- #1
Soren4
- 128
- 2
Homework Statement
Calculate the equivalent capacity of the capacitor system between points ##x## and ##y##
Homework Equations
##C=q/V##
The Attempt at a Solution
I'm asking about this problem because I know how to calculate ##V_x-V_y##, but I don't know how can I extract the equivalent capacity from there.
Using Kirchoff Laws I get the following equations
- $$V_x-V_y=q_1/C_1+q_2/C_2+q_3/C_3 $$
- $$V_x-V_y=q_4/C_4+q_3/C_3 $$
- $$V_x-V_y=q_1/C_1+q_5/C_5 $$
- $$0=-q_1/C_1-q_2/C_2+q_4/C_4 $$
- $$0=q_2/C_2+q_3/C_3-q_5/C_5 $$
I can solve for ##V_x-V_y##, but then here it comes the doubt.
As said I have three different ways to go from ##x## to ##y## (the first three equations are there because of that), I highlighted them in the picture (path ##A## is equation ##1##, path ##B## is equation ##2##, path ##C## is equation ##3##)
So can I express the equilvalent capacitance of the system as the ratio of the charge met following the particular path and ##V_x-V_y##? That would be
$$C_{equivalent}=(q_1+q_2+q_3)/V_x-V_y=(q_4+q_3)/V_x-V_y=(q_1+q_5)/V_x-V_y$$
This does not look correct at all, but it is the only reasonable way to use the potential difference to calculate capacity.
So I ask: what is the right way to link ##V_x-V_y## to ##C_{equivalent}##? That is : how to calculate ##C_{equivalent}## knowing ##V_x-V_y##?
I would like to use this method (using ##V_x-V_y##) and not other ones (if there are any) because I would like to understand how it should work, so please help me figure out the link between ##V_x-V_y## and ##C_{equivalent}## in cases like this one.