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Archived What is the equivalent capcitance of this circuit! - Figured it out.

What is the equivalent capcitance of this circuit! -- Figured it out.

I have a Physics final exam in 2 hours and I'm re-reviewing things....

This problem DOESN'T make any sense!!

HELP_zps7fc6f1a6.png


17. If C1 = 2 F, C2 = 3 F, C3 = 5 F, C4 = 6 F, the equivalent capacitance of the circuit is:


A. 8.5 F is supposed to be the answer - that makes NO sense
B. 16 F
C. 5.5 F
D. 10 F

I didn't get ANY of those numbers...
This is what I did.

1) Determine PARALLEL capacitance of C1 and C2

Ceq = C1 + C2
Ceq = 2 +3
Ceq = 5

2) Determine the SERIES capacitance of the whole circuit now

1/Ceq = 1/5 + 1/C3 + 1/C4
1/Ceq = 1/5 + 1/5+ 1/6
1/Ceq = 0.567
Ceq = 1/0.567
Ceq = 1.763 :(:(:(:(!!!!!


Edit:

Oh wait nvm.

I figured it out.

the upper arm (C net) is- 1/C net = 1/5 + 1/C3 = 1/5 + 1/5 =2/5
=>C net = 5/2 = 2.5
now 2.5 and 6 in parallel- 2.5+6 = 8.5
 
Last edited:

Bystander

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Claims he's figured it out. Looks good to me.
 

epenguin

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Stream of consciousness thinking, charges in parallel add up so capacitances in parallel add up. So the capacitance of the combination C1 and C2 is 2 + 3 = 5. They've made it easy for you, you have two equal capacitances in series. So the voltage across one of them Is half the total voltage. So the charge on this one is 5/2 = 2.5. Add that to the 6 on C4 and you have got 8.5. Quite fast.

I haven't put in pedantries like units :oldbiggrin: because this is stream of consciousness thinking for a quiz.

Just remember - it makes sense I hope - that charges in parallel add up whereas the voltages in series and up. The charge on the leftmost plate in series equals that on the rightmost plate (though opposite sign) and that is the total charge on the equivalent capacitor. The internal plates not connected to conductor have total zero charge (though separated into a positive and negative one).

Because of this if you have say equal capacitances in series the total capacitance is less than that of any one of them. Not realising that is why you thought the given answer didn't make any sense.
 
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I know you exam is over but if you are interested.....



BIG EASY HINT: to add L and C values..

You would total L's just like Resistors...

Totaling C's are opposite of resistors.. ( parallel C's you add like series R's, and series C's you combine (add) like parallel R's..1/1/C1+1/C2+1/C3.....

IOW: You would total parallel C's as you would add series R's in a network. and when you total series C's in series, total them like you would total parallel R's.



Again,

- Treat the totaling of parallel capacitor values, opposite mathematically, as if they were like series resistor values, but only when 'totaling' their values, not their AC reactances!1.
- Treat the totaling of series or parallel Inductor values as if they were resistors.


img527.png






I always suggested to start with parallel capacitors because it's easy to see in your mind's eye...:oldwink:

Slide C2 up to the 'plates' of C1 note the combined total capacitance doubles in value, in plate area so to speak.


What you are doing is actually, or electronically is combining the plate area twice, as to increase the capacitance twice..:oldwink:

Hence, L's are like R's and C's are opposite of R's, when totaling.

1 This is a completely different story for a different time.
 

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