The average value of the wave-form shown in figure is.

Question

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6ee206cb753f6450d753fd77417f00bc-.png" /> The average value of the wave-form shown in figure is. (A) 15 √2 (B) 10 √2 (C) 10 (D) 15

Tardigrade Admin · Accepted Answer

Equation of line AB is given by y=10+((10/T)) t y av =(1/T) ∫ limits0T y d t =(1/T) ∫ limits0T(10+(10/T) t) d t =(1/T)|10 t+(5 t2/T)|0T=15

Q.
The average value of the wave-form shown in figure is.

$152$

$102$

10

15

Equation of line $A B$ is given by $y = 10 + (\frac{10}{T}) t$
$y_{a v} = \frac{1}{T} 0 \int T y d t$
$= \frac{1}{T} 0 \int T (10 + \frac{10}{T} t) d t$
$= \frac{1}{T} ∣ ∣ 10 t + \frac{5 t ^{2}}{T} ∣ ∣_{0}^{T} = 15$

Q. The average value of the wave-form shown in figure is.

152​

102​

10

15

Equation of line AB is given by y=10+(T10​)t yav​=T1​0∫T​ydt =T1​0∫T​(10+T10​t)dt =T1​∣∣​10t+T5t2​∣∣​0T​=15

Q.
The average value of the wave-form shown in figure is.

$152$

$102$

Equation of line $A B$ is given by $y = 10 + (\frac{10}{T}) t$
$y_{a v} = \frac{1}{T} 0 \int T y d t$
$= \frac{1}{T} 0 \int T (10 + \frac{10}{T} t) d t$
$= \frac{1}{T} ∣ ∣ 10 t + \frac{5 t ^{2}}{T} ∣ ∣_{0}^{T} = 15$