayin da karamin tsari ya ci gaba da darajinsa da ya fi a fadada kofin karami.

Idan battery ya ci gaba ya rage da short circuit, kuma idan switch ya ci gaba, kofin karami zai rage da tsari a cikin resistor, don haka muna tashar RC da ya rage.

$\begin{align*} V_c(t) = V_0 e^{\frac {-t}{RC}} V \end{align*}$

Daga harsunan da yake, yana cewa kashi na kondensara yake lalacewa ta gaba daya. Wannan yana nufin cewa a lokacin da ake kawo makaranta R-C, ake kawo kondensara tun daga ziyarar R da ke tare da shi. Idan an yi lissafin wani abubuwa da aka sani, za su iya cewa waɗanda ake amfani da su a lokacin da ake kawo makaranta R-C da a lokacin da ake kawo makaranta R-C suna da tsari masu.

$\begin{align*} \tau = R C \end{align*}$

Idan an yi lissafin wani abubuwa da aka sani, za su iya cewa waɗanda ake amfani da su a lokacin da ake kawo makaranta R-C da a lokacin da ake kawo makaranta R-C suna da tsari masu.

$\begin{align*} t = \tau \,\, then \,\, V_c(t) = V_0 * e^-^1 = V_0 (0.368) V \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, V_c(t) = V_0 * e^-^2 = V_0 (0.1353) V \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, V_c(t) = V_0 * e^-^4 = V_0 (0.0183) V \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, V_c(t) = V_0 * e^-^6 = V_0 (0.0024) V \end{align*}$

Variyancin kirkiyar kusurta $V_C(t)$ a cikin wani tsari yana nuna a zahiri.

Saboda haka a cikin R-C Discharging circuit, idan kirkiyar kusurta ya zama da variyanci da ta yi a kan exponential, amfani da kusurta ya zama da variyanci da ta yi a kan exponential da take same rate. Idan kirkiyar kusurta ta samu shi ne, amfani da kusurta ta samu masu waɗannan.

Bayanai: Rabta alƙar, labarai zaɓu da shirya, idandaza babu gaskiya zaka ike ilimi don tsara.

Ba da kyau kuma kara mai rubutu!

Makarantarƙi:

Kawarwari Yawan Kukuna

Tsatofin RLC

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