- #1

- 7

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter chrisking2021
- Start date

In summary, the conversation discusses a problem related to finding power in an electrical circuit and how to express it in terms of voltage, resistance, and trigonometric functions. There is some confusion about notation and the use of trig identities to solve the problem. Ultimately, the conversation concludes that the average power can be expressed as P=V^2/2R.

- #1

- 7

- 0

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

Mathematics news on Phys.org

- #2

Science Advisor

Homework Helper

- 43,008

- 974

If it is really P= V

If it is really P= v

- #3

Science Advisor

- 2,797

- 21

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

- 7

- 0

I did think i should be using trig identities . 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

Share:

- Replies
- 1

- Views
- 785

- Replies
- 1

- Views
- 839

- Replies
- 16

- Views
- 3K

- Replies
- 7

- Views
- 1K

- Replies
- 9

- Views
- 783

- Replies
- 3

- Views
- 865

- Replies
- 4

- Views
- 649

- Replies
- 8

- Views
- 912

- Replies
- 16

- Views
- 841

- Replies
- 9

- Views
- 754