How Do Capacitors Behave in Series and Parallel Configurations?

[Rblswimmer456]

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
Answer in units of μF.

PART B: Determine the capacitance of the smaller capacitor.
Answer in units of μF.

Homework Equations

[tex]\frac{1}{C_s}[/tex] = [tex]\frac{1}{C₁}[/tex] + [tex]\frac{1}{C₂
}[/tex]

The Attempt at a Solution

1) [tex]\frac{1}{(\frac{1}{4}C_s}[/tex] = [tex]\frac{1}{C₁}[/tex] + [tex]\frac{1}{C₂
}[/tex]

2) C₂ = 2.6-C₁

3) [tex]\frac{1}{(\frac{1}{4}C_s}[/tex] = [tex]\frac{1}{C₁}[/tex] + [tex]\frac{1}{2.1-C₁
}[/tex]

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

 

[gneill]

"...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 C_s in your equation for the equivalent capacitance of the series connection. With C_s out of the picture you should be able to solve for C₁ and C₂.

 

[SammyS]

$$C_s= \frac{1}{4}\ C_1$$

∴ Your equation (3) should be:

$$\frac{1}{\frac{1}{4}C_1}=\frac{1}{C_1}+\frac{1}{2.6-C_1}$$

Of course, $$\frac{1}{\frac{1}{4}C_1}=\frac{4}{C_1}$$