Calculating Power in Electrical Circuits: Help Needed

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The discussion focuses on calculating power in electrical circuits using the formula P = V²/R. The user seeks clarification on how to express power in terms of voltage (V), resistance (R), and the cosine of double angles (cos 2t) when voltage is sinusoidal (v = V cos t). Key points include the distinction between constant and sinusoidal voltage cases, leading to different power expressions. The average power for an AC circuit is derived using trigonometric identities, resulting in P_avg = V²/2R. The conversation highlights the importance of precise notation and understanding trigonometric relationships in electrical calculations.
chrisking2021
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i am just about to start an electrical course at college and I've been reading up on some mathematics i will need when i start my problem is i don't know much about compound angles and trig and i want to complete some examples within a book the following problem is the one i need help with

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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First, clarify your notation. Is the "V" in P= V2r the "V" or "v" in v= V cos t?

If it is really P= V2/R, there is nothing to be done, P is already in terms of V and R (and cos 2t is irrelevant).

If it is really P= v2/R, then v= Vcos t so v2= V2 cos2 t and you can use the trigonometric identity cos 2t= cos2 t- sin2 t= cos2 t- (1- cos2 t)= 2cos2 t- 1 so that 2cos2 t= cos 2t+ 1 and then
cos2t= (cos(2t)+ 1)/2.

v2= V2cos2 t= V2(cos(2t)+ 1)/2.

Now, if P= v2/R, then P= V2(cos(2t)+ 1)/(2R).

A third, though unlikely, possibility is that P= V2/R and v= Vcos t but you meant "in terms of v, R, and cos 2t. In that case, V= v/cos t so cos2 t belongs in the denominator.
In that case, P= 2v2/(R(cos(2t)+ 1)).
 
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,

P = v^2 / R

For the “DC case” where the voltage is a constant, v = V, so P=V^2/RFor the “AC case” where the voltage is sinusoidal, v = V cos(t), so

P(t) = V^2 cos^2(t) / R

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,

2 \cos^2(t) = 1 + \cos(2t),

to write the AC power as,

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

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

P_{av} = V^2 / 2R
 
Last edited:
Thankyou Hallsofivy

I did think i should be using trig identities . I was a little confused: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 :approve:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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