QUESTION: In the image they take the polarity of vo opposite to vi . .why?
ANSWER: Hello Muhammad,
I think there must be some information I am missing. Plus and minus signs on the capacitor indicate that it is charged such that the left end is positive with respect to the right end. But that would make the terminal at the top left (near the Vi symbol) more positive than the terminal at the top right (near the Vo symbol). And the plus and minus signs next to the Vi and Vo symbols indicate the opposite.
It's not clear how the capacitor got charged. There is a vague indication that there is at least 1 diode involved here, so I suspect that is involved in how the capacitor got its charge - but that does not explain the polarity discrepancy. I don't see an indication of where the diode was connected. I was unfamiliar with the initials RB. I will assume RB stands for reversed bias? But that assumption does not help.
The best I can suggest is that perhaps the plus and minus signs next to the Vi and Vo symbols indicate the polarity for the value of whatever voltage is found on Vi and Vo. (Similar to the sign on the value of the acceleration due to gravity. In some solutions it is advantageous to make upward be the positive direction and in other solutions it is advantageous to make downward be the positive direction.) So Vi would have a positive value only if the terminal at top left is negative with respect to the terminal at the bottom left. With that interpretation, and with the capacitor charged as indicated in the sketch, both Vi and Vo would have negative values.
If you can send a follow-up to this question to help me understand the situation, I will see if I can be more helpful to you.
---------- FOLLOW-UP ----------
QUESTION: Sir ,its a clamper circuit,RB is actually reverse biase,in this step the diode act as open so he defined it by removing so that it will be clear
It appears that an AC signal is applied to Vi. Ignoring the diode for a moment, the AC signal would pass through the capacitor (assuming the frequency is high enough) but would lose its DC reference. Go to this web site:
Scroll down about 40% of the way to "Negative Clamper". Notice that the waveshape at Vo has not changed but it obtained a negative DC shift. And notice that the +- polarity indication next to the capacitor's plates agrees with the polarity indication on the capacitor on your image. That DC voltage across the capacitor could be measured with a DC voltmeter.
I believe your image is a snapshot of a time when the AC input is negative. That would be why it shows - on top at the Vi end. That is the input to the clamp. The output of the clamp at the right end, labeled Vo. We still have the disagreement that the web site's plot of the signal at Vo has negative polarity - opposite that indicated on your image. Or is that what they are trying to indicate? Check the polarity symbols near the Vo terminals on the image you sent me. Perhaps the +- closest to the terminals defines the polarity of how to connect those terminals to a meter, or to more of the system. (One being 'signal' and the other 'ground'.) I am suggesting that the - at lower right indicates that this is the ground terminal. There is also a dash(?) just under the "Vo". Perhaps the dash indicates that at the moment that Vi is negative, Vo is also negative. You will have to judge if that explanation makes sense in your context.
I hope this helps,