difference between circuits with capacitors in series and parallel?

Question

CAPACITORS IN DC CIRCUITS, WORK OUT VOLTAGE STORED, AND VALUE OF CAPACITANCE, ETC?

Answer 1

Ware has it 1/2 right, caps in parallel add for total capacitance, but in series capacitance is reduced in exactly the same way as resistances are reduced when in parallel. the formula for caps in series is:

C-total= 1/(1/C1 + 1/C2 + 1/C3 +...)

For 2 caps in series, you can use:

C-total= C1*C2/(C1+C2)

If all the caps are equal, then in series their total capacitance is the capacitance of one cap divided by the number of caps.

To see why this is so, consider that the capacitance of a cap is inversely proportional to the thickness of the insulator. A string of 3 equal caps in series looks like this:

-------| D |--------| D |--------| D |--------

(Let "D" represent the dielectric insulator) Now notice that the right side of the first cap is connected directly to the left side of the second cap. As well the right side of the second cap to the left side of the third. So you could simply replace the string of three with a single cap of dielectric thickness of 3:

----------| DDD |--------

Since capacitance is inversely proportional to dielectric thickness, this equivalent cap has 1/3 the capacitance of one of the original 3 caps, right? On the plus side, since you've divided the supply voltage across 3 equal caps, the voltage rating for the 3 is 3 times higher than that of one of the caps - in other words, you can use 3 15-uF, 5-volt caps in series for a 15-volt supply, but the total capacitance will only be 5uF*.

Is cool?


*Actually, in practice you should derate cap voltages by some amount, 33-50% is pretty common, so for the above example, plan on perhaps 3 7-volt caps if a single 20-30V 5uF is not available.

Answer 2

I must correct a previous answer.

Two caps in series

If value of caps is x F
Series value is y
1/y = 1/x + 1/x

So if caps are 1 uF , (10^-6 F)

1/y = 1/(10^-6) + 1/(10^-6)
1/y = 10^6 + 10^6
1/y = 2 x 10^6

y = 1/(2 X 10^6)
y = 0.5 X 10^-6


value is 0.5uF

Answer 3

Caps in series:
Voltage is ADDED
Capacitance is lowest of the two

Caps in parallel:
Voltage is lowest of the two
Capacitance is ADDED

I.E. ----------------------------------------------
two 1uF, 10v caps in series:
Voltage is 20v
Capacitance is 1uF

two 1uF, 10v caps in parallel:
Voltage is 10v
Capacitance is 2uF

Answer 4

a conductor is working as switch ...
and no diffrence