Question 10.5: In the circuit shown in Fig. 10.13, a load having an impedan...

a) The line and load impedances are in series across the voltage source, so the load current equals the voltage divided by the total impedance, or

Because the voltage is given in terms of its rms value, the current also is rms. The load voltage is the product of the load current and load impedance:

= 234.36 \angle -3.18° V (rms).

b) The average and reactive power delivered to the load can be computed using Eq. 10.29 = V_{eff}I^{*}_{eff} . Therefore

= 975 + j650 VA.

Thus the load is absorbing an average power of 975 W and a reactive power of 650 VAR.

c) The average and reactive power delivered to the line are most easily calculated from Eqs. 10.33 P = \left|I_{eff}\right|^{2}R =\frac{1}{2}I^{2}_{m} R, and 10.34 Q= \left|I_{eff}\right|^{2}X =\frac{1}{2}\pmb{I}^{2}_{m} X, because the line current is known. Thus

P = \left(5\right)^{2}\left(1\right) = 25 W,

Q = \left(5\right)^{2}\left(4\right) = 100 VAR.

Note that the reactive power associated with the line is positive because the line reactance is inductive.

d) One way to calculate the average and reactive power delivered by the source is to add the complex power delivered to the line to that delivered to the load, or

S = 25 + j100 + 975 + j650

= 1000 + j 750 VA.

The complex power at the source can also be calculated from Eq. 10.29 = V_{eff}I^{*}_{eff}.

S_{s} = -250\pmb{I}^{*}_{L} .

The minus sign is inserted in Eq. 10.29 = V_{eff}I^{*}_{eff} whenever the current reference is in the direction of a voltage rise. Thus

S_{s} = -250\left(4 + j3\right) = -\left(1000 + j 750\right) VA.

The minus sign implies that both average power and magnetizing reactive power are being delivered by the source. Note that this result agrees with the previous calculation of S, as it must, because the source must furnish all the average and reactive power absorbed by the line and load.