Calculating Power in Electrical Circuits: Help Needed

Click For Summary
SUMMARY

The power \( P \) in an electrical circuit is defined by the formula \( P = \frac{V^2}{R} \) for DC circuits, where \( V \) is the voltage and \( R \) is the resistance. In AC circuits, the voltage is expressed as \( v = V \cos(t) \), leading to the instantaneous power formula \( P(t) = \frac{V^2 \cos^2(t)}{R} \). The average power can be calculated using the trigonometric identity \( 2 \cos^2(t) = 1 + \cos(2t) \), resulting in \( P_{av} = \frac{V^2}{2R} \). Clarification on notation and proper use of trigonometric identities is essential for accurate calculations.

PREREQUISITES
  • Understanding of electrical power formulas, specifically \( P = \frac{V^2}{R} \)
  • Familiarity with AC and DC circuit concepts
  • Knowledge of trigonometric identities, particularly \( \cos(2t) \) and \( \cos^2(t) \)
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of power formulas in AC circuits
  • Learn about trigonometric identities and their applications in electrical engineering
  • Explore the differences between instantaneous and average power calculations
  • Review examples of power calculations in resistive loads
USEFUL FOR

Students beginning their electrical engineering courses, educators teaching electrical circuit theory, and anyone looking to deepen their understanding of power calculations in both AC and DC circuits.

chrisking2021
Messages
7
Reaction score
0
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
 
Mathematics news on Phys.org
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 grammar as i was rushing before work.

Thanks again

Chris :approve:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K