## Equivalent capacitance in series/parallel

1. The problem statement, all variables and given/known data

What is the equivalent capacitance for the following schematic?

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

2. Relevant equations

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

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

3. 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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Homework Help
 Quote by exi 1. The problem statement, all variables and given/known data What is the equivalent capacitance for the following schematic? Note that C1 is 11 µF, and C2 is 3 µF. 2. Relevant equations $$C_{parallel} = C_1 + C_2 + C_3 + ... + C_n$$ $$C_{series} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}$$ 3. 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

 Quote by Andrew Mason 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):

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

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

$$C_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{7.8824} + \frac{1}{3}} = 1.5147 \mu F$$

## Equivalent capacitance in series/parallel

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
 hi plz anser this this is not good see next post Attached Thumbnails
 plz answer this regards Attached Thumbnails

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