Balancing Power Delivered with Power Absorbed in an ac Circuit
a) Calculate the total average and reactive power delivered to each impedance in the circuit shown in Fig.10.17.
b) Calculate the average and reactive powers associated with each source in the circuit.
c) Verify that the average power delivered equals the average power absorbed,and that the magnetizing reactive power delivered equals the magnetizing reactive power absorbed.
a) The complex power delivered to the \left(1+j2\right) \Omega impedance is
S_{1} =\frac{1}{2} \mathbf{V_{1} I^{\ast }_{1}} =P_{1} +jQ_{1}=\frac{1}{2}\left(78-j104\right)\left(-26+j52\right)
=\frac{1}{2}\left(3380+j6760\right)
=1690+j3380 \mathbf{ VA} .
Thus this impedance is absorbing an average power of 1690 W and a reactive power of 3380 VAR .The complex power delivered to the \left(12-j16\right) \Omega impedance is
S_{2} =\frac{1}{2} \mathbf{V_{2} I^{\ast }_{x}} =P_{2} +jQ_{2}=\frac{1}{2}\left(72+j104\right) \left(-2-j6\right)
=240-j320 \mathbf{VA}
Therefore the impedance in the vertical branch is absorbing 240 W and delivering 320 VAR.The complex power delivered to the \left(1+j3\right) \Omega impedance is
S_{3} =\frac{1}{2} \mathbf{V_{3} I^{\ast }_{2}} =P_{3} +jQ_{3}=\frac{1}{2}\left(150-j130\right) \left(-24-j58\right)
=1970+j5910VA
This impedance is absorbing 1970 W and 5910 VAR.
b) The complex power associated with the independent voltage source is
S_{s} =-\frac{1}{2} \mathbf{V_{s} I^{\ast }_{1}} =P_{s} +jQ_{s}=-\frac{1}{2}\left(150\right) \left(-26-j52\right)
=1950-j3900 VA.
Note that the independent voltage source is absorbing an average power of 1950 W and delivering 3900 VAR.The complex power associated with the current-controlled voltage source is
S_{x} =\frac{1}{2} \left(39\mathbf{I_{x}} \right) \left( \mathbf{I^{\ast }_{2}} \right) =P_{x} +jQ_{x}=\frac{1}{2}\left(-78+j234\right) \left(-24+j58\right)
=-5850-j5070 VA.
Both average power and magnetizing reactive power are being delivered by the dependent source.
c) The total power absorbed by the passive impedances and the independent voltage source is
P_{absorbed}=P_{1}+P_{2}+P_{3}+P_{s}=5850 WThe dependent voltage source is the only circuit element delivering average power.Thus
P_{delivered}=5850 WMagnetizing reactive power is being absorbed by the two horizontal branches.Thus
Q_{absorbed} =Q_{1}+Q_{3}=9290 VARMagnetizing reactive power is being delivered by the independent voltage source,the capacitor in the vertical impedance branch, and the dependent voltage source.Therefore
Q_{delivered}=9290VAR