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Question
Three capacitors of capacitance `C_1 = 3muf` , `C_2 = 6muf` , `C_3 = 10muf` , are connected to a 10V battery as shown in figure 3 below :
Calculate :
(a) Equivalent capacitance.
(b) Electrostatic potential energy stored in the system
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Solution
The equivalent capacitance of the series combinationf of `C_1` and `C_2`, is
`C_12 = (C_1.C_2)/(C_1+C_2)`
= `(3 xx 6)/((3 + 6)) muF = 2muF`
This is in parallel with `C_3`.
(a) So the equivalent capacitance of the combination of all the three capacitors, is
`C_(eq) = C_12 + C_3 = (2 + 10) = 12muF`
(b) Electrostatic PE stored in the system, is
`E = 1/2 C_(eq) . V^2`
= `1/2 xx 12muF xx (10V)^2 = 600muJ = 6 xx 10^-4 "J"`
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