100 Single-phase a.c. circuits
In general, then, in a single-phase sinusoidal a.c. circuit for which the rms
value of the supply voltage is V, the rms value of the supply current is I and the
phase angle of the circuit is th, the power is given by
P = V/cos 4) (4.44)
For a purely resistive circuit ~b = 0 and cos 4~ = 1 so that the power is VI, which
agrees with the result obtained previously. For a purely reactive circuit ~b = 90 ~
and cos ~b = 0 so that the power is zero which agrees with the results obtained
previously.
Power components
Fig. 4.40 shows the phasor diagram for a circuit having a lagging phase angle of
~. The current I is shown to have two components at right angles. These are
I cos 4~ in phase with V and /sin 4~ lagging V by 90 ~ If we multiply all three
Vl~o~o (p)
I cos r V -~VI sin 4) VI
! sin r I (Q) (S)
Figure 4.40 Figure 4.41
currents by V we obtain the phasor diagram of Fig. 4.41 and, in this diagram: VI
is a phasor representing the so-called apparent power (symbol S) which is
measured in volt-amperes (VA); VI cos 05 is a phasor representing the real
power (symbol P) which is measured in watts (W); and VI sin 4~ is a phasor
representing the reactive power (symbol Q) which is measured in volt-amperes
reactive (Var).
In complex form we have, for a lagging phase angle
S = P -jQ (4.45)
For a leading phase angle
S = P + jQ (4.46)
The magnitude of S is obtained from
S- ~/(P: + Q2) (4.47)The phase angle is obtained from
ch = tan-'(Q/P) (4.48)
Also
ch = sin-~(Q/S) (4.49)