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**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.

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