The network in Fig. 12.32 has the following parameters:

$V _{S}=120\underline{ /0^{\circ}} V , \quad G=0.01 S$
$C=600 \mu F , \quad \text { and } \quad L=120 mH$

If the source operates at the resonant frequency of the network, compute all the branch currents.