Labari game da Resistor a cikin jerin
Idan an haɗa resistors ɗin kamar yadda aka nuna a cikin hoton, resistors ɗin za a jera su a jere. Resistor ko juriyar lantarki da ake magana a kai na iya kasancewa a cikin nau'in abubuwan resistor, fitilu, ko wasu juriyar lantarki.
Cajin wutar lantarki yana motsawa ta hanyar juriya 1 (R)1) = da cajin lantarki yana motsawa ta hanyar juriya 2 (R)2) = cajin lantarki yana motsawa ta hanyar juriya 3 (R3). Wutar lantarki (I) cajin lantarki ne wanda ke gudana a cikin wani takamaiman lokaci (I = Q / t), don haka wutar lantarki ta hanyar juriya 1 (I1) = wutar lantarki ta hanyar juriya 2 (I2) = wutar lantarki ta hanyar juriya 3 (I3). A lissafi, jimlar wutar lantarki (I) = I1 = Ni2 = Ni3.
Akasin haka, ƙarfin lantarki (V) yana raguwa lokacin da cajin lantarki ke motsawa ta kowace resistor. Ƙarfin lantarki wanda kuma ake kira bambancin ƙarfin lantarki, makamashi ne na ƙarfin lantarki a kowace raka'a na cajin lantarki. Ƙarfin lantarki yana raguwa saboda ana amfani da wutar lantarki a kowace juriyar lantarki. Don haka, jimlar ƙarfin lantarki (V) daidai yake da adadin wutar lantarki a kowace resistor. A lissafi, jimlar ƙarfin lantarki (V) = V1 +V2 +V3V = IR don haka daidaiton ƙarfin lantarki shine V = IR1 + IR2 + IR3Wutar lantarki da ke gudana akan kowace resistor daidai take don haka wannan lissafin ya canza zuwa V = I (R1 + R2 + R3).
Dangane da lissafin da ke sama, za a iya kammala da cewa jimlar juriyar lantarki (R) ko kuma daidai juriyar juriyar lantarki
wanda aka haɗa a jere daidai yake da jimlar kowace juriyar lantarki, a lissafi R = R1 + R2 + R3Idan akwai resistor guda biyu kawai a jere, resistor ɗin daidai (R) = R1 + R2Idan akwai resistor guda huɗu a jere, resistor ɗin daidai (R) = R1 + R2 + R3 + R4Haka kuma, idan akwai resistor biyar ko fiye da biyar da aka haɗa a jere.
Misalin matsala ta 1:
Idan an sani, R 1 = 2 Ω, R 2 = 3 Ω da R 3 = 4 Ω. Duk resistor guda uku an haɗa su a jere. Menene ƙimar resistor ɗin daidai? (Ω = Ohm ).
Magani:
R = R 1 + R 2 + R 3 = 2 + 3 + 4 = 9 Ω.
Wannan sakamakon yana nuna cewa juriyar da ta yi daidai ta fi ƙimar juriyar da aka haɗa a jere.
Misalin matsala ta 2:
An haɗa resistors guda biyu R 1 = 50 Ω da 50 Ω a jere kuma an haɗa su da batirin Volt 12. Ƙayyade
(a) Daidaiton juriya
(b) Wutar lantarki ta kowace resistor
Magani:
(a) R = R 1 + R 2 = 50 Ω + 50 Ω = 100 Ω.
(b) I = V / R = Volts 12 / 100 Ω = 0.12 A
