Haka kuma a yi nazarin misalan resistor masu layi daya da misalan resistor masu layi.
Jerin Kayan Resistance Resistance
Idan an haɗa resistor kamar yadda aka nuna a cikin zane da ke sama, ana haɗa su a jere. Resistor ɗin da ake magana a kai na iya zama abubuwan resistor, fitilu, ko wasu na'urorin rufe wutar lantarki.
Cajin wutar lantarki (Q) yana nan a tsaye don haka cajin wutar lantarki yana tafiya ta juriya 1 (R1 ) = cajin wutar lantarki yana tafiya ta juriya 2 (R2 ) = cajin wutar lantarki yana tafiya ta juriya 3 (R3 ) . Wutar lantarki (I) ita ce cajin wutar lantarki da ke gudana a cikin wani lokaci (I = Q/t), saboda haka wutar lantarki da ke wucewa ta juriya 1 (I1 ) = wutar lantarki da ke wucewa ta juriya 2 (I2 ) = wutar lantarki da ke wucewa ta juriya 3 (I3 ) . A lissafi, jimlar wutar lantarki (I) = I1 = I2 = I3.
A gefe guda kuma, ƙarfin lantarki (V) yana raguwa lokacin da cajin lantarki ke motsawa ta kowace juriya. Ƙarfin lantarki, wanda kuma ake kira bambancin ƙarfin lantarki, shine makamashin ƙarfin lantarki a kowace raka'a na cajin lantarki. Ƙarfin lantarki yana raguwa saboda ana amfani da makamashin lantarki a kowace juriyar lantarki. Don haka jimlar ƙarfin lantarki (V) daidai yake da jimlar ƙarfin lantarki akan kowace juriya. A lissafi, jimlar ƙarfin lantarki (V) = V 1 + V 2 + V 3. V = IR don haka jimlar daidaiton ƙarfin lantarki V = IR 1 + IR 2 + IR 3. Ƙarfin lantarki da ke gudana a cikin kowane juriya iri ɗaya ne don haka wannan daidaiton yana canzawa zuwa V = I (R 1 + R 2 + R 3 ).
Dangane da lissafin da ke sama, an kammala da cewa jimlar juriyar lantarki (R) ko kuma juriyar maye gurbin juriyar lantarki da aka haɗa a jere daidai yake da jimlar kowace juriyar lantarki, a lissafi R = R 1 + R 2 + R 3. Idan akwai juriya guda biyu kawai da aka haɗa a jere, ƙimar juriyar maye gurbin (R) = R 1 + R 2. Idan akwai juriya guda huɗu da aka haɗa a jere, ƙimar juriyar maye gurbin (R) = R 1 + R 2 + R 3 + R 4. Haka kuma, idan akwai juriya guda biyar ko fiye da biyar da aka haɗa a jere.
Misalin matsalar resistor na series:
1. An ba da R 1 = 2 Ω, R 2 = 3 Ω da R 3 = 4 Ω. An haɗa resistor guda uku a jere. Menene darajar resistor ɗin maye gurbin? (Ω = Ohm).
Tattaunawa
R = R 1 + R 2 + R 3 = 2 + 3 + 4 = 9 Ω.
Wannan sakamakon ya nuna cewa ƙimar juriyar maye gurbin ta fi ƙimar juriyar kowace juriya da aka haɗa a jere.
2. An haɗa resistors guda biyu R 1 = 50 Ω da R 2 = 50 Ω a jere kuma an haɗa su da batirin Volt 12. Kayyade (a) juriyar da ta yi daidai (b) wutar lantarki da ke ratsa kowace resistor.
Tattaunawa
(a) R = R 1 + R 2 = 50 Ω + 50 Ω = 100 Ω.
(b) I = V / R = Volts 12 / 100 Ω = 0,12 Ampere
