# Equivalent capacitance in series/parallel

## 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

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

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

## 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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Andrew Mason
Homework Helper

## 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

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

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

## 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

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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.

hi plz anser this
this is not good see next post

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