Recent content by Rohit Mallya

http://www.circuitstoday.com/wp-content/uploads/2009/10/Biased-Positive-Clipper.jpg I meant as per the diagram we use a dc supply to clip at higher levels. Is there a way to clip at higher level say 3.5V without using the dc supply?

Good Day everyone. Is it possible to design a clipper circuit to bias at higher voltage level without using bias voltage? For example. to clip at 3.5V for a input signal of 10V sine without using a BIAS VOLTAGE??

Thank you for the animation Lance... But I want the P-V graph...

[FONT="Comic Sans MS"]Can anyone please show the P-V diagram of 2-stroke Petrol engine?:confused:

Sir may I know the structures of A-Carotene...??

Yes you are correct snshusat161..!

Oh Borek Sir I didn't mean to say that...! Very Sorry Sir :( One double bond is equivalent to 2 pi electrons so 3[double bonds]*2[electrons] = 6 pi electrons

and in Huckle's rule n=1.

6 pi electron in Benzene..!

Yes you are correct Ghostbuster25 the object should be placed in between the convex lens and its focal point.

[FONT="Garamond"][SIZE="5"]Yeah you are absolutely right..!

yeah...Mark Sir is correct... Eqn(1) y=kx3/y2 Eqn(2) y'=k(76.7)3/(145.5)2 Eqn(2)/Eqn(1)=> y'=21.37y

Oops...Sorry Sir... its 76.7...

x decreases BY 23.3% means x=100-23.3=76.6% So you have to sub. x=76.6. But if they had given decreases to 23.3% then x=23.3 z=145.5 substituting this in the equation with k=1 we get x=21.37 in the answer which you have mentioned they have divided it by 100...

Is the answer 21.37%?