This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Huu hapa ni ufumbuzi wa maswali yaliyoonekana kwenye picha. Swali (b): Taja sababu mbili zinazoathiri upinzani wa kondakta ya metali. (alama 1) • Urefu wa kondakta. • Eneo la msalaba (cross-sectional area) la kondakta. • Aina ya nyenzo (resistivity) ya kondakta. • Joto la kondakta. Swali (c): Tumia mzunguko katika kielelezo cha 7 hapa chini kujibu maswali yanayofuata. Swali (c)(i): Kokotoa jumla ya upinzani katika mzunguko. Step 1: Kokotoa upinzani wa kwanza wa sambamba (R_p1). Kuna vipingamizi vitatu vilivyounganishwa sambamba: 10 \, , 20 \, , na 20 \, . (1)/(R_p1) = (1)/(10 \, ) + (1)/(20 \, ) + (1)/(20 \, ) (1)/(R_p1) = (2)/(20 \, ) + (1)/(20 \, ) + (1)/(20 \, ) = (4)/(20 \, ) = (1)/(5 \, ) R_p1 = 5 \, Step 2: Kokotoa upinzani wa pili wa sambamba (R_p2). Kuna vipingamizi viwili vilivyounganishwa sambamba: 10 \, na 5 \, . (1)/(R_p2) = (1)/(10 \, ) + (1)/(5 \, ) (1)/(R_p2) = (1)/(10 \, ) + (2)/(10 \, ) = (3)/(10 \, ) R_p2 = (10)/(3) \, Step 3: Kokotoa jumla ya upinzani (R_total). Vipingamizi R_p1, 3 \, , na R_p2 vimeunganishwa kwa mfululizo. R_total = R_p1 + 3 \, + R_p2 R_total = 5 \, + 3 \, + (10)/(3) \, R_total = 8 \, + (10)/(3) \, Ili kuongeza, tunahitaji denominator ya kawaida: R_total = (24)/(3) \, + (10)/(3) \, R_total = (34)/(3) \, R_total ≈ 11.33 \, Jumla ya upinzani katika mzunguko ni 11.33 . Swali (c)(ii): Tambua mkondo wa umeme unaopita kwenye kipingamizi cha 5 \, . Step 1: Kokotoa jumla ya mkondo wa umeme (I_total) kutoka chanzo. Tumia Sheria ya Ohm, I = (V)/(R). Voltage ya chanzo ni 20 \, V. I_total = (20 \, V)/(34)3 \, I_total = (20 × 3)/(34) \, A = (60)/(34) \, A = (30)/(17) \, A Step 2: Kokotoa voltage kwenye kipingamizi cha sambamba cha R_p2. Mkondo wa jumla unapita kwenye R_p2. V_p2 = I_total × R_p2 V_p2 = (30)/(17) \, A × (10)/(3) \, V_p2 = (300)/(51) \, V = (100)/(17) \, V Step 3: Kokotoa mkondo wa umeme unaopita kwenye kipingamizi cha 5 \, . Kwa kuwa vipingamizi vya 10 \, na 5 \, vimeunganishwa sambamba, voltage kwenye kila kipingamizi ni V_p2. I_5 = V_p25 \, I_5 = (100)/(17) \, V5 \, I_5 = (100)/(17 × 5) \, A = (20)/(17) \, A I_5 ≈ 1.18 \, A Mkondo wa umeme unaopita kwenye kipingamizi cha 5 \, ni 1.18 A. 3 done, 2 left today. You're making progress.