Question

What is the equivalent capacitance of the three capacitors in Figure EX26.27?

Short answer

The equivalent capacitance of the three capacitors is$7.5\mu F$.

Step-by-step solution

Given information

We need to find that the equivalent capacitance of thre capacitors.

Simplify 

Consider that the capacitance of the two capacitors connected in parallel, to the left-bottom: their common capacitance will be

$C_b=20+10=30\mu F$.

From this, our problem is simplified to a $30\mu F$capacitor connected in series with a $7.5\mu F$one. That means, the total capacitance is

$C=\frac{30·{10}^{-6}·10·{10}^{-6}}{30·{10}^{-6}+10·{10}^{-6}}=7.5\mu F$