Intuitive Math – Understanding Angular Bisectors.

Here’s a small article on the real logic behind angular bisection which no teacher would have told you

Link : http://iamsuhasm.wordpress.com/tutsproj/intuitive-math-angular-bisectors/

Proof for the Maximum Power Transfer Theorem

I was reading a copy of Art of Electronics that i had borrowed from the IIWC librar when i saw a very interesting problem of which i had never thought of :

Prove that maximum power is transferred to the load only when source impedance is equal to load impedance.

A.K.A. the maximum power transfer theorem.

Instead of blindly googling the proof , i decided to find the proof myself. Thankfully , AOE gave a hint that the proof involved calculus. Otherwise , it is doubtful if my math challenged mind would have been able to work this out.

Not finding any ‘noob – understandable’ proofs for the MPTT on the internet , i have posted the proof here.

The problem involves finding when maximum power is delivered to load L.

It is pretty clear that power delivered to the load will be zero when both Rl is 0 and infinity. So , the maximum power should be transferred somewhere in between.

At the point where the power transfer peaks , the tangent to that point will be horizontal. Therefore , the derivative at that point will be zero.

So , we can find the point of maximum power transfer by finding the maxima of the derivative of the power function.

First step : Find the Power Function

The power transferred to the load will be equal to the voltage across it times the current in it.

Rl and Rs form a voltage divider , hence the voltage across Rl will be

The current through Rl will be

So , the power dissipated by Rl will be the product of the above two equations bcoz P=IV.

Second Step : Differentiate the Power Function.

if we try to differentiate this entire equation , it will result in a reasonably complex derivative for my noob brain due to the presence of the variables V and Rs.

Instead , to simply the process , i have taken V and Rs as two constants and then have found the derivative.

I have taken V = 5V and  Rs = 7Ω.

So , we get the Power as :

When this is differentiated , you get the derivative as :

Third Step : Find the Maxima

Setting the derivative to zero to find the maxima :

The solution to this equation is 7 !! Which is precisely the same as the value of Rs !!!

If you find the solution to the equation by taking Rl and V differently , you will still find that Rl = Rs each and everytime!!

So , it is indeed true that maximum power is transferred when source resistance = load resistance.

Finding the area of a circle by Integration

Every calculus tutorial that i have read , shows the derivation of  πr^2 by either adding together tiny rings or by cutting up the circle into triangles and integrating. So , i thought why not derive the formula , by actually integrating the equation of a circle? And voila to my surprise , it actually worked (I’m still  a calculus noob)

So , first we need the function for a circle. Every point on the circle is equidistant from the origin. Hence the equation is clearly x^2 + y^2 = r^2 , where r is the required radius.

Rearranging to get a proper function ,

Now , if i take this function and plot it , it should ideally give me a circle. And it does. Unfortunately , for reasons unknown , Microsoft math is seeming to clip off the negative half of the circle. I dont know why. But that really doesnt matter  as i will only be taking the part of the circle in the 1st quarant and will be finding its area. Then we can multiply the area by 4 to get the area of the whole circle.

Now since i have my function , i can find the area of the circle by integrating the function between o and r. This integration will give me the area of a quarter of a circle. We can multiply that area by 4 to get the full area.

Ok. To integrate , we need to find the antiderivative of the circle function. Microsoft math tells me that it is :

The Anti derivative

Now , let me call this function F(x).

Finding F(r) – F(0) should give me the area.

F(0) is very clearly 0 , as both the terms in the equation become zero.

F(r) is

as arctan of  (x/sqrt(0)) is apparently 90 = π/2 radians. (I dont know why. Maybe by taking the limit of the function?)

and there you have it folks , ( πr^2 /2) , which is (πr^2/4)  !!

Which is precisely the area of a quarter of a circle.

So , now if we multiply it by 4 , we get the area of the whole circle = πr^2 !!!