Answer:
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:
[tex]P_a_v_g = \frac{V_m*I_m}{2}*cos ( \beta )[/tex]
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
[tex]P_a_v_g = \frac{6*\sqrt{117} }{2}*cos ( 0.98279 )\\\\P_a_v_g = 18 W[/tex]
Answer: The average power is 18 W.
Answer:
18W
Explanation:
To find the average power of the electrical device you use the following formula:
[tex]P(t)=\frac{1}{T}\int_0^{T}i(t)v(t)dt\\\\i(t)=6sin(100\pi t)+9cos(100\pi t)\\\\v(t)=6sin(100\pi t)[/tex]
T: period of the oscillation
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{100\pi}=\frac{1}{50}s[/tex]
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:
[tex]P=\frac{1}{1/50}\frac{9}{25}=18W[/tex]
hence, the power of the device is 18W
