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Here's my circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I want to find the voltage over the 20uF capacitor; however, what I come up with is 50V which does not seem correct based on the voltage source.

Here's how I got 50V: combine the six caps (the three parallel on each side) to get a circuit as shown:

schematic

simulate this circuit

So, $$C_{eq}={1\over{ {1\over60} + {1\over20} + {1\over30} }} + 40=10+40=50uF$$ Here is where I think I'm doing something wrong: $$Q=CV=C_{eq}V=(0.00005)(20)=0.001C\;\;\;\;\;\;\;\;\;V_{20}={Q\over C_{20}}={0.001\over{0.00002}}=50V$$ That is how I got my 50V over the 20uF cap. Was that the correct way of doing this? Is that the right answer? If not what did I do wrong?

Thank you for any help! Hopefully, that is enough work shown.

EDIT: If the 40uF cap can be ignored (makes sense), then,

$$C_{eq}=10uF=0.00001F\;(found\;above)\;\;\;\;\;Q=C_{eq}*V=(0.00001)(20)=0.0002C$$ Then, $$V_{20}={Q\over C_{20}}=0.0002/0.00002=10V$$

My final answer is V=10V, is that correct? Seems a lot more reasonable. This should be right since after finding the other capacitors voltages and then doing KVL, the sum of the voltages across the caps comes out to be zero.

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    \$\begingroup\$ You can actually ignore the 40 μF capacitor entirely; nothing it does is going to affect the 20 μF one. \$\endgroup\$ – Hearth Oct 23 '18 at 23:07
  • \$\begingroup\$ Oh, then I can get the equivalent capacitance, use Q=Ceq*V to get the charge through that chain/branch, and then use that Q and C20 to find the voltage. I'll make an edit. \$\endgroup\$ – JustHeavy Oct 23 '18 at 23:10
  • \$\begingroup\$ @Felthry, can you verify my solution? (In the edit) \$\endgroup\$ – JustHeavy Oct 23 '18 at 23:23
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    \$\begingroup\$ The series capacitance of 60µF and 30µF is (60*30)/(60+30) µF= 20µF. So, it's 20µF and 20µF in series in the branch, and your 20µF cap sees half the source voltage. \$\endgroup\$ – Janka Oct 23 '18 at 23:38
  • \$\begingroup\$ Ah, good, so 20/2=10V, as I found. Thanks. \$\endgroup\$ – JustHeavy Oct 23 '18 at 23:39

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