Equivalent capacitance in series/parallel

AI Thread Summary
The discussion focuses on calculating the equivalent capacitance of a circuit with capacitors in series and parallel. The initial attempt incorrectly combined capacitors, leading to an unexpected low result of 1.8051 µF. A correct approach identifies the right-most three capacitors as being in series, yielding an equivalent capacitance of 3.8824 µF, which is then added in parallel with a 4.0 µF capacitor for a total of 7.8824 µF. Finally, this value is used in series with other capacitors to find the overall equivalent capacitance of 1.5147 µF. The calculations confirm that the answer, though surprisingly low, is accurate.
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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

Homework Equations



C_{parallel} = C_1 + C_2 + C_3 + ... + C_n

\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}

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:

C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{47} + \frac{1}{3}} = 1.8051 \mu F

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

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

Homework Equations



C_{parallel} = C_1 + C_2 + C_3 + ... + C_n

C_{series} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}

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:

C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{47} + \frac{1}{3}} = 1.8051 \mu F

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:
Andrew Mason said:
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):

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

C_{rt parallel} = 3.8824 + 4 = 7.8824 \mu F

C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{7.8824} + \frac{1}{3}} = 1.5147 \mu F
 
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looks good to me
 
Thanks for the look-over; 1.5147 is the answer.

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

h
Untitled.png
 
hi please anser this
this is not good see next post
 

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please answer this



regards
 

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