Introduction to Electric Circuits

4.7 Power in single-phase a.c. circuits 101

and

= cos-'(P/S)

Since V- IZ and cos & = R/Z then

n = VI cos 4)= (IZ)I(R/Z) = I2R watts

Also, since sin & = X/Z then Q = VI sin & = (IZ)I(X/Z) and

Q - I2X volt-amperes reactive

(4.50)

(4.51)

(4.52)

Power factor

The real power (P) in a circuit is obtained by multiplying the apparent power
(S) by a factor cos ~b which is called the power factor of the circuit.

P = S cos ~ (4.53)

Power factor = cos 4' = P/S (4.54)

Example 4.22

The circuit of Fig. 4.42 is fed from a 12 V, 50 Hz supply. Calculate (1) the
current, (2) the reactive power, (3) the power factor.

._~__! lOft 20mH

I

v

Figure 4.42

Solution

1 The impedance is

Z- R + jXL -- 10 + jZ'nfL - 10 + jZ~r 50 • 20 • 10 -3- (10 + j6.28)l)

Z- V/(102 + 6.282) - 11.8a

The current is I- V/Z - 12/11.81 - 1.02 A.

2 From Equation (4.52) the reactive power is
Q = IZXL = 1.022 • 6.28 = 6.53 Var
3 The power factor is cos ~b - R/Z - 10/11.81 - 0.846 lagging.