# Capacitors in a series

## Homework Statement

When two capacitors are connected parallel,
the equivalent capacitance is 2.6 μF. If the
same capacitors are reconnected in series, the
equivalent capacitance is one-fourth the ca-
pacitance of one of the two capacitors.

PART A: Determine the capacitance of the larger capacitor

PART B: Determine the capacitance of the smaller capacitor.

## Homework Equations

$$\frac{1}{Cs}$$ = $$\frac{1}{C1}$$ + $$\frac{1}{C2 }$$

## The Attempt at a Solution

1) $$\frac{1}{(\frac{1}{4}Cs}$$ = $$\frac{1}{C1}$$ + $$\frac{1}{C2 }$$

2) C2 = 2.6-C1

3) $$\frac{1}{(\frac{1}{4}Cs}$$ = $$\frac{1}{C1}$$ + $$\frac{1}{2.1-C1 }$$

..and I have no idea how to go from here?

gneill
Mentor
"...the equivalent capacitance is one-fourth the capacitance of one of the two capacitors."

You need to pick one of the capacitors, C1 or C2 (it doesn't matter which) and take 1/4 of it, to replace Cs in your equation for the equivalent capacitance of the series connection. With Cs out of the picture you should be able to solve for C1 and C2.

SammyS
Staff Emeritus
$$C_s= \frac{1}{4}\ C_1$$
$$\frac{1}{\frac{1}{4}C_1}=\frac{1}{C_1}+\frac{1}{2.6-C_1}$$
Of course, $$\frac{1}{\frac{1}{4}C_1}=\frac{4}{C_1}$$