Question

A 3.00-µF and 4.00-µF capacitor are connected in series, and this combination is connected in parallel with a 2.00-µF capacitor. (a) Show the illustration (b) What is the net capacitance?

Answer 1

Solution;

The two in series;

$\frac{1}{R_T}=\frac 13+\frac14=\frac{7}{12}$

$R_T=\frac{12}{7}u F$

They are then connected in parallel with 2uF;

$R_{net}=\frac{12}{7}+2=3.174uF$