Answer to Question #201803 in Electrical Engineering for weam

Question

 A sinusoidal voltage having a maximum amplitude of 625 V is applied to the terminals of a 50 Ω

 The average power delivered to the resistor is:

 Select one:

 a 19531.25 W

 b 6510.42 W

 c 3906.25 W

 d 9765.63 W

 e 4882.61 W

Answer 1

Given that

$$v(t)=v_m cos(\omega t)=625cos(\omega t) \space V$$

where

$$v_m =625 \space V- maximum \space voltage \space amplitude$$ $$\omega=\frac {2 \pi} T, \space T-harmonic \space period$$

from Ohm's law

$$i(t)=\frac {v(t)}{R}$$

the instantaneous power p(t) absorbed by an element is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it

$$p(t)=v(t) \space i(t)=\frac {v^2 (t)}{R}$$

Thus, the average power is given by

$$P=\frac 1 T \int_0^T p(t)dt=\frac 1 T \int_0^T \frac{v^2(t)}{R}dt$$

for a sinusoidal voltage with ω=2π/T

$$P=\frac {v_m^2}{2R}=\frac {625V \cdot 625V}{2 \cdot50\Omega}=3906.25 \space W$$

Answer: c) 3906.25 W