# Answer to Question #201803 in Electrical Engineering for weam

Question #201803

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

1
2021-06-04T16:12:02-0400

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"

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