Question

Consider the figure below. C1 0.300 μF (a) Find the charge stored on each capacitor in the figure shown above (C1 = 18.7 μF, C2 = 8.72 μF, and C3 = 0.300 μF) when a 1.69 V battery is connected to the combination e. (b) What energy is stored in each capacitor? E1 = E2 =

Answer 1

$C_{12}= C_1+C_2= 27.42\: \mu F$

$C_{eq}= \frac {C_{12}C_3}{C_{12}+C_3} = \frac {27.42\times 0.300}{27.72}\: \mu F = 0.296\: \mu F$

$Q_{eq}= C_{eq}V = (0.296\times10^{-6}\times 1.69)\:C = 5.00\times 10^{-7}\: C$

$Q_{eq}= Q_3= 5.00\times 10^{-7}\: C$

$V_{3}= \frac {Q_3}{C_3} = \frac {5.00\times 10^{-7}}{0.300\times 10^{-6}}\: V = 1.67\: V$

$V_{12}= (1.69-1.67)\: V=0.02 \: V$

$Q_{1}=C_1V_{12}= 3.47\times 10^{-7}\: C$

$Q_{2}=C_2V_{12}= 1.74\times 10^{-7}\: C$

(b)

$E_1 = \frac {1}{2}C_1V_{12}^2 = 3.74\times 10^{-9}\: J$

$E_2 = \frac {1}{2}C_2V_{12}^2 = 1.74\times 10^{-9}\: J$

$E_3 = \frac {1}{2}C_3V_3^2 = 4.18\times 10^{-9}\: J$