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That's not always the case. One formula for power dissipated is:
P = IR
So, a higher resistance means that more power is dissipated - if the current is the same. The reason for this is precisely that resistance is related to the conversion of electrical energy into heat.
However, if you put a higher resistance across a specific voltage, you'll get less power dissipation, not more, since less current will flow at a higher resistance.
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What is the formula for exponential decay?
y=a(1-r) to the t power
Find the area of the largest square incribed in a circle?
A circle of radius, r, has area pi r2. The largest square inside it is made up of four isosceles triangles whose equal sides are of length, r, and thus has area 1/2 r2. Thus the area of the contained square is 2 r2.
Calculate the resolving power if the wavelength is 600 nm and the numerical apertures are 0 0.2 0.4 0.6 0.8 and 1.0?
Use the Equation, Resolving Power=lambda/2(Numerical Aperture). So, given the values for Numerical Aperture(NA): If NA=0, then R=0, NA=0.2, then R=1500, NA=0.4, then R=750, etc. Simply solve the equation substituting the provided Numerical Aperture (NA) values in.
Why power waveform is not symmetrical to X-Axis in R-L circuit?
Because power dissipated in rl circuit is given by p= 1/2(Li²)+ i²R which will give a curve and not a linear graph. Secondly the graph is a cosine curve ,with a phase difference between current and voltage. Hence the waveform is not symmetrical to x-axis .
How do you find the radius when the circumfrence is given?
The circumference (not circumfrence!) is given by the equation C = 2*pi*r where r is the radius. So r = C/(2*pi) where pi is the irrational number that is approx 3.14159 or even more approximately, 3.14.
