Tusaale Su'aalaha Doodda Korontada ee Tooska ah ee Hadda
Korontada Tooska ah (DC) waa dhacdo caan ah oo ku jirta injineernimada iyo fiisigiska. Korontada Tooska ah waa socodka elektaroonada hal jiho u socda iyada oo loo marayo koontaroole, badanaa laga bilaabo terminaalka togan ilaa terminaalka taban ee wareegga. Maqaalkan, waxaan ku sahamin doonnaa dhowr dhibaato oo tusaale ah iyo dooddooda la xiriirta korontada tooska ah si aan u horumarinno fahamkeenna mowduucan.
1. Wareegga Taxanaha
Su'aal 1aad:
Marka la eego wareeg taxane ah oo ka kooban saddex iska caabin ah, mid walbana leh qiimayaal ah 4Ω, 6Ω, iyo 10Ω, oo ku xiran ilo danab oo ah 20V. Xisaabi hadda socda wareegga.
Dood:
Marka hore, waxaan u baahanahay inaan xisaabino isku-darka iska caabinta ee wareegga taxanaha ah. Wareegga taxanaha ah, isku-darka iska caabinta (R_total) waa wadarta iska caabinta shaqsiyeed kasta.
\[ R_{\text{wadarta}} = R_1 + R_2 + R_3 \]
\[ R_{\text{total}} = 4Ω + 6Ω + 10Ω = 20Ω \]
Intaa ka dib, waxaan isticmaalnaa sharciga Ohm si aan u xisaabino hadda. Sharciga Ohm wuxuu dhigayaa in \( V = I \times R \). Sidaa darteed, hadda (I) waxaa lagu xisaabin karaa:
\[ I = \frac{V}{R_{\text{total}}} \]
\[ I = \frac{20V}{20Ω} = 1A \]
Markaa, qulqulka hadda socda ee wareegga waa 1 amperes.
2. Wareegga Isbarbardhigga ah
Su'aal 2aad:
Saddex iska caabin oo leh iska caabin 3Ω, 6Ω, iyo 12Ω ayaa isku xiran oo ku xiran isha danabka 12V. Xisaabi hadda socda oo dhex maraya iska caabin kasta.
Dood:
Wareeg barbar socda, danab (V) oo ku yaal iska caabin kasta waa isku mid oo la mid ah danab isha. Marka hore, waxaan xisaabinaynaa koronto kasta annagoo adeegsanayna sharciga Ohm.
\[ I_1 = \frac{V}{R_1} = \frac{12V}{3Ω} = 4A \]
\[ I_2 = \frac{V}{R_2} = \frac{12V}{6Ω} = 2A \]
\[ I_3 = \frac{V}{R_3} = \frac{12V}{12Ω} = 1A \]
Korontada dhex marta iska caabin kasta waa 4A, 2A, iyo 1A.
Intaa waxaa dheer, waxaan xisaabin karnaa wadarta guud ee hadda ka soo qulqulaya isha annagoo adeegsanayna sharciga hadda jira ee Kirchoff kaas oo sheegaya in wadarta guud ee hadda soo galaysa dhibic ay la mid tahay tan hadda ka baxaysa:
\[ I_{\text{total}} = I_1 + I_2 + I_3 = 4A + 2A + 1A = 7A \]
3. Isku-darka Wareegyada Taxanaha iyo Isbarbardhigga
Su'aal 3aad:
Afar iska caabin ah oo leh qiimayaal 4Ω, 6Ω, 12Ω, iyo 12Ω ayaa lagu habeeyay wareeg isku dhafan, kuwaas oo kala ah 4Ω iyo 6Ω ayaa loo habeeyay taxane, ka dibna natiijada waxaa loo habeeyay si barbar socda 12Ω, ugu dambeyntiina waxaa loo habeeyay taxane iyadoo la adeegsanayo 12Ω saddexaad. Haddii isha danabku tahay 24V, go'aami wadarta guud ee qulqulka wareegga.
Dood:
Tallaabada ugu horreysa waa in la xisaabiyo iska caabinta wareegga taxanaha koowaad.
\[ R_{\text{series}} = 4Ω + 6Ω = 10Ω \]
Kadib, waxaan natiijooyinka kor ku xusan ku darnaa 12Ω oo ah habayn is barbar socda.
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{10Ω} + \frac{1}{12Ω} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60} \]
\[ R_{\text{isbarbardhig}} = \frac{60}{11}Ω \qiyaastii 5.45Ω \]
Hadda, natiijadan waxaan ku soo bandhignay taxane iyadoo la adeegsanayo 12Ω ee ugu dambeeyay.
\[ R_{\text{total}} = R_{\text{parallel}} + 12Ω \]
\[ R_{\text{wadarta}} = 5.45Ω + 12Ω = 17.45Ω \]
Si loo helo wadarta guud ee hadda, waxaan isticmaalnaa sharciga Ohm:
\[ I_{\text{total}} = \frac{V}{R_{\text{total}}} \]
\[ I_{\text{total}} = \frac{24V}{17.45Ω} \qiyaastii 1.38A \]
4. Awoodda Wareegga
Su'aal 4aad:
Marka la eego iska caabin leh qiime 5Ω ah iyo hadda dhex marta iska caabinta 2A, xisaabi awoodda uu kala diray iska caabinta.
Dood:
Awoodda (P) waxaa lagu xisaabin karaa qaacidada:
\[ P = I^2 \times R \]
\[ P = (2A)^2 \jeer 5Ω \]
\[ P = 4 \ jeer 5 = 20W \]
Markaa, awoodda uu kala dirayo iska caabintu waa 20 watts.
5. Suurtagalnimada Dhibcaha Wareegga
Su'aal 5aad:
Wareeggu wuxuu ka kooban yahay laba iska caabin oo 10Ω iyo 20Ω ah mid walbana wuxuu ku xiran yahay taxane iyadoo la adeegsanayo isha danab ee 30V. Xisaabi kartida barta u dhaxaysa labada iska caabin.
Dood:
Marka hore, waxaan xisaabineynaa qulqulka hadda socda ee wareegga.
\[ R_{\text{wadarta}} = 10Ω + 20Ω = 30Ω \]
\[ I = \frac{30V}{30Ω} = 1A \]
Awoodda barta u dhaxaysa laba iska caabin waxaa lagu heli karaa iyadoo la xisaabinayo hoos u dhaca danabka ee ku dhaca iska caabinta koowaad.
\[ V_{10Ω} = I \jeer R_1 = 1A \jeer 10Ω = 10V \]
Sidaas darteed, awoodda barta u dhaxaysa labada iska caabinta waa ( 30V – 10V = 20V \).
Gabagabo
Maqaalkani wuxuu ka hadlay dhowr tusaale oo dhibaatooyin ah iyo xalalkooda la xiriira wareegyada korantada ee tooska ah. Kuwaas waxaa ka mid ah wareegyada taxanaha ah, kuwa barbar socda, iyo kuwa isku dhafan, iyo sidoo kale xisaabinta awoodda iyo kartida meelo kala duwan. Faham adag oo ku saabsan fikradahan aasaasiga ah ayaa lagama maarmaan u ah qof kasta oo baranaya korontada iyo elektarooniga. Layliga joogtada ah ee noocyada kala duwan ee dhibaatooyinka ayaa si weyn uga caawin doona barashada agabkan.
