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Equivalent capacitance in series/parallel

  • Thread starter exi
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  • #1
exi
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Homework Statement



What is the equivalent capacitance for the following schematic?

Note that C1 is 11 µF, and C2 is 3 µF.

http://img511.imageshack.us/img511/8051/questionzi5.png [Broken]

Homework Equations



[tex]C_{parallel} = C_1 + C_2 + C_3 + ... + C_n[/tex]

[tex]\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}[/tex]

The Attempt at a Solution



I tried the following:

1: Considered the four right-hand caps to be in parallel and add them as above, yielding 47 µF, and:

2: Added that 47 µF cap in series with the remaining 5 µF and 3 µF caps like so:

[tex]C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{47} + \frac{1}{3}} = 1.8051 \mu F[/tex]

But no go (and appropriately so, since the number seems a bit small). Where am I making a mistake?
 
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Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,560
323

Homework Statement



What is the equivalent capacitance for the following schematic?

Note that C1 is 11 µF, and C2 is 3 µF.

http://img511.imageshack.us/img511/8051/questionzi5.png [Broken]

Homework Equations



[tex]C_{parallel} = C_1 + C_2 + C_3 + ... + C_n[/tex]

[tex]C_{series} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}[/tex]

The Attempt at a Solution



I tried the following:

1: Considered the four right-hand caps to be in parallel and add them as above, yielding 47 µF, and:

2: Added that 47 µF cap in series with the remaining 5 µF and 3 µF caps like so:

[tex]C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{47} + \frac{1}{3}} = 1.8051 \mu F[/tex]

But no go (and appropriately so, since the number seems a bit small). Where am I making a mistake?
The right-most three capacitors are in series. Find the equivalent capacitance of those three. That capacitance is then in parallel with the 4.0 uf capacitor. Find the equivalent capacitance for the parallel part. That is in series with the other two.

AM
 
Last edited by a moderator:
  • #3
exi
85
0
The right-most three capacitors are in series. Find the equivalent capacitance of those three. That capacitance is then in parallel with the 4.0 uf capacitor. Find the equivalent capacitance for the parallel part. That is in series with the other two.

AM
That makes a hell of a lot of sense.

I've got one remaining shot at this problem, so I would much appreciate it if someone could double-check me on this conceptually (oddly, the number still seems low at first glance, but I'm completely new to caps):

[tex]C_{rt series} = \frac{1}{\frac{1}{24} + \frac{1}{11} + \frac{1}{8}} = 3.8824 \mu F[/tex]

[tex]C_{rt parallel} = 3.8824 + 4 = 7.8824 \mu F[/tex]

[tex]C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{7.8824} + \frac{1}{3}} = 1.5147 \mu F[/tex]
 
Last edited:
  • #4
350
0
looks good to me
 
  • #5
exi
85
0
Thanks for the look-over; 1.5147 is the answer.

Surprised to see it so low, but it's the answer.
 
  • #6
somebody help me solve this question please
find eq capacitance when each capacitoras capaccitance C

h Untitled.png
 
  • #7
hi plz anser this
this is not good see next post
 

Attachments

  • #8
plz answer this



regards
 

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