Question

Three capacitors are connected as follows: 2.8 F capacitor and 5.57 F capacitor are connected in series, then that combination is connected in parallel with a capacitor of 7.13 F. What is the capacitance of the total combination?

Answer 1

ANSWER

8.99 F

EXPLANATION

We know that two capacitors of capacitances 2.8 F and 5.57 F are connected in series, while a third capacitor of capacitance 7.13 F is connected in parallel to that combination,

The capacitance works similarly to the resistance, except that when capacitors are connected in parallel, their capacitances add up, while when they are connected in series, the equivalent capacitance is like we were finding the equivalent resistance of resistors connected in parallel,

$C_{eq}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}+C_3=\frac{1}{\frac{1}{2.8F}+\frac{1}{5.57F}}+7.13F\approx8.99F$

Hence, the total capacitance is 8.99 F, rounded to the nearest hundredth.