1 Crore+ students have signed up on EduRev. Have you? 
7. Power in an AC circuit
In case of a steady current the rate of doing work is given by,
P = Vi
In an alternating circuit, current and voltage both vary with time, so the work done by the source in time interval dt is given by
dW= Vidt
Suppose in an ac, the current is leading the voltage by an angle φ. Then we can write,
V = V_{0} sinωt
and i = i_{0} sin(ωt + φ)
dW = V_{0}i_{0} sin ωt sin (ωt + φ) dt
= V_{0} i_{0} (sin^{2} ωt cos f + sinωt cos ωt sin φ) dt
The total work done in a complete cycle is
W = V_{0}i_{0} cos + V_{0}i_{0}sin
+ =
The average power delivered by the source is, therefore,
P = = =
= V_{rms} i_{rms} cos φ
or <P>_{one cycle} = V_{rms} i_{rms} cos φ
Here, the term cos φ is known as power factor.
It is said to be leading if current leads voltage, lagging if current lags voltage. Thus, a power factor of 0.5 lagging means current lags the voltage by 60° (as cos^{1}0.5 = 60°). The product of V_{rms} and i_{rms} gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor cos φ. Thus,
and apparent power = V_{rms} × i_{rms}
True power = apparent power × power factor
For φ= 0°, the current and voltage are in phase. The power is thus, maximum (V_{rms}× i_{rms}). For
φ = 90°, the power is zero. The current is then stated as wattless. Such a case will arise when resistance in the circuit is zero. The circuit is purely inductive or capacitive.
157 videos452 docs213 tests

157 videos452 docs213 tests
