Capacitors in a series

  • #1

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

The Attempt at a Solution



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

2) C2 = 2.6-C1

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

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

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
"...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.
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,394
1,045
[tex]C_s= \frac{1}{4}\ C_1[/tex]

Your equation (3) should be:

[tex]\frac{1}{\frac{1}{4}C_1}=\frac{1}{C_1}+\frac{1}{2.6-C_1}[/tex]

Of course, [tex]\frac{1}{\frac{1}{4}C_1}=\frac{4}{C_1}[/tex]
 

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