# A 7.0 µF and a 3.0 µF capacitor are connected in series and this combination is connected in parallel with a 3.9 µF capacitor. What is the net capacitance? in µF

May 16, 2014

To add this combination of capacitors, you first have to add the two series capacitors together using the equation

$\frac{1}{{C}_{s e r i e s}} = \frac{1}{{C}_{1}} + \frac{1}{{C}_{2}} + \ldots$

so you have:

1/(C_(series)) = 1/(7 µF) + 1/(3 µF) = 3/(21 µF) + 7/(21 µF) = 10/(21 µF)

so C_(series) = (21 µF)/10 = 2.1 µF

Then you add this series combination to the parallel capacitor using the equation:

${C}_{p a r} = {C}_{1} + {C}_{2} + \ldots$

so

C_(eq) = 2.1 µF + 3.9 µF = 6 µF