If the current and voltage of an electrical device (in the passive reference configuration) are v \left (t \ right) v (t) = 6 \sin\left(100\pi t\right)sin(100πt) V and i\left(t\right)i(t) = 6 \sin\left (100 \ pi t \ right) sin (100πt) + 9 \cos \left (100\pi t\right) cos (100πt) A. Then the average power delivered to the device (computed to one decimal place) is given by:.

The average power is 18 W

Explanation:

Solution:-

If the current and voltage of an electrical device (in the passive reference configuration) are:

                         v(t) = 6 sin(100πt) V  

                         i(t) = 6 sin(100πt) + 9 cos(100πt) Amp

- First we will find the harmonic form of the current function i(t):

                 a sin ( wt ) + b cos ( wt ) ≡ R*sin ( wt + α )

Where,

                 R = √( a^2 + b^2 )

                 α = arctan ( b / a )  

- Transforming the function i (t) into the harmonic form:

                a = 6 , b = 9

                 R = √( a^2 + b^2 ) = √( 6^2 + 9^2 ) = √( 36 + 81 )

                R = √117

                 α = arctan ( b / a ) = arctan ( 9 / 6 )

                 α = 0.98279 rads

Hence,

                  i(t) = 6 sin(100πt) + 9 cos(100πt) ≡ √117*sin (100πt + 0.98279 )

- The average power is given by the following formula:

Where,

                 Vm: The mean voltage

                 Im : The mean current

                 β : The phase difference

- Using the given functions we have:

                Vm = 6 V , Im = R = √117 Amps , β = 0.98279

The average power is 18 W.

18W

Explanation:

To find the average power of the electrical device you use the following formula:

T: period of the oscillation

i(t): current

v(t): voltage

By replacing i and v in the integral and solve it you obtain:

The solution of the integral is attached below:

By replacing the value of the integral (9/25) you obtain:

hence, the power of the device is 18W

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