Sunday, February 28, 2016

A.C Fundamentals (Solved Examples)

Problem 1


aa)      Prove that the R.M.S of,
V = k + A Sin wt   is,


Where, v1 = A Sin wt   and k is a D.C Component (constant).

bb)      Find the R.M.S Voltage of
V = 10 + 5 Sin (300t + 600 )  

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Answers

aa)      
Equation for the R.M.S is,

 



Since,




And
Cos 2π – Cos 0  =  1 – 1  = 0




bb)      This is a sin wave with a D.C Component. So We can use the equation we derived above.
Pabindu Lakshitha
B.Sc(Engineer) 
Electrical and Information Engineering. 



Fundamentals of Alternative Currents


In this chapter, methods of finding R.M.S & Average values of Alternating Voltages and currents are discussed with examples.
Note: Generation of Alternative currents and Alternative voltages will not be discussed here.

Equations for alternative currents and voltages

An Alternative voltage can be represented by,

E(t) = Em Sin wt

Where ‘E(t)’ is the Voltage at any instant, ‘Em’ is the Peak voltage.
Similarly an Alternative current can be represented by,

I(t) = Im Sin wt

Where ‘I(t)’ is the Current at any instant, ‘Im’ is the Peak Current.

Root Mean Squared (R.M.S) Value


The average value of a Sine Wave (For a cycle) is zero. Hence, Method of R.M.S is used.
Equation for the R.M.S voltage is,


Or in terms of phase,


Where, ‘v’ is the instantaneous voltage.
For a Sine Wave form,


It can be proved that the R.M.S of a Sine wave is,

. 
where vm is the peak (or the maximum) value. 

You can use the above equation anywhere (Only for Sine waves) without proving. 

Average value


Since the average value for a full cycle of a sine wave-form is zero, we calculate the average value for a half cycle.

Or in terms of phase,


Where, ‘v’ is the instantaneous voltage.
All the other terms have their usual meanings.


This is the End of the quick revision. See you soon in the next article with solved examples. 

Pabindu Lakshitha
B.Sc(Engineer) 
Electrical and Information Engineering. 

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