Find the charge stored on each capacitor in Figure 18.24 (C_1 = 10.0 mu F, C_2 =

Question

Find the charge stored on each capacitor in Figure 18.24 (C_1 = 10.0 mu F, C_2 = 6.00 mu F) when a 1.54 V battery is connected to the combination. C_1__________c C_2___________c 0.300 mu F capacitor_________c What energy is stored in each capacitor? C_1___________J C_2__________J 0.300 mu F capacitor___________J Additional Materials What total energy is stored in the capacitors in Figure 18.25 (C_1 = 0.600 mu F, C_2 = 22.0 mu F) if 1.80 times 10^-4 J is stored in the 2.50 muF capacitor?__________J

Explanation / Answer

1.)

C1 and C2 are connected in parallel

so net capacitance = C1 + C2 = 16 uF

16 uF and 0.3 uF are connected in series

so net capacitace = 16 uF x 0.3 uF / ( 16 + 0.3)uF = 0.294 uF

Net Charge (Q) = C/V = 0.294 uF / 1.54 V = 1.90x10^-7 C

charge on 0.3uF = 1.90 x 10^-7 C

charge on C1 (Q1) = (Q1,2)C1/(C1+C2) = 1.90 x 10^-7 C x 10 x10^-6 / (16 x 10^-6 )

Q1 = 1.1875 x10^-7 C

charge on capacitance C2 (Q2) = (Q1,2)C2/(C1+C2) = 1.90 x 10^-7 C x 6 x10^-6 / (16 x 10^-6 ) = 1.125x10^-7 C

Energy stored in capacitor

U = 0.5 CV^2 = 0.5 (Q1)^2 / C1

U1 = 0.5 * (1.1875 x10^-7)^2 / 10^-5 = 1.75 x10^-9 J

U2 = 0.5 CV^2 = 0.5 (Q2)^2 / C2

U1 = 0.5 * (1.125x10^-7)^2 / 6x10^-6 = 1.30 x10^-9 J

energy in 0.3 uF capactior

U3 = 0.5 CV^2 = 0.5 (Q3)^2 / C3

U1 = 0.5 * (1.90 x 10^-7 C)^2 / 0.3x10^-6 = 6.016 x10^-8 J

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