Find Equivalent Capacitance for Circuit with Five Capacitors | Homework Question

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SUMMARY

The discussion focuses on calculating the equivalent capacitance for a circuit with five capacitors: C1 = C5 = 4.1 μF, C2 = 3 μF, C3 = 6.8 μF, and C4 = 3.4 μF, connected to a 12 V battery. The user correctly identifies that the equivalent capacitance between points a and b is calculated using the formula C234 = C4 + (1/C23), where C23 is the combined capacitance of C2 and C3 in parallel. The confusion arises from the treatment of C5, which can be neglected in this specific calculation as it does not influence the capacitance between points a and c due to the current's clockwise flow.

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Homework Statement



A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 4.1 μF, C2 = 3 μF, C3 = 6.8 μF, and C4 = 3.4 μF. The battery voltage is V = 12 V.

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I need to find the equivalent capacitance between points a and c.

Homework Equations



I have the answer but I am a bit confused as to how it is arrived at. See below.

The Attempt at a Solution



I found the equivalent capacitance between a and b as c234=(c4)+(1/c23)

I see next that c234 is in series with c1 and c5. Why does the equation then become Cac=(1/((1/C1)+(1/C234)) and not (1/((1/C1)+(1/C5)+(1/C234))?
 
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You can neglect c5 since it doesn't affect the capacitance at a or b since the current is flowing clockwise.
 
Apparently they're only looking for the capacitance to the right of the points a and c. This is not clear in your problem description, so perhaps the problem is stated differently in the original?
 
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gneill said:
Apparently they're only looking for the capacitance to the right of the points a and c. This is not clear in your problem description, so perhaps the problem is stated differently in the original?

This was exactly the way the problem was written, so it left me a little confused as well
 
iRaid said:
You can neglect c5 since it doesn't affect the capacitance at a or b since the current is flowing clockwise.

Current direction doesn't affect equivalent capacitance! Not unless there's a nonlinear circuit element like an open switch or a diode involved that would prevent "seeing" the capacitance.
 

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