# Calculating Power in Electrical Circuits: Help Needed

• chrisking2021
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.

#### chrisking2021

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

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 $$P=V^2/R$$

For 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,

$$P_{av} = V^2 / 2R$$

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Thankyou Hallsofivy

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 