Power in A.C. Circuits

The power(P) dissipated in an alternating current circuit is given by
P = V_rms I_rms cosΦ where cosΦ is the power factor. [You should remember that Φ is the phase difference between current and voltage].
Power in an A.C. circuit is also given by P = (I_rms)²R. Note that power is dissipated in resistance only.
Another form of the power relation is P = (I_rms)²Z cosΦ, which follows from cosΦ = R/Z

Some of you may be wondering which relation is to be used for calculating power. Well, you may use any one of them which is convenient to you in the context of the problem. Now, consider the following MCQ:
A voltage v = V₀ cosωt is applied in series with an inductance L and a resistance R. The average power dissipated per cycle in the circuit is
(a) V₀² /R (b) V₀²/2R (c) V₀²/(R+L) (d) V₀²R/[(R²+ L²ω ²)] (e) V₀²R/[2(R²+ L²ω ²)]
We shall use the relation, P = (I_rms)²R. Since I_rms = V_rms/Z = (V₀/√2)/[√(R²+ L²ω ²)], we obtain P = V₀²R/[2(R² + L²ω ²)].
You could have used the other relations (for power) also to obtain the result.
Consider the folowing MCQ which appeared in C.E.T. J&K 2000 question paper:
In an A.C. circuit V and I are given by
V = 100 sin (1000t) volt
I = 1000 sin (1000t + π/3 ) mA.
The power dissipated in the circuit is
(a) 10⁴ W (b) 25 W (c) 10 W (d) 250 W
Here current, voltage and the phase difference (between current and voltage) are given so that it is convenient to use the relation P = V_rms I_rms cosΦ. [Note that the current has a phase lead of π/3 and is given in mA].
Therefore, P = (100/√2) × (1000×10 –3/√2) ×cos (π/3) = 25 W