# Circuit capacitance question

Given the following circuit, what ways there are to calculate its total capacitance?

Thing is this is not a simple case, and I can't really decide which part is parallel and which is a series, since this all seems to be rather intertwined, and nor do I see the order in which this should be solved.

• Have you tried tracing the current flow with your finger? Mar 18, 2014 at 21:52
• I had, and there are quite a few options for the current here to go. Mar 18, 2014 at 21:52
• "Options" means "parallel". Mar 18, 2014 at 21:53
• So you are saying there are indeed 3 such options? Mar 18, 2014 at 21:54
• There is one unambiguous way to resolve the circuit. Mar 18, 2014 at 21:54

The total capacitance can be calculated in just a few steps using the basic rules for combining capcitors in series and parallel.

To get you started, three 6 uF capcitors in series is equivalent to a 2 uF capacitor.

And a 2 uF capacitor in parallel with a 4 uF capacitor is equivalent to a 6 uF capacitor.

simulate this circuit – Schematic created using CircuitLab

$C_7$, $C_9$, $C_8$ - are in series, thus: $$\frac{1}{C_7} + \frac{1}{C_9} + \frac{1}{C_8} = \frac{1}{C_a}$$

simulate this circuit

$C_6$ is parallel to $C_a$: $$C_6 + C_a = C_b$$

simulate this circuit

$C_4$, $C_b$, $C_5$ - are in series, thus: $$\frac{1}{C_4} + \frac{1}{C_b} + \frac{1}{C_5} = \frac{1}{C_c}$$

Repeat until done.

Actually it's not hard to see that the answer is 2 µF in a few seconds just from inspection. The three caps at the right end are in series, so 2 µF total. That is in parallel with the forth cap from the end to make 6 µF. Now the remaining right end is the same problem you just solved, which means the 4 caps at right resolve to 6 µF. That yields three 6 µF caps in series for a total of 2 µF.

Yes, it really is that easy.

I can't really decide which part is parallel and which is a series

Series connected circuit elements have identical current through, parallel connected circuit elements have identical voltage across.

Observe that the three right-most $6 \mu F$ capacitors have, by KCL, identical current through so they are in series. Combine the three series connected capacitors into one capacitor with an equivalent capacitance.

Then, observe that this equivalent capacitor is connected across the right-most $4\mu F$ capacitor so they have identical voltage across. Combine these parallel connected capacitors into one capacitor with an equivalent capacitance.

Now, this equivalent capacitor is clearly series connected with the two middle $6\mu F$ capacitors...

Do you see what to do and how to further proceed?

I find that for circuit problems like this, if you don't have the intuition yet for how to start subdividing the circuit to calculate the capacitance, it helps greatly to redraw the circuit in a way that makes it more clear.

Here is for example one way to redraw it.

As you can see, the most logical way to start is to combine C3, C6, and C9 into one equivalent capacitor using the series rule

$$C_{eq} = \frac{1}{ \dfrac{1}{C_3} + \dfrac{1}{C_9} + \dfrac{1}{C_{10}} }$$

in this case because the values are the same, it is 1/3 of 6uF = 2uF.

Then you combine that new capacitance (call it C10) with C5, and so on.

• I find the redrawn circuit much more confusing :D
– Rev
Oct 14, 2014 at 12:24