Misali na Tambayoyin Tattaunawa kan Wutar Lantarki Kai Tsaye
Wutar lantarki ta kai tsaye (DC) wani abu ne da aka saba gani a fannin injiniyanci da kimiyyar lissafi. Wutar lantarki kai tsaye ita ce kwararar electrons zuwa hanya ɗaya ta hanyar mai jagora, yawanci daga tashar mai kyau zuwa tashar mara kyau a cikin da'ira. A cikin wannan labarin, za mu binciki misalai da dama na matsaloli da tattaunawarsu da ta shafi wutar lantarki kai tsaye don taimakawa wajen inganta fahimtarmu game da wannan batu.
1. Da'irar Jeri
Tambaya ta 1:
Idan aka ba da da'irar jerin da ta ƙunshi resistors guda uku, kowannensu yana da ƙimar 4Ω, 6Ω, da 10Ω, an haɗa su da tushen ƙarfin lantarki na 20V. Lissafa wutar da ke gudana ta cikin da'irar.
Tattaunawa:
Da farko, muna buƙatar ƙididdige jimlar juriya a cikin da'irar jerin. A cikin da'irar jerin, jimlar juriya (R_total) shine jimlar kowace juriya ta mutum ɗaya.
\[ R_{\text{jimlar}} = R_1 + R_2 + R_3 \]
\[ R_{\text{total}} = 4Ω + 6Ω + 10Ω = 20Ω \]
Bayan haka, muna amfani da dokar Ohm don ƙididdige halin yanzu. Dokar Ohm ta bayyana cewa \( V = I \times R \). Don haka, ana iya ƙididdige halin yanzu (I) ta hanyar:
\[ I = \frac{V}{R_{\text{total}}} \]
\[ I = \frac{20V}{20Ω} = 1A \]
Saboda haka, ƙarfin wutar lantarki da ke gudana ta cikin da'irar shine 1 amperes.
2. Layi Mai Layi
Tambaya ta 2:
An haɗa resistors guda uku masu juriya na 3Ω, 6Ω, da 12Ω a layi ɗaya kuma an haɗa su da tushen ƙarfin lantarki na 12V. Lissafa wutar da ke gudana ta kowace resistor.
Tattaunawa:
Ga da'irar layi daya, ƙarfin lantarki (V) a kan kowace resistor iri ɗaya ne kuma daidai yake da ƙarfin lantarki na tushe. Da farko, muna ƙididdige kowace wutar lantarki ta amfani da dokar Ohm.
\[ I_1 = \frac{V}{R_1} = \frac{12V}{3Ω} = 4A \]
\[ I_2 = \frac{V}{R_2} = \frac{12V}{6Ω} = 2A \]
\[ I_3 = \frac{V}{R_3} = \frac{12V}{12Ω} = 1A \]
Wutar lantarki da ke gudana ta kowace resistor ita ce 4A, 2A, da 1A.
Bugu da ƙari, za mu iya ƙididdige jimillar kwararar wutar lantarki daga tushe ta amfani da dokar Kirchoff ta yanzu wadda ta bayyana cewa jimillar kwararar wutar lantarki da ke shiga wani wuri daidai yake da kwararar wutar da ke barin ta:
\[ I_{\text{total}} = I_1 + I_2 + I_3 = 4A + 2A + 1A = 7A \]
3. Haɗakar Da'irori Masu Layi da Masu Layi
Tambaya ta 3:
An shirya resistors guda huɗu masu ƙimar 4Ω, 6Ω, 12Ω, da 12Ω a cikin da'irar haɗin gwiwa, wato 4Ω da 6Ω an shirya su a jere, sannan sakamakon ya kasance a layi ɗaya da 12Ω, sannan a ƙarshe a jeri tare da 12Ω na uku. Idan tushen ƙarfin lantarki shine 24V, a ƙayyade jimlar wutar lantarki da ke gudana a cikin da'irar.
Tattaunawa:
Mataki na farko shine a ƙididdige juriyar da'irar jerin farko.
\[ R_{\text{series}} = 4Ω + 6Ω = 10Ω \]
Sannan, muna haɗa sakamakon da ke sama da 12Ω a cikin tsari mai layi ɗaya.
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{10Ω} + \frac{1}{12Ω} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60} \]
\[ R_{\text{parallel}} = \frac{60}{11}Ω \approx 5.45Ω \]
Yanzu, mun sanya wannan sakamakon a jere tare da 12Ω na ƙarshe.
\[ R_{\text{total}} = R_{\text{parallel}} + 12Ω \]
\[ R_{\text{jimlar}} = 5.45Ω + 12Ω = 17.45Ω \]
Don samun jimlar wutar lantarki, muna amfani da dokar Ohm:
\[ I_{\text{total}} = \frac{V}{R_{\text{total}}} \]
\[ I_{\text{total}} = \frac{24V}{17.45Ω} \approx 1.38A \]
4. Wutar Lantarki a Da'ira
Tambaya ta 4:
Idan aka ba da resistor mai ƙimar 5Ω da kuma wutar lantarki da ke ratsa resistor na 2A, ƙididdige ƙarfin da resistor ɗin ya wargaza.
Tattaunawa:
Ana iya ƙididdige ƙarfi (P) ta amfani da dabarar:
\[ P = I^2 \sau R \]
\[ P = (2A)^2 \sau 5Ω \]
\[ P = sau 4 5 = 20W \]
Saboda haka, ƙarfin da resistor ke watsawa shine watts 20.
5. Yiwuwa a Maki a cikin Da'ira
Tambaya ta 5:
Da'ira tana ƙunshe da resistors guda biyu na 10Ω da 20Ω kowannensu an haɗa shi a jere tare da tushen ƙarfin lantarki na 30V. Lissafa ƙarfin da ke tsakanin resistors guda biyu.
Tattaunawa:
Da farko, muna ƙididdige wutar lantarki da ke gudana ta cikin da'irar.
\[ R_{\text{jimlar}} = 10Ω + 20Ω = 30Ω \]
\[ I = \frac{30V}{30Ω} = 1A \]
Ana iya samun damar da ke tsakanin resistor guda biyu ta hanyar ƙididdige raguwar ƙarfin lantarki a kan resistor na farko.
\[ V_{10Ω} = I \sau R_1 = 1A \sau 10Ω = 10V \]
Saboda haka, yuwuwar da ke tsakanin resistors guda biyu shine ( 30V – 10V = 20V \).
Kammalawa
Wannan labarin ya tattauna misalai da dama na matsaloli da mafita da suka shafi da'irar lantarki kai tsaye. Waɗannan sun haɗa da da'irori masu layi, masu layi ɗaya, da kuma waɗanda aka haɗa, da kuma lissafin iko da ƙarfin aiki a wurare daban-daban. Fahimtar waɗannan mahimman ra'ayoyi yana da mahimmanci ga duk wanda ke nazarin wutar lantarki da na'urorin lantarki. Yin aiki akai-akai tare da nau'ikan matsaloli daban-daban zai taimaka sosai wajen ƙware wannan kayan.
