Q: What is the Purpose of a Capacitance Bridge & How Does it Impact Output Q?

[casanova2528]

What is the purpose of a capacitance bridge?

let's say 4 capacitors are arranged

...../.|.\
.....c1..|..c2
...____ /...0...\ _____
...|...\...|.../...|
...|...c3..|..c4...|
...|...\..|../...|
...|......|
...|......|

0 indicates the voltmeter.

if 0 indicates a potential of 0, we get this unique situation where v1 = v 3 and v2 = v4

what is the difference between this arrangement and a circuit with a set c1c2 (c1 and c2 in series) parallel with set c3c4 (c3 and c4 in series)??

which arrangement will produce more total Q output from a battery?

 

[NoTime]

Your ascii art didn't survive posting well.

If I read it right though...
Aren't c1c2 and c3c4 in parallel in your drawing?
The distinction in your question eludes me.

 

[casanova2528]

Your ascii art didn't survive posting well.

If I read it right though...
Aren't c1c2 and c3c4 in parallel in your drawing?
The distinction in your question eludes me.

yes...sets c1c2 is parallel to c3c4. c1 and c2 are in series. c3 and c4 are in series.

the capacitance bridge is different from the above arrangement. the capacitance bridge arrangement is drawn as seen where 0 indicates the voltmeter.

 

[NoTime]

How is the addition of the voltmeter supposed to change anything?
Particularly if c1=c3 and c2=c4.

 

[casanova2528]

How is the addition of the voltmeter supposed to change anything?
Particularly if c1=c3 and c2=c4.

the addition of a voltmeter in the middle is the key to a capacitance bridge. if there is a potential difference in the middle, the problem becomes very complex. the fact that the voltmeter reads zero makes v1 = v3 and v2 = v4.

remember, there can be a capacitor in the area where the voltmeter is placed...that is equal to having a potential difference on the voltmeter.

 

[berkeman]

What is the purpose of a capacitance bridge?

let's say 4 capacitors are arranged

...../.|.\
.....c1..|..c2
...____ /...0...\ _____
...|...\...|.../...|
...|...c3..|..c4...|
...|...\..|../...|
...|......|
...|......|

0 indicates the voltmeter.

if 0 indicates a potential of 0, we get this unique situation where v1 = v 3 and v2 = v4

what is the difference between this arrangement and a circuit with a set c1c2 (c1 and c2 in series) parallel with set c3c4 (c3 and c4 in series)??

which arrangement will produce more total Q output from a battery?

I would think the arrangement is analogous the the resistive Wheatstone bridge:

http://en.wikipedia.org/wiki/Wheatstone_bridge

It gives you a way to make more sensitive measurements of an impedance, IIRC.

 

[casanova2528]

I would think the arrangement is analogous the the resistive Wheatstone bridge:

http://en.wikipedia.org/wiki/Wheatstone_bridge

It gives you a way to make more sensitive measurements of an impedance, IIRC.

yes, this configuration is analogous to the wheatstone bridge. I haven't gotten there yet...next section. thus, i need some assistance in understanding this configuration in relation to total charge output by the battery compared to a similar series and parallel configuration.

does anybody here have any field experience with a wheatstone bridge or a capacitance bridge? I wonder how this configuration will affect the battery's output of current or charge.

 

[NoTime]

the addition of a voltmeter in the middle is the key to a capacitance bridge. if there is a potential difference in the middle, the problem becomes very complex. the fact that the voltmeter reads zero makes v1 = v3 and v2 = v4.

remember, there can be a capacitor in the area where the voltmeter is placed...that is equal to having a potential difference on the voltmeter.

The voltmeter, particularly the perfect kind assumed in the classroom, isn't going to make any difference to the battery's output of current or charge.
Substituting a resistor or capacitor will in the case where v1 not equal v3, but not in the case where v1 = v3.

I think what you are looking for here is an understanding of Thevenin and Norton equivalent circuits.
You may wish to do some reading on these subjects.

PS: Thanks for fixing up the artwork.

 

[casanova2528]

The voltmeter, particularly the perfect kind assumed in the classroom, isn't going to make any difference to the battery's output of current or charge.
Substituting a resistor or capacitor will in the case where v1 not equal v3, but not in the case where v1 = v3.

I think what you are looking for here is an understanding of Thevenin and Norton equivalent circuits.
You may wish to do some reading on these subjects.

PS: Thanks for fixing up the artwork.

you may be missing the point. it's true that the addition of a voltmeter is not going to alter the battery's output, but you may be missing the uniqueness of the configuration of a wheatstone bridge type set up of capacitors. if there is a variable capacitor on anyone of the capacitors, then you can adjust the capacitance so that the voltmeter in the middle is reading zero. then, you have the unique V1 = V3 and v2=v4 as opposed to V1 + V2 = V3 +V4. so, the question remains...what changes will these two different types of configuration create in a battery in terms of total charge output?

 

[NoTime]

you may be missing the point. it's true that the addition of a voltmeter is not going to alter the battery's output, but you may be missing the uniqueness of the configuration of a wheatstone bridge type set up of capacitors. if there is a variable capacitor on anyone of the capacitors, then you can adjust the capacitance so that the voltmeter in the middle is reading zero. then, you have the unique V1 = V3 and v2=v4 as opposed to V1 + V2 = V3 +V4. so, the question remains...what changes will these two different types of configuration create in a battery in terms of total charge output?

Yes, I'm very much failing to see a point here.
The only unique thing about the balanced condition is that it does not matter if the connection labeled o is an open, shorted or any other state.
There will be no change in charge.
That is, of course, an important property of the wheatstone bridge circuit in general.

 

[zeitghost]

So far as I'm aware, bridges with reactive elements are energised with alternating current rather than d.c. (as per the battery).

The detector will then need to be sensitive to A.C.

There are lots of weird & wonderful bridge circuits, the Maxwell Bridge, the Hay's Bridge, Owen's Bridge, Anderson's Bridge, Heaviside's Bridge, the Schering Bridge, & last but not least, the Wein Bridge, so beloved of audio oscillator designers.


All the above are variations on the theme by Wheatsone... with the disadvantage that the maths gets more complex because it's got phase & amplitude in it, rather than nice d.c.