Answer to Question #302753 in Electric Circuits for ilovephysicssolman

Question

Three capacitors are connected in series: C₁ = 28 μF, C ₂ = 42 μF and C₃ = 64 μF. What is the total capacitance?

Answer 1

Answer

total capacitance

$C=\frac{ C_1* C_2*C_3} {C_2*C_3+C_1*C_3+C_1*C_2}\\=\frac{28*42*64*10^{-18}}{(28*42+42*64+64*28) *10^{-12}}\\=13.31*10^{-6}F$