Tusaalaha Su'aalaha Doodda Soo-jeedinta
Inductor waa qayb aan firfircoonayn oo inta badan loo isticmaalo wareegyada elektaroonigga ah si loogu kaydiyo tamarta qaab goob birlab ah. In kasta oo mabaadi'da aasaasiga ah ee inductor-ku ay aad u fudud yihiin, fahamka iyo xisaabinta dhaqankeeda codsiyada kala duwan ee wax ku oolka ah waxay noqon kartaa mid adag. Maqaalkani wuxuu higsanayaa inuu ka hadlo dhowr dhibaato oo tusaale ah iyo doodo ku saabsan inductor-yada si loo caddeeyo fikradda iyo codsigeeda injineernimada korontada.
Fikradda Aasaasiga ah ee Wax-soo-saarayaasha
Inductor-ka, oo badanaa ah gariirad ama gariirad fiilooyin ah, wuxuu awood u leeyahay inuu iska caabiyo isbeddellada ku yimaada qulqulka korantada ee dhex mara. Tani waxay sabab u tahay mabda'a Faraday ee kicinta korantada. Marka qulqulka korontada uu isbeddelo inductor-ka, goobta birlabta ee uu soo saaro hirarkaas ayaa sidoo kale isbeddela, taas oo iyaduna soo saarta emf la kiciyey (xoog koronto) oo ka soo horjeeda isbeddelka hadda.
Qaacidada aasaasiga ah ee inta badan loo isticmaalo in lagu qeexo inductor-ka wareegga korantada waa:
\[ V = L \frac{di}{dt} \]
Halkee:
– \( V \) waa danabka ku wareegsan inductor-ka (volts),
– \( L \) waa soo-gelinta inductor-ka (Henry),
– \(\frac{di}{dt} \) waa isbeddelka hadda socda waqtiga (amperes halkii ilbiriqsi).
Hadda aan aragno sida loo isticmaalo inductor-yada dhibaatooyinka qaarkood.
Tusaale 1: Danabka ku wareegsan Inductor-ka
Su'aal:
Inductor leh induction 2 H ayaa la dhex maraa isbeddel hadda ah oo ku socda xawaare ah 3 A/s. Waa maxay danabka ku wareegsan inductor-ka?
Dood:
Isticmaal qaacidada asaasiga ah ee inductor-ka:
\[ V = L \frac{di}{dt} \]
Waa la ogyahay:
– \( L = 2 \) H
– \ (\frac{di}{dt} = 3 \) A/s
\[ V = 2 \ jeer 3 \]
\[ V = 6 \]
Markaa, danab ku wareegsan inductor-ka waa 6 V.
Su'aal Tusaale 2aad: Tamarta Lagu Kaydiyo Inductor-ka
Su'aal:
Immisa tamar ayaa lagu kaydiyaa inductor-ka 4 H marka uu hadda dhex maraa yahay 5 A?
Dood:
Tamarta ku kaydsan inductor-ka waxaa lagu xisaabin karaa qaacidada:
\[ E = \frac{1}{2} LI^2 \]
Halkee:
– \( E \) waa tamar (joules),
– \( L \) waa soo-jiidashada (Henry),
– \( I \) waa hadda (ampers).
Waa la ogyahay:
– \( L = 4 \) H
– \( I = 5 \) A
\[ E = \frac{1}{2} \jeer 4 \jeer 5^2 \]
\[ E = 2 \ jeer 25 \]
\[ E = 50 \]
Markaa, tamarta ku kaydsan inductor-ka waa 50 joules.
Tusaale ahaan Dhibaatada 3aad: Wareegga Taxanaha RL
Su'aal:
Wareegga taxanaha RL wuxuu ka kooban yahay iska caabin 10 Ω ah iyo inductor 2 H ah. Haddii la isticmaalo ilo danab oo 20 V ah, waa maxay haddalka xaaladda deggan ee ku socda wareegga?
