71It is certainly the potential across R.
In the adjoining figure, If the INPUT is a Square Wave, the Output wave will look like :
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12 Answers
49Are you considering the V' as output?
is it potential across R??
1Should it look like a square output (in the same phase) as well????
71Well, I don't know. Anyone who sees the problem would say that only. But It doesn't satisfies the mind. Kindly present the logic please (if any)
49Well.. i dont agree with the above post..
I ll post something to give u the feel of the problem after a few mins.. a bit busy now...
49Look it is a problem of transients... we are charging and discharging the C periodically....
Thus accordingly, the R will have some potential differential across it..
another thing to realise is the output wave form is exponential and not square...
I guess this basic background (in "somewhat" layman language) is enough?
1Yes....shubhomoy is probably right...my fault...
The graph is exponential...but the amplitude is less than the supply voltage as there is going to be a drop across the capacitor equal to magnitude I*XC.
1Perhaps u have already given the final ans. in ur previous post ... Vivek!!!!