Integral Calculus Homework: Mean & RMS Voltage

[maali5]

Homework Statement

A generated AC voltage has a value given by V = 4cos2θ.

You will need to use the identity cos^2⁡θ = 1/2(1+cos2θ)


a) Find the mean value of the voltage over a full cycle (0 ≤ θ ≤ 2π) using integral calculus

b) Find the r.m.s. value of the voltage over a full cycle (0 ≤ θ ≤ 2π) using integral calculus

Homework Equations

The Attempt at a Solution

4∫cos2x dx u = 2x du= 2dx du/2 = dx


4∫ cos u du

2 sin 2x +c



What is next now?

 

[SammyS]

Homework Statement

A generated AC voltage has a value given by V = 4cos2θ.

You will need to use the identity cos^2⁡θ = 1/2(1+cos2θ)

a) Find the mean value of the voltage over a full cycle (0 ≤ θ ≤ 2π) using integral calculus

b) Find the r.m.s. value of the voltage over a full cycle (0 ≤ θ ≤ 2π) using integral calculus

Homework Equations

The Attempt at a Solution

4∫cos2x dx u = 2x du= 2dx du/2 = dx

[STRIKE]4[/STRIKE] 2∫ cos u du

2 sin 2x +c

What is next now?

Find the average over a full cycle.

The average (mean) value of function f(x) over the interval [a, b] is given by

$\displaystyle f_\text{Mean}=\frac{\displaystyle \int_a^b{f(x)\,dx}}{b-a}\ .$

 

[maali5]

Find the average over a full cycle.

The average (mean) value of function f(x) over the interval [a, b] is given by

$\displaystyle f_\text{Mean}=\frac{\displaystyle \int_a^b{f(x)\,dx}}{b-a}\ .$

Where is the interval?

Is this f(x)? = 2 sin 2x +c

 

[Yukoel]

Where is the interval?

Is this f(x)? = 2 sin 2x +c

Hello maali5,
The interval has been quoted in your question right?
(0 ≤ θ ≤ 2π)
You have to have a definite integral right?So you will have to plug in these limits.And then the use the expression quoted by SammyS with these limits.

regards
Yukoel

 

[Nickie]

I would first confirm the formula being used for the generated AC voltage and its units. Assuming the units are in volts, the formula V = 4cos2θ would represent a sinusoidal AC voltage with an amplitude of 4 volts and a frequency of 2 cycles per unit of θ (radians or degrees).

a) To find the mean value of the voltage over a full cycle, we need to integrate the voltage equation over the range of 0 to 2π and divide by the total range. This can be expressed as:

Mean value = (1/2π) ∫Vdθ

Substituting the given voltage equation, we get:

Mean value = (1/2π) ∫4cos2θ dθ

Using the identity cos^2⁡θ = 1/2(1+cos2θ), this becomes:

Mean value = (1/2π) ∫4(1/2)(1+cos2θ) dθ

= (1/4π) ∫(1+cos2θ) dθ

= (1/4π) (θ + (1/2)sin2θ) + C

= (1/4π) (2π + 0) + C

= 1 + C

Therefore, the mean value of the voltage over a full cycle is 1 volt.

b) To find the r.m.s. value of the voltage over a full cycle, we need to first square the voltage equation and then integrate it over the range of 0 to 2π. This can be expressed as:

R.m.s. value = √((1/2π) ∫V^2dθ)

Substituting the given voltage equation, we get:

R.m.s. value = √((1/2π) ∫(4cos2θ)^2dθ)

= √((1/2π) ∫16cos^2⁡2θdθ)

= √((1/2π) ∫16(1/2)(1+cos4θ)dθ)

= √((1/4π) ∫(1+cos4θ)dθ)

= √((1/4π) (θ + (1/4)sin4θ) + C)

= √((1/4π) (2