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Question
Two capacitors of capacitances 4⋅0 µF and 6⋅0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery.
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Solution
Given :
`C_1 = 4 "uF"`
`C_2 = 6 "uF"`
`V = 20 V`
Now,
The equivalent capacitance is given by
`C_(eq) =( C_1C_2)/(C_1+C_2) = (4 xx 6)/(4+6) = 2.4 "uF"`
The energy supplied by the battery is given by
`E = C_(eq) V^2`
= `(2.4 "uF") xx (20)^2`
= `(2.4 "uF") xx 400 = 960 "uJ"`
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