Introduction to Electric Circuits

v,i,p

0

P

4.7 Power in single-phase a.c. circuits 99

Figure 4.39

Purely capacitive circuits

For the circuit of Fig. 4.11 the voltage, current and power waveforms are given

in Fig. 4.39. If the voltage is given by v - Vm sin wt then the current, being 90 ~

ahead of the voltage, will be given by i - Im COS cot and the instantaneous power

is

p = Vmlm sin ~ot cos wt (4.43)

Since Equations (4.42) and (4.43) are identical mathematically then calculation

of the mean power will yield the same result as for the purely inductive circuit.

The mean power in a purely capacitive circuit is therefore zero.

Resistive-reactive circuits
For circuits which contain resistance together with inductive and/or capacitive
reactance there will be a phase angle ~b which in general lies between 0 ~ and 90 ~
In these cases, if the voltage is represented by v = Vm sin ~ot then the current
will be given by i - Im sin (wt + q~) where ~b can be positive or negative. The
instantaneous power is then p = vi = V m sin o~t I m sin (~ot + ~b).
The average power is
2~ to
P - (oJ/2 I7") f Vm sin o~t lm sin (~ot + 4~)dt
0
2 "rr/o~

• 
  
  
  
  • Vm&~/2 ~r J" (sin ~ot[sin ~ot cos ~ + cos 6ot sin ~])dt
    0
    2Tr,, ~n
  
  
  
  


• Vmlm~O/2Tr f (sin 2 ~ot cos q~ + sin ~ot cos ~ot sin 4~])dt
  0
  27r, oJ 2Trio
  = Vmlm~/2"n- f (sin 2 oJt cos ~b)dt + Vmlm~O/2"rr f (sin cot cos ~ot sin q~)dt
  0 0
  From the analysis of the purely resistive circuit, we see that the first term of
  the right-hand side of this equation reduces to VI cos q~. From the analysis of
  purely reactive circuits and Equation (4.42) we see that the second integral is
  zero. The average power is therefore given by P - VI cos 4~.