Question 5.7: AC Power Calculations Compute the power and reactive power t...

To find the power and reactive power for the source, we must first find the power angle which is given by Equation 5.62:

\theta =\theta _v-\theta _i          (5.62)

The angle of the source voltage is θ_v = -90°, and the angle of the current delivered by the source is θ_i = -135°. Therefore, we have

\theta =-90^\circ –(-135^\circ)=45^\circ

The rms source voltage and current are

V_{s\mathrm{rms}}=\frac{\left|\mathrm{V}_s\right| }{\sqrt2}=\frac{10}{\sqrt{2} }  =7.071\mathrm{~V}

I_{\mathrm{rms}}=\frac{\left|\mathrm{I}\right| }{\sqrt2}=\frac{0.1414}{\sqrt{2} }  =0.1\mathrm{~A}

Now, we use Equations 5.60 and 5.63 to compute the power and reactive power delivered by the source:

P=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos{(\theta)}    (5.60)

Q=V_{\mathrm{rms}}I_{\mathrm{rms}}\sin{(\theta)}   (5.63)

P=V_{s\mathrm{rms}}I_{\mathrm{rms}}\cos{(\theta)}=7.071\times 0.1\cos {(45^\circ )}=0.5\mathrm{~W}

Q=V_{s\mathrm{rms}}I_{\mathrm{rms}}\sin{(\theta)}=7.071\times 0.1\sin {(45^\circ )}=0.5\mathrm{~VAR}

An alternative and more compact method for computing P and Q is to first find the complex power and then take the real and imaginary parts:

\mathrm{S}=\frac{1}{2}\mathrm{V}_s \mathrm{I}^*=\frac{1}{2}\left(10\angle -90^\circ \right)\left(0.1414\angle 135^\circ \right)=0.707\angle 45^\circ =0.5+j0.5

P=\mathrm{Re(S)}=0.5\mathrm{~W}

Q=\mathrm{Im(S)}=0.5\mathrm{~VAR}

We can use Equation 5.70 to compute the reactive power delivered to the inductor, yielding

Q=I^2_{\mathrm{rms}}X       (5.70)

Q_L=I^2_{\mathrm{rms}}X_L=\left(0.1\right)^2\left(100\right)=1.0\mathrm{~VAR}

For the capacitor, we have

Q_C=I^2_{C\mathrm{rms}}X_C=\left(\frac{0.1}{\sqrt{2} } \right)^2\left(-100\right)=-0.5\mathrm{~VAR}

Notice that we have used the rms value of the current through the capacitor in this calculation. Furthermore, notice that the reactance X_C of the capacitance is negative. As expected, the reactive power is negative for a capacitance. The reactive power for the resistance is zero. As a check, we can verify that the reactive power delivered by the source is equal to the sum of the reactive powers absorbed by the inductance and capacitance. This is demonstrated by

Q = Q_L + Q_C

The power delivered to the resistance is

P_R=I^2_{R\mathrm{rms}}R=\left(\frac{\left|\mathrm{I}_R\right| }{\sqrt{2} } \right)^2R=\left(\frac{0.1}{\sqrt{2} } \right) ^2100=0.5\mathrm{~W}

The power absorbed by the capacitance and inductance is given by

P_L=0

P_C=0

Thus, all of the power delivered by the source is absorbed by the resistance.