capacitors Calc's

[ICE]

Four capacitors of 5, 6,10, and 12 microfarads respectively are connected in series. What is the combined capacitance? Show your working


The capacitances in the previous question are now connected in parallel. What is the combined capacitance? Show your working


Any help would be appreciated.

Cheers

 

[telectrix]

in parallel, just add them up 5+6+10+12=33mF

in series its (C1 XC2 XC3 XC4)/ (C1+C2+C3+C4) = 3600/33= 109.0909

opposite to resistors in series/parallel

 

[accordfire]

Four capacitors of 5, 6,10, and 12 microfarads respectively are connected in series. What is the combined capacitance? Show your working


The capacitances in the previous question are now connected in parallel. What is the combined capacitance? Show your working


Any help would be appreciated.

Cheers

Okay, with workings:

Series Capacitance.....

1/(1/5 + 1/6 +1/10 +1/12) =

1/(0.2+0.16+0.1+0.08)=

1/0.54 =

Capacitance of 1.85uF.

Parallel Capacitance....

Add the values - 5+6+10+12 = 31uF.

The simple rule with capacitors is that if there are more than one in series, the TOTAL capacitance will be LESS than the value of any one alone. For capacitance in parallel, the total will be GREATER.

This is the reverse of what is true for resistors. The very simplest way of understanding this is to consider that while resistors make it harder for electrons to travel (and hence, adding them in series will lower the overall number of electrons flowing) capacitors store electrons, and by connecting them in series, you effectively slow them down and lower the number - think of a queue of people getting through a turnstile.

 

[accordfire]

in parallel, just add them up 5+6+10+12=33mF

in series its (C1 XC2 XC3 XC4)/ (C1+C2+C3+C4) = 3600/33= 109.0909

opposite to resistors in series/parallel

Yes and no mate......

The calc for series capacitance is

1
---------------
1/a + 1/b + 1C

Divide 1 by each value first - then add those totals up. Then divide 1 by that total.