Su'aalaha Tusaalaha Kondenser-ka
Kaambareetarku waa qayb koronto oo kaydin karta tamarta qaab goob koronto. Kaambareetarku wuxuu leeyahay codsiyo kala duwan oo ku jira wareegyada elektaroonigga ah, sida shaandheeyayaasha, kaydinta tamarta ku meel gaarka ah, iyo matoorada calaamadaha. Maqaalkani wuxuu ka hadli doonaa tusaalooyin dhowr ah oo ku saabsan dhibaatooyinka kaambareetar, oo ay weheliso xalal iyo sharraxaad, si loo caawiyo fahamka fikradda aasaasiga ah ee kaambareetarrada fiisigiska.
1. Awoodda Kondensarka
Su'aal Tusaale 1aad:
Kaaliyaha saxanka fidsan wuxuu leeyahay bedka saxanka \(A = 2 \, m^2 \) iyo masaafada u dhaxaysa saxanka \(d = 0.01 \, m\). Haddii joogtada dielectric ee hawada \(\epsilon_0 = 8.85 \times 10^{-12} \, F/m\), xisaabi awoodda kaaliyaha.
Xalka:
Awoodda (C) ee capacitor-ka saxanka fidsan waxaa lagu xisaabin karaa qaacidada:
\[
C = \frac{\epsilon_0 A}{d}
\]
Ku beddel qiimayaasha la bixiyay:
\[
C = \frac{(8.85 \jeer 10^{-12} \, F/m) \jeer 2 \, m^2}{0.01 \, m}
\]
\[
= \frac{17.7 \jeer 10^{-12} \, F}{0.01}
\]
\[
= 1.77 \jeer 10^{-9} \, F
\]
\[
= 1.77 \, nF
\]
Markaa, awoodda kaansadu waa 1.77 nanofarads (nF).
2. Tamarta lagu kaydiyo Kondensarka
Su'aal Tusaale 2aad:
Kaademiye awooddiisu tahay \(C = 5 \, \mu F\) wuxuu ku xiran yahay isha danab ee \(V = 12 \, V\). Xisaabi tamarta ku kaydsan kaademiyaha.
Xalka:
Tamarta (E) ee ku kaydsan kapastarka waxaa lagu xisaabin karaa qaacidada:
\[
E = \frac{1}{2} CV^2
\]
Ku beddel qiimayaasha la bixiyay:
\[
E = \frac{1}{2} \jeer 5 \jeer 10^{-6} \, F \jeer (12 \, V)^2
\]
\[
= \frac{1}{2} \jeer 5 \jeer 10^{-6} \jeer 144
\]
\[
= 2.5 \jeer 10^{-6} \jeer 144
\]
\[
= 360 \jeer 10^{-6} \, J
\]
\[
= 0.36 \, mJ
\]
Markaa, tamarta ku kaydsan capacitor-ka waa 0.36 millijoules (mJ).
3. Kondenser-yada Taxanaha iyo Wareegyada Isbarbardhigga ah
Su'aal Tusaale 3aad:
Saddex kapasitar, mid walba oo leh awoodo \(C_1 = 2 \, \mu F\), \(C_2 = 3 \, \mu F\), iyo \(C_3 = 6 \, \mu F\), ayaa ku xiran wareeg:
a) Taxane
b) Isbarbar socda
Xisaabi awoodda u dhiganta labada qaabaynba.
Xalka:
a) Wareegga Taxanaha:
Kondenser-yada ku xiran taxanaha, awoodda u dhiganta (\(C_{series}\)) waxaa lagu xisaabin karaa qaacidada:
\[
\frac{1}{C_{taxane}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
\]
Ku beddel qiimayaasha la bixiyay:
\[
\frac{1}{C_{taxane}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}
\]
\[
\frac{1}{C_{taxane}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6}
\]
\[
\frac{1}{C_{taxane}} = 1
\]
\[
C_{taxane} = 1 \, \mu F
\]
Sidaas darteed, awoodda u dhiganta ee habaynta taxanaha waa 1 microfarad (\(\mu F\)).
b) Wareegga Isbarbardhigga ah:
Kondenser-yada isku xiran ee is barbar socda, kondenser-ka u dhigma (\(C_{parallel}\)) waxaa lagu xisaabin karaa qaacidada:
\[
C_{isbarbardhig} = C_1 + C_2 + C_3
\]
Ku beddel qiimayaasha la bixiyay:
\[
C_{isbarbardhig} = 2 + 3 + 6
\]
\[
C_{isbarbardhig} = 11 \, \mu F
\]
Sidaas darteed, awoodda u dhiganta ee habaynta is barbar socota waa 11 microfarads (\(\mu F\)).
4. Kondensarka Dielectric
Su'aal Tusaale 4aad:
Kaaliyaha saxanka fidsan oo leh awood \(C_0 = 8 \, pF\) waxaa ka buuxa walxo dielectric ah oo leh joogto dielectric ah \(k = 4\). Xisaabi awoodda cusub ee kaaliyaha.
Xalka:
Awoodda cusub (C) ee kapasitor-ka leh dielectric waxaa lagu xisaabin karaa qaacidada:
\[
C = k \times C_0
\]
Ku beddel qiimayaasha la bixiyay:
\[
C = 4 jeer 8, pF
\]
\[
= 32 \, pF
\]
Markaa, awoodda cusub ee capacitor-ka waa 32 picofarads (pF).
5. Dalacsiinta iyo Bixinta Kondenser-yada
Su'aal Tusaale 5aad:
Kaademiye leh awood u dhiganta \(C = 10 \, \mu F\) wuxuu ku xiran yahay iska caabin ah \(R = 2 \, k\Omega\) oo ku jirta wareegga dallacaadda. Xisaabi wakhtiga loo baahan yahay si loogu dallaco kaademiyaha ilaa 63% danabkiisa ugu sarreeya.
Xalka:
Waqtiga loo baahan yahay in lagu dallaco kaabsoosha ilaa 63% ee danabka ugu badan waxaa loo yaqaan joogtada waqtiga (τ), kaas oo lagu xisaabin karo qaacidada:
\[
\tau = R \times C
\]
Ku beddel qiimayaasha la bixiyay:
\[
\tau = 2 \jeer 10^3 \, \Omega \jeer 10 \jeer 10^{-6} \, F
\]
\[
= 2 jeer 10^{-2} \, s
\]
\[
= 20 \, ms
\]
Markaa, waqtiga loo baahan yahay in lagu dallaco capacitor-ka ilaa 63% ee danabka ugu badan waa 20 milliseconds (ms).
6. Kondenser-yada ku jira Wareegyada AC
Su'aal Tusaale 6aad:
Kaaliyaha awoodda u leh \(C = 5 \, \mu F\) wuxuu ku xiran yahay isha danabka AC oo leh soo noqnoqosho ah \(f = 50 \, Hz\). Xisaabi falgalka awoodda ee kaaliyaha.
Xalka:
Falcelinta awoodda (X_C) waxaa lagu xisaabin karaa qaacidada:
\[
X_C = \frac{1}{2 \pi f C}
\]
Ku beddel qiimayaasha la bixiyay:
\[
X_C = \frac{1}{2 \pi \times 50 \times 5 \times 10^{-6}}
\]
\[
= \frac{1}{2 \pi \times 250 \times 10^{-6}}
\]
\[
= \frac{1}{1.57 \jeer 250 \jeer 10^{-6}}
\]
\[
= \frac{1}{392.5 \jeer 10^{-6}}
\]
\[
= 2550 \, \Omega
\]
Markaa, falcelinta awoodda ee capacitor-ka waa 2550 ohms (\(\Omega\)).
Gabagabo
Maqaalkan, waxaan ka wada hadalnay dhowr dhibaato oo tusaale ah oo ku lug leh kaabsoodhada qaabaynta iyo xaaladaha kala duwan, laga bilaabo awoodda aasaasiga ah, kaydinta tamarta, qaababka taxanaha ah iyo kuwa isbarbar socda, saameynta walxaha dielectric, ilaa jawaabta kaabsoodhada wareegyada AC. Fahmidda fikradaha iyo xisaabinta la xiriirta kaabsoodhada ayaa muhiim u ah elektaroonigga iyo fiisigiska, maadaama ay yihiin qaybo aasaasi ah oo ku jira codsiyo badan. Waxaan rajeyneynaa, dhibaatooyinkan tusaalaha ah waxay kaa caawin doonaan inaad si qoto dheer u fahamto fikradda kaabsoodhada.