Dood:
Wareegga RL ee taxanaha ah, hadda taagan ee xaaladda deggan waxaa lagu xisaabin karaa iyadoo la adeegsanayo sharciga Ohm sababtoo ah xaaladda deggan ee inductor-ku wuxuu u dhaqmaa sida silig caadi ah oo gudbiya (eber impedance).
\[ V = IR \]
Waa la ogyahay:
– \( V = 20 \) V
– \( R = 10 \) Ω
\[ I = \frac{V}{R} \]
\[ I = \frac{20}{10} \]
\[ I = 2 \]
Markaa, hadda joogtada ah ee ku socda wareegga waa 2 A.
Tusaale 4: Soo noqnoqoshada dhawaaqa ee Wareegga RLC ee Taxanaha ah
Su'aal:
Wareegga RLC ee taxanaha ah wuxuu leeyahay iska caabin 5 Ω ah, inductor 150 mH ah, iyo capacitor 100 μF ah. Waa maxay soo noqnoqoshada resonant-ka ee wareegga?
Dood:
Soo noqnoqoshada resonant \( f_0 \) ee wareegga RLC ee taxanaha ah waxaa lagu xisaabin karaa qaacidada:
\[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \]
Halkee:
– \( L \) waa soo-jiidashada (Henry),
– \( C \) waa awoodda korantada (farads).
Waa la ogyahay:
– \( L = 150 \) mH = 0.15 H
– \( C = 100 \) μF = 100 × 10^-6 F
\[ f_0 = \frac{1}{2 \pi \sqrt{0.15 \times 100 \times 10^{-6}}} \]
\[ f_0 = \frac{1}{2 \pi \sqrt{0.15 \times 10^{-4}}} \]
\[ f_0 = \frac{1}{2 \pi \sqrt{0.15 \times 10^{-4}}} \]
\[ f_0 = \frac{1}{2 \pi \sqrt{0.000015}} \]
\[ f_0 = \frac{1}{2 \pi \times 0.00387} \]
\[ f_0 = \frac{1}{0.0243} \]
\[ f_0 \ qiyaastii 41.15 \]
Markaa, soo noqnoqoshada resonant ee wareegga RLC ee taxanaha ah waa qiyaastii 41.15 Hz.
Su'aal Tusaale ah 5: Ku-meel-gaarnimada Wareegyada RL
Su'aal:
Wareegga RL wuxuu ka kooban yahay iska caabin 8 Ω ah iyo inductor 100 mH ah. Marka danab tallaabo ah oo 24 V ah la isticmaalo, intee in le'eg ayay qaadataa in hadda ay gaarto 63.2% qiimaheeda ugu dambeeya?
Dood:
Waqtiga loo baahan yahay in la gaaro 63.2% qiimaha kama dambaysta ah ee wareegga RL waa joogtada waqtiga \( \tau \), halkaasoo:
\[ \tau = \frac{L}{R} \]
Waa la ogyahay:
– \( L = 100 \) mH = 0.1 H
– \( R = 8 \) Ω
\[ \tau = \frac{0.1}{8} \]
\[ \tau = 0.0125 \, s \]
Markaa, waqtiga loo baahan yahay in hadda la gaaro 63.2% qiimaheeda kama dambaysta ah waa 0.0125 ilbiriqsi.
Gabagabo
Tusaalooyinka kor ku xusan, waxaan ka wada hadalnay dhinacyo kala duwan oo la xiriira inductor-yada, oo ay ku jiraan danabka ku dhex jira inductor-ka, tamarta la keydiyay, dhaqankeeda wareegyada RL, iyo soo noqnoqoshada resonant-ka ee wareegyada RLC. Faham qoto dheer oo ku saabsan fikradahan iyo xisaabinta ayaa lagama maarmaan u ah qof kasta oo raadinaya xirfad ku saabsan injineernimada korontada ama elektaroonigga. Inductor-yadu waxay door muhiim ah ka ciyaaraan codsiyo badan, oo ay ku jiraan shaandheeyayaasha, wareegyada oscillator-ka, iyo beddelayaasha awoodda. Marka aan fahanno sida ay u shaqeeyaan iyo sida loo xisaabiyo xuduudahooda, waxaan naqshadeyn karnaa wareegyo waxtar badan leh oo shaqeynaya.
