What is the Correct Rule for Voltage in a Parallel Circuit?

[Daniiel]

Homework Statement

[PLAIN]http://img37.imageshack.us/img37/2208/43252578.jpg

Ok so
i did this questions and was satisfied with my answer, but some guy rekons i did it wrong so i just want to double check

First i found the total capacitance, so i got 1/12 + 1/12 = 1/6 = 6 (for the series in the paralell)
then added that series to the parallel capacitor, 6+12 = 18
then added that in series to the final capacitor 1/18+1/12 = blah = 7.2
so the total capacitance is 7.2, to find the charge i times it by the voltage
7.2 x 12 = 86.4
then with this i found the voltage through the parallel circuit, 86.4/18 = 4.8
then the same with the lone capacitor 86.4 / 12 = 7.2
7.2 + 4.8 = 12
so voltage in = voltage out... is that a correct rule ? that's what he said i did wrong
that voltage in doesn't = voltage out
but with that
i halfed 4.8 because they're parallel , so 2.4 on the top part and 2.4 on the bottom part of the parallel circuit
and the lone capacitor has a V = 7.2

does that all seem right? or is voltage in not supposed to equal voltage out?
the guy got like 6 6 3 3 for his voltages which is 18V total
i dono it doesn't seem right to me at all but I am probably wrong

 

[kuruman]

Homework Statement

[PLAIN]http://img37.imageshack.us/img37/2208/43252578.jpg

Ok so
i did this questions and was satisfied with my answer, but some guy rekons i did it wrong so i just want to double check

First i found the total capacitance, so i got 1/12 + 1/12 = 1/6 = 6 (for the series in the paralell)
then added that series to the parallel capacitor, 6+12 = 18
then added that in series to the final capacitor 1/18+1/12 = blah = 7.2
so the total capacitance is 7.2, to find the charge i times it by the voltage
7.2 x 12 = 86.4
then with this i found the voltage through the parallel circuit, 86.4/18 = 4.8
then the same with the lone capacitor 86.4 / 12 = 7.2
7.2 + 4.8 = 12
so voltage in = voltage out... is that a correct rule ? that's what he said i did wrong
that voltage in doesn't = voltage out

I am not sure what you mean by "voltage in = voltage out". Nevertheless, you did this right. The 7.2 μF equivalent capacitance will have 4.8 V across it and the lone 12 μF capacitor will have 7.2 V across it. It is correct that the sum of potential differences should add up to the overall difference of 12 V.

but with that
i halfed 4.8 because they're parallel ,

This is where you went wrong. Given that the equivalent, three-capacitor, combination has 4.8 V across it, you can't just say that half the voltage is across each branch. "Parallel" means that each branch has the same voltage across it. In other words, the 12 μF capacitor in the left branch has the same voltage across as the two 12 μF series combination. So how much voltage is across each capacitor in the combination?

 

[Daniiel]

2.4? cause if they have 2.4 it all adds up to 12
i keep getting confused
and would the two capcitors in series (of the three combination) have the same potential across them?

 

[kuruman]

Look at your drawing. Add point "C" between the lone capacitor and the combination. This will help you become unconfused. You have already established that V_BC = 7.2 V and that V_CA = 4.8 V. What is the potential difference across the one capacitor (left branch) in the combination?

 

[Daniiel]

7.2 + 4.8 = 12
is that what you mean?
so doesn't the capacitor in the left branch of the combination have to be <4.8 because if it was bigger then the total potential will be >12
i thought when a voltage goes into a parallel circuit its halved, or 1/3 depending on how many parallells there are
like equally split across the paralells?

 

[kuruman]

You are confused. The parallel combination has 4.8 V across it. "Parallel" means that the each branch in the combination has the same voltage across it. So the branch with the one capacitor (left) has 4.8 V across it and the branch with the two capacitors (right) also has 4.8 V across it. So

The voltage across the single capacitor on the left is (you guessed it) 4.8 V.
The voltage across the two capacitors combined is also 4.8 V, but the voltage across each of the two is not. Since they are in series they share the same charge. Now Q = CV which means that you can write Q = C₁V₁ = C₂V₂. In other words, C₁V₁ = C₂V₂. If it so happens that C₁ = C₂, then you can say that the 4.8 V is equally split between the two because that implies that C₁V₁ = C₂V<sub>2[/SUB1 = V2. That's the case here, but in general it is not.

The voltage is not "equally split between the parallels." If you have to memorize a rule, "the voltage is split equally among the series capacitors but only if they all have the same capacitance."</sub>

 

[Daniiel]

oh mad and when you do it
with the math it works out aswell
q = 4.8 x 6
28.8/12 = 2.4
so you wouldn't add it up like 2.4 + 2.4 + 4.8 + 7.2
itd just be 7.2 + 4.8 because the 4.8 includes everything in that circuit?
sorry i must of listened to someone say the wrong rule and believed it was true since i heard it
is current split equally through paralells or same thing?
thanks a lot

 

[kuruman]

oh mad and when you do it
with the math it works out aswell
q = 4.8 x 6
28.8/12 = 2.4
so you wouldn't add it up like 2.4 + 2.4 + 4.8 + 7.2
itd just be 7.2 + 4.8 because the 4.8 includes everything in that circuit?

Yes. Or you could go from a to b adding voltages across individual capacitors following one of the following paths.

Left path 4.8 + 7.2 = 12
Right path 2.4 + 2.4 + 7.2 = 12

You see how it works now?

sorry i must of listened to someone say the wrong rule and believed it was true since i heard it
is current split equally through paralells or same thing?
thanks a lot

There is no current here because you have fully charged capacitors, but yes, if you have resistors instead of capacitors, the current is split between two resistors in parallel according to I₁R₁ = I₂R₂. It will be equally split only when R₁ = R₂.

 

[Daniiel]

ohh i understand
so that's what i got it confused with
thanks a lot for all your help