# Answer in Electric Circuits for Ulrich #159156

Explanations & Calculations

a) Circuit diagram

b)

• Since “S_2” is kept open, no current flow through “small 10Omega” resistor.
• What happens after “S_1”set-off is some amount of charge pass-through the capacitors & saturates (assume saturates) & after no charge pass-through.
• To calculate that amount, consideration of the equivalent capacitance would be easy.

“qquadqquadnbegin{aligned}nsmall Equivalent &= small frac{5mu Ftimes(2mu F+3mu F)}{5mu F+(2mu F+3mu F)}\n&= small frac{5}{2}mu F\n&= small bold{2.5mu F}nend{aligned}”

• By “small Q=CV”

• Since the equivalent of the shunted two is also “small 5mu F”, that charge is qually distributes between the shunted two & “small 5mu F”.
• Therefore, the charge stored in “small 5mu F” = “small bold{125mu F}”
• The charge stored in “small 2mu F” = “small 125mu F times large frac{2}{(3+2)}=small bold{50mu F}”
• In “small 3mu F” = “small 125-50=bold{75mu F}”

c)

• The energy on “small 5mu F” = “large frac{Q^2}{2C}=frac{(125times10^{-6}C)^2}{2(5times10^{-6}F)}=small bold{1.563times10^{-3}J}”
• On “small 2 mu F” =“large frac{(50times10^{-6} C)^2}{2(2times10^{-6}F)}=small bold{0.625times10^{-3}J}”
• On “small 3mu F” = “large frac{(75times 10^{-6}C)^2}{2(3times 10^{-6}F)}=small bold{0.938times 10^{-3}J}”

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