(a) INITIALLy total charge , Q = C V = 0.6 x 27
Qi = 16.2 uC
finally, Qf = Q1 + Q2 = (0.6 x 9) + (9 C)
= 5.4 + 9 C
Applying charge conservation,
16.2 = 5.4 + 9 C
C = 1.2 uF .........Ans
(B) Ui = (0.6 x 10^-6)(27^2)/2
Ui = 2.187 x 10^-4 J
UF = (0.6 x 10^-6 + 1.2 x 10^-6) (9^2) / 2
Uf = 7.29 x 10^-5 J
Energy lost = Ui - Uf = 1.46 x 10^-4 J
= 146 uJ
A 0.6 pF capacitor is charged to 27 V. After charging, the capacitor is disconnected from...
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