Recent content by user1296

Ooh ok ok , didn't think of this that way Thank you!

Ohh okay ..Thank you👍

Thank you for your help! :) I just had a small doubt in the last circuit that I drew in the earlier post(#15) , If V is at a greater potential than V1 (since initially also current was flowing that way) then why does the current flow back from point V1 to V again from the right side of the loop ?

Yes I redrew it and I think I got the answer...is this the correct method ?

Ohh okayy I think I understood Q20 C option...Thank you But what about option A ? I just know what currents through the inductors won't change suddenly when the switch is opened, but I don't know how to proceed & find the potential difference... Please help

Yes yes, I've learned about LC oscillations

I'm not really sure why... I just thought maybe after infinite time the inductors and resistors will lose all energy through heat and so then there will be no current left... Just guessed not sure

Yes thank you , I understood your hint& tried redrawing the circuit and attached it in the above post... Hope I've got it right this time...For Q 20, After a long time I'm not really sure but I guess current will be 0? So then the voltmeter will also measure 0...

Oh yes yes thank you. I redrew the circuit... Now is it correct? Since the voltmeter is ideal ,current flowing through it will be 0( so in my diagram i1 = 0) , so can I say that the voltmeter is connected across the 2R resistor only and so will measure 'E ' as the potential difference?

I did try redrawing the circuit at steady state , but I'm not really sure. I have attached the circuit that I tried drawing, I assumed the branch with the capacitors to be absent at steady state since current won't flow through them anyway. With this diagram I get the correct answer for Q 19 ...