Equivalent resistors/capacitors etc. tricky problem

[bobthebanana]

Homework Statement

Find equivalent capacitance between terminals

Homework Equations

C_eq = C_1 + C_2 for parallel
C_eq = 1/((1/C_1) + (1/C_2))

The Attempt at a Solution

I can't tell what's in parallel and what's in series. That conductor in the center is throwing me off.

Any help por favor?
You guys might not be able to see the image because it's pending approval. I uploaded it to imageshack:
http://img377.imageshack.us/img377/5690/53235366dg9.png

 

[tiny-tim]

Hi!

I can't tell what's in parallel and what's in series. That conductor in the center is throwing me off.

Then pluck it out!

Hint: Draw it as a figure-of-eight …

does that have the same potentials?

 

[baba89]

I can understand your confusion with the conductor in the center. However, this is a common problem in equivalent capacitance calculations and can be solved using the equations you have provided.

First, we need to identify which capacitors are in parallel and which are in series. In this case, the two capacitors on the left are in parallel and the two on the right are in series. This means we can use the equation C_eq = C_1 + C_2 to find the equivalent capacitance for the parallel capacitors and the equation C_eq = 1/((1/C_1) + (1/C_2)) for the series capacitors.

For the parallel capacitors, we get C_eq = 20μF + 20μF = 40μF.

For the series capacitors, we get C_eq = 1/((1/10μF) + (1/10μF)) = 5μF.

Now, we can combine the equivalent capacitance for the parallel capacitors (40μF) with the equivalent capacitance for the series capacitors (5μF) using the equation C_eq = C_1 + C_2, which gives us a final equivalent capacitance of 45μF between the terminals.

I hope this helps clarify the problem for you. Keep in mind that in more complex circuits, it may be helpful to redraw the circuit and identify which components are in parallel and which are in series before solving for the equivalent capacitance.