Question

Two capacitors 3uf and 6uf arranged in series are connected in parallel with a third capacitor 4uf the arrangement is connected to a 6v battery calculate the total energy stored in the capacitors

Answer 1

Answer:

$c = \ \frac{c1 \times c2}{c1 + c2}$

$c = \frac{3 \times 6}{3 + 6} = 2uF \:$

$Ct = C + C3 = 2 + 4 = 6uF. \:$

$Total Energy \: = 0.5C \ \times { \: E }^{2}$

$0.5 \times 6 \times {6}^{2} = 108J. \:$