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.
ScanSolve AI Answer
a) Matumizi moja ya capacitor ni: • Kuhifadhi nishati ya umeme. b) i) Uwezo jumla (Effective capacitance): Step 1: Tafuta uwezo jumla wa capacitors zilizounganishwa sambamba (2 µF na 8 µF). C_p = C_1 + C_2 C_p = 2 \, + 8 \, C_p = 10 \, Step 2: Tafuta uwezo jumla wa capacitors zilizounganishwa mfululizo (5 µF, C_p (10 µF), na 3.2 µF). (1)/(C_eff) = (1)/(C_a) + (1)/(C_p) + (1)/(C_b) (1)/(C_eff) = (1)/(5 \, ) + (1)/(10 \, ) + (1)/(3.2 \, ) (1)/(C_eff) = (1)/(5) + (1)/(10) + (10)/(32) (1)/(C_eff) = (16)/(80) + (8)/(80) + (25)/(80) (1)/(C_eff) = (16 + 8 + 25)/(80) (1)/(C_eff) = (49)/(80) C_eff = (80)/(49) \, C_eff ≈ 1.63 \, Uwezo jumla ni 1.63 . ii) Chaji kwenye 3.2 µF: Step 1: Tafuta chaji jumla kwenye mzunguko. Katika mzunguko wa mfululizo, chaji jumla ni sawa na chaji kwenye kila capacitor. Q_total = C_eff × V_total Q_total = (80)/(49) \, × 12 \, V Q_total = (960)/(49) \, Q_total ≈ 19.59 \, Step 2: Chaji kwenye capacitor ya 3.2 µF ni sawa na chaji jumla. Q_3.2 = Q_total Q_3.2 ≈ 19.59 \, Chaji kwenye 3.2 µF ni 19.59 . iii) Tofauti ya uwezo (P.d) kwenye 5 µF: Step 1: Tumia formula V = (Q)/(C) kupata tofauti ya uwezo. Chaji kwenye 5 µF ni Q_total. V_5 = Q_totalC_5 V_5 = (960)/(49) \, 5 \, V_5 = (960)/(49 × 5) \, V V_5 = (960)/(245) \, V V_5 = (192)/(49) \, V V_5 ≈ 3.92 \, V Tofauti ya uwezo kwenye 5 µF ni 3.92 V. Tuma swali linalofuata 📸