A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of begin quantity omega times t end quantity, where V is the maximum voltage, omega is the angular frequency, and t is the time. The current supplied by this source that flows through this resistor is described with the function i of t is equal to I times cosine of begin quantity omega times t end quantity, where I is the maximum current. What is the average power supplied by this AC source?

Answers

Answer 1

Answer:

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=(1)/(T)\intop_(0)^(T)p(t)dt$

But:

$v(t)=v_(m)cos(\omega t) \n \n i(t)=i_(m)cos(\omega t)$

So:

$P=(1)/(T)\intop_(0)^(T)v_(m)cos(\omega t)i_(m)cos(\omega t)dt \n \n P=(v_(m)i_(m))/(T)\intop_(0)^(T)cos^(2)(\omega t)dt \n \n But: cos^(2)(\omega t)=(1+cos(2\omega t))/(2)$

$P=(v_(m)i_(m))/(T)\intop_(0)^(T)((1+cos(2\omega t))/(2) )dt \n\nP=(v_(m)i_(m))/(T)\intop_(0)^(T)[(1)/(2)+(cos(2\omega t))/(2)]dt \n\nP=(v_(m)i_(m))/(T)[(1)/(2)(t)\right|_0^T +(sin(2\omega t))/(4\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2T)[(t)\right|_0^T +(sin(2\omega t))/(2\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2)$

In terms of RMS values:

$V_(RMS)=V=(v_(m))/(√(2)) \n \n I_(RMS)=I=(i_(m))/(√(2)) \n \n Then: \n \n P=VI$

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Answers

Answer:

The correct option is D

Explanation:

Integumentary system consists of organs including the skin which is used in the excretion of sweat.

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Answer: D

Explanation:

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa.
B. 651.605 kPa.
C. 277.280 kPa.
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Answers

By definition we have to:

Where,

Pabs: absolute pressure

Patm: atmospheric pressure

Pg: gage pressure

The atmospheric pressure is constant and its value is:

Then, by clearing gage pressure we have:

Substituting values we have:

Answer:

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is:

D. 448.955 kPa

The answer is D. 448.955 kPa

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Answers

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Answers

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Answers

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