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30.62. Model: Assume the battery is an ideal battery. Visualize: de=4 uF L + GQ 7 + CG Ov Tour L Y Ov Tsar [Cq23 C= 6 uF The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C; and C; are in series, 1 1 1 1 1 10 1 = = 4 = (iF) > CL, = BF = 24 Gan GG ap OnE 34 (HF) ‘as = RMP = 24 uF e923 Cqq23 and C; are in parallel, so Coq = Cogs + Cy = 2.4 ME +5 uF = 7.4 JE A potential difference of AV¢ = 9 V across a capacitor of equivalent capacitance 7.4 uF produces a charge QO = CuAVe = (7.4 HE)(9 V) = 66.6 uC Because Cyg is a parallel combination of C; and Cyq2s, these capacitors have AV; = AV,,,; =AV, =9 V. Thus the charges on these two capacitors are Q1=(6 HEYOV)=45 LC Oagrs = (2.4 HE) V) = 21.6 uC Because Q,_ 5 is due to a series combination of C) and C3, 0; = Q; = 21.6 uC. This means ane Se =54V AY, Bae n36V In summary, Q, = 45 “C, V, =9 V; Q) = 21.6 4, V = 5.4 V; and Q; = 21.6 uC, V3 =3.6V. 30.68. Model: Assume the battery is an ideal battery. Visualize: Please refer to Figure P30.68. The battery is connected to two series capacitors C; and Cy. Solve: The equivalent capacitance is Because the charge on capacitor C is 450 wC, the charge on C,, and C, is also 450 wC. We have Qa 450 _ (450 wC)(12 wF+C,) (450 wC)(12uF) Mac? OV=CeTELG) (2u)G 720 - 45000 uF Assess: Note that capacitors connected in series have the same charge. 
