How Does Charge Flow in a C-R Circuit After Closing the Switch?

[thunderhadron]

Hi friends,
I have a big issue in a simple C-R circuit. Thank you in advance if anyone would help me in this case.
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In this problem we have to find out after closing the switch 's' how much charge will flow from Y to X.

Well I have got the answer also but I have confusion in that only.
The doubt is as follows.

Initially I found the potential difference across 3μF, 6μF capacitors and the 3Ω
and 6Ω resistor. The readings are as follows respectively - 6 V, 3 V, 3 V and 6 V. According to these the charge on each capacitor is 18μC cause both are in series. But after closing the switch 6μF and 6Ω comes in parallel. The P.D. across and 6Ω resistor is 6 V and hence the P.D across 6μF becomes 6 V. And hence the P.D. across 3μF capacitor becomes 3 V. Now according to the relation "Q = CV" the new charges on 6μF and 3μF are 36μC & 9μC. Hence the amount of charge from Y to X is (36μC - 9μC = 27μC).

Friends my confusion is after being in parallel with 6Ω resistor why P.D. across 6μF becomes 6 V why not some other value. Someone said it is due to 6Ω is at higher potential. If that so why this theory is not applied with 3μF and 3Ω resistor, here 3μF is at higher potential 6 V.

 

[anathema]

Thank you for sharing your problem with us. I understand your confusion regarding the potential difference across the 6μF capacitor after closing the switch. Let me try to clarify this for you.

When the switch is closed, the 6μF capacitor and the 6Ω resistor are in parallel. This means that they have the same potential difference across them. Since the potential difference across the 6Ω resistor is 6V, the potential difference across the 6μF capacitor must also be 6V. This is because in a parallel circuit, the potential difference across each component is the same. So, the potential difference across the 6μF capacitor cannot be any other value than 6V in this case.

As for your question about why this theory is not applied to the 3Ω resistor and the 3μF capacitor, it is because they are in series. In a series circuit, the potential difference is divided across each component based on their resistance. So, in this case, the 3μF capacitor will have a potential difference of 3V because it has half the resistance of the 6Ω resistor. This is why the potential difference across the 3μF capacitor is not 6V, even though it is at a higher potential compared to the 6Ω resistor.

I hope this helps to clear up your confusion. If you have any further questions, please do not hesitate to ask. Good luck with your circuit!