- #1

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the power P in an electrical circuit is given by P=V²/R

find the power in terms of V, R and cos2t when v= Vcost

any help would be appreciated

chris

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- Thread starter chrisking2021
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- #1

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the power P in an electrical circuit is given by P=V²/R

find the power in terms of V, R and cos2t when v= Vcost

any help would be appreciated

chris

- #2

HallsofIvy

Science Advisor

Homework Helper

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If it is really P= V

If it is really P= v

- #3

uart

Science Advisor

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Yes Chis your notation is a little confused, but I suppose that’s to be expect since you are also a little confused by the area you’re studying. Still what about some punctuation, there’s no excuse for that!

The general statement for power in a resistive load is that the instantaneous power is,

[tex]P = v^2 / R[/tex]

For the “DC case” where the voltage is a constant, [tex]v = V[/tex], so [tex]P=V^2/R[/tex]

For the “AC case” where the voltage is sinusoidal, [tex]v = V cos(t)[/tex], so

[tex]P(t) = V^2 cos^2(t) / R[/tex]

Now often we are interested primarily in just the average value of the power. In this case it is convenient to use the trig identity,

[tex]2 \cos^2(t) = 1 + \cos(2t)[/tex],

to write the AC power as,

[tex]P(t) = \frac{V^2}{2R} \left( 1 + \cos(2t) \right)[/tex]

Note here that the cos(2t) part has zero average. Hence for the AC case the average power is,

[tex]P_{av} = V^2 / 2R[/tex]

The general statement for power in a resistive load is that the instantaneous power is,

[tex]P = v^2 / R[/tex]

For the “DC case” where the voltage is a constant, [tex]v = V[/tex], so [tex]P=V^2/R[/tex]

For the “AC case” where the voltage is sinusoidal, [tex]v = V cos(t)[/tex], so

[tex]P(t) = V^2 cos^2(t) / R[/tex]

Now often we are interested primarily in just the average value of the power. In this case it is convenient to use the trig identity,

[tex]2 \cos^2(t) = 1 + \cos(2t)[/tex],

to write the AC power as,

[tex]P(t) = \frac{V^2}{2R} \left( 1 + \cos(2t) \right)[/tex]

Note here that the cos(2t) part has zero average. Hence for the AC case the average power is,

[tex]P_{av} = V^2 / 2R[/tex]

Last edited:

- #4

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I did think i should be using trig identities:surprised . I was a little confused because the question was taken directly from a book it is written exactly as i wrote it in my request.

P.S sorry about my grammer as i was rushing before work.

Thanks again

Chris

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