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Equivalent capacitance in series/parallel 
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#1
Jul2707, 03:37 PM

P: 85

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 [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] 3. The attempt at a solution I tried the following: 1: Considered the four righthand 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? 


#2
Jul2707, 04:20 PM

Sci Advisor
HW Helper
P: 6,671

AM 


#3
Jul2707, 04:33 PM

P: 85

I've got one remaining shot at this problem, so I would much appreciate it if someone could doublecheck 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] 


#4
Jul2707, 06:25 PM

P: 356

Equivalent capacitance in series/parallel
looks good to me



#5
Jul2707, 07:05 PM

P: 85

Thanks for the lookover; 1.5147 is the answer.
Surprised to see it so low, but it's the answer. 


#6
Jun511, 11:15 AM

P: 1

somebody help me solve this question please
find eq capacitance when each capacitoras capaccitance C h 


#7
Mar1912, 01:15 AM

P: 2

hi plz anser this
this is not good see next post 


#8
Mar1912, 01:17 AM

P: 2

plz answer this
regards 


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