Su'aalo Tusaale ah oo Ka Hadlaya Kondenser-yada Saxanka Isbarbar socda
Pendahuluan
Kondenser-yadu waa qaybo muhiim ah oo elektaroonik ah oo kaydiya oo sii daaya tamarta qaab koronto ah. Kondenser-yada laalaaban ee laalaaban waa nooca ugu fudud uguna ballaaran ee la isticmaalo. Maqaalkani wuxuu dabooli doonaa dhowr tusaale iyo doodo la xiriira kondenser-yada laalaaban si loo helo faham qoto dheer oo ku saabsan fikradooda iyo adeegsigooda.
Fahmidda Kondenser-yada Saxanka Isbarbardhigga ah
Kaaliyaha saxanka barbar socda wuxuu ka kooban yahay laba saxan oo wax qabta oo ay kala soocaan dielectric, oo ah walxo dahaar ah oo kordhiya awoodda kaydinta dallacaadda korontada. Awoodda (C) ee kaaliyaha saxanka barbar socda waxaa lagu xisaabin karaa qaacidada soo socota:
\[ C = \frac{\varepsilon A}{d} \]
Halkee:
– \( \varepsilon \) waa ogolaanshaha walxaha dielectric-ka,
– \( A \) waa bedka dusha sare ee bacda,
– \( d \) waa masaafada u dhaxaysa laba qaybood.
Qaaciddadani waxay muujinaysaa in awoodda kaabsoosha saxanka barbar socda ay si toos ah ugu dhigantaa aagga saxanka iyo ogolaanshaha dielectric, iyo inay si liddi ku ah ugu dhigantaa masaafada u dhaxaysa saxanka.
Su'aalo iyo Doodo Tusaale ah
Su'aal Tusaale 1aad: Xisaabinta Awoodda
Su'aal:
Laba saxan oo bir ah oo midkiiba leh bedka dusha sare ee 0.02 m² waxaa kala fogeeya masaafada 0.001 m iyadoo la isticmaalayo hawada dielectric ahaan (ogolaanshaha \(\varepsilon_{0} = 8.85 \times 10^{-12} \, F/m\)). Xisaabi awoodda kaabsootarka.
Dood:
Isticmaal qaacidada capacitance-ka si aad u hesho capacitor saxan oo barbar socda.
\[ C = \frac{\varepsilon_{0} A}{d} \]
Ku beddel qiimayaasha la yaqaan:
\[ \varepsilon_{0} = 8.85 \jeer 10^{-12} \, F/m \]
\[ A = 0.02 \, m² \]
\[ d = 0.001 \, m \]
\[ C = \frac{(8.85 \jeer 10^{-12} \, F/m) \jeer 0.02 \, m²}{0.001 \, m} \]
\[ C = \frac{1.77 \jeer 10^{-13} \, F}{0.001 \, m} \]
\[ C = 1.77 \jeer 10^{-10} \, F \]
Markaa, awoodda kaabsoosha saxanka barbar socda waa (1.77 \times 10^{-10} \, F \) ama 177 pF (picofarads).
Su'aal Tusaale 2: Xisaabinta Tamarta Kaydka ah
Su'aal:
Haddii kalkulator-ka Tusaalaha Su'aasha 1aad lagu dallaco awood dhan 50 V, intee in le'eg ayaa tamar lagu kaydiyaa kalkulator-ka?
Dood:
Tamarta (\(U\)) ee ku kaydsan kapastarka waxaa lagu xisaabin karaa qaacidada:
\[ U = \frac{1}{2} CV^2 \]
Ku beddel qiimayaasha la yaqaan:
\[ C = 1.77 \jeer 10^{-10} \, F \]
\[ V = 50 \, V \]
\[ U = \frac{1}{2} \jeer 1.77 \jeer 10^{-10} \, F \jeer (50 \, V)^2 \]
\[ U = \frac{1}{2} \jeer 1.77 \jeer 10^{-10} \, F \jeer 2500 \, V^2 \]
\[ U = \frac{1.77 \jeer 10^{-10} \, F \jeer 2500 \, V^2}{2} \]
\[ U = \frac{4.425 \jeer 10^{-7} \, J}{2} \]
\[ U = 2.2125 \jeer 10^{-7} \, J \]
Markaa, tamarta ku kaydsan kaabsootarka waa ( 2.2125 \times 10^{-7} \, J \) ama 221.25 nJ (nanoujoules).
Tusaale 3: Xisaabinta Isbeddelka Awoodda
Su'aal:
Kaaliyaha saxanka barbar socda wuxuu leeyahay bedka saxanka oo ah 0.01 m² waxaana u kala fog masaafada 0.002 m. Walxaha dielectric-ka ee la isticmaalay waa mica oo leh ogolaansho ah \( \varepsilon = 6 \times \varepsilon_{0} \). Xisaabi awoodda kaaliyaha.
Dood:
Ogolaanshaha walxaha mica dielectric waa:
\[ \varepsilon = 6 \jeer \varepsilon_{0} \]
Isticmaal qaacidada capacitance-ka si aad u hesho capacitor saxan oo barbar socda:
\[ C = \frac{\varepsilon A}{d} \]
Ku beddel qiimayaasha la yaqaan:
\[ \varepsilon_{0} = 8.85 \jeer 10^{-12} \, F/m \]
\[ A = 0.01 \, m² \]
\[ d = 0.002 \, m \]
\[ \varepsilon = 6 \jeer 8.85 \jeer 10^{-12} \, F/m = 53.1 \jeer 10^{-12} \, F/m \]
\[ C = \frac{53.1 \jeer 10^{-12} \, F/m \jeer 0.01 \, m²}{0.002 \, m} \]
\[ C = \frac{5.31 \jeer 10^{-13} \, F}{0.002 \, m} \]
\[ C = 2.655 \jeer 10^{-10} \, F \]
Markaa, awoodda kapasitor-ka oo leh mica oo ah maaddada dielectric waa (2.655 jeer 10^{-10} \, F \) ama 265.5 pF.
Su'aal Tusaale ah 4: Xisaabinta Awoodda Isutagga
Su'aal:
Laba kaabayaal oo is barbar socda oo leh awood koronto oo ah 100 pF iyo 200 pF, ayaa isku xiran taxane ahaan. Waa maxay awoodda korontada oo dhan?
Dood:
Qaacidada guud ee awoodda korantada ee capacitors-ka ku xiran taxanaha waa:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Ku beddel qiimayaasha la yaqaan:
\[ C_1 = 100 \, pF = 100 \jeer 10^{-12} \, F \]
\[ C_2 = 200 \, pF = 200 \jeer 10^{-12} \, F \]
\[ \frac{1}{C_{\text{total}}} = \frac{1}{100 \times 10^{-12}} + \frac{1}{200 \times 10^{-12}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{1}{100 \times 10^{-12}} + \frac{1}{200 \times 10^{-12}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{2}{200 \times 10^{-12}} + \frac{1}{200 \times 10^{-12}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{2 + 1}{200 \times 10^{-12}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{3}{200 \times 10^{-12}} \]
\[ C_{\text{total}} = \frac{200 \times 10^{-12}}{3} \]
\[ C_{\text{total}} = 66.67 \jeer 10^{-12} \, F \]
Markaa, awoodda guud ee labada kapasitor ee ku xiran taxanaha waa \( 66.67 \times 10^{-12} \, F \) ama 66.67 pF.
Gabagabo
Maqaalkan, waxaan ku soo qaadanay dhowr tusaale oo dhibaatooyin iyo doodo ah oo la xiriira capacitors-ka barbar socda. Waxaan ku soo qaadanay xisaabinta awoodda, tamarta la keydiyay, iyo awoodda guud ee capacitors-ka oo ku xiran taxane. Fahmidda mabaadi'da aasaasiga ah iyo sida loo xisaabiyo xuduudahan kala duwan ayaa muhiim u ah codsiyada wax ku oolka ah ee elektaroonigga. Waxaan rajeyneynaa in dooddani ay kaa caawin doonto inaad si fiican u fahamto oo aad u isticmaasho fikradaha aad baratay.
