Question

011 10.0 points Consider the capacitor circuit 5.7 uF 9.7xF 9.7 uF 42.7H well.2 HE 9.7 uFl15.7 F 69.7 F EB Find the equivalent capacitance between points a and c for the group of capacitors connected as shown. Answer in units of F.

Answer 1

when n are capacitore , having capacitance C1, C2-..cn connected in series then their equivalent given by, can be capacitance G + + +1 Ch ㅓㅏ I ca Ceq G Ch E If they are connected in parallel, G Ceg - Cat Cat + Ch (2 HE Cn HH € band ca are NOW: Between both capacitors [ 9.7 uF) connected in parallel so their equivalent Capacitance, Сьс 9.7 t 9. 7 Cbc = 19.4 F a and Between 9.7 elf) equivalent are b, upper capacitors (5.7 uF and connected in series. So their Capacitance, ata 1.7 +5.7 = + 2 Cabi 5.7 9.7 9. 5.7 * 9.7 Cabi 5.7 x 9. 7 5.7 + 9.7 Cabl = 3.59 ef similarly for lower capacitances (9.4 wf and S. 4 uf). Cabe 3.59 MF Now he have circuit 3

Cabia = 3.59 uf ㅓㅏ 2.7 46 th b Cbc = 19.4 MG a HI Cab2 = 3.59 MF EB Now a ( canla between and b , three capacitances Cabz and 2.7 46) 2.7 uF) are connected in parallel. equivalent capacitance, Their Caь - Cabi + Cana + 2.7 -3.594 3.59 of 2.7 Cab = 9.88 uf We will have now circuit Сьс Cab th Ah NEO С then, Equivalent of capacitance circuit is Caь + Сьс I + of = 1 Cho Cab Сqь Сьс C = Caь Сьс 9.88x19.4 9.88 + 19.4 Cara Coc C = 6.55 uf

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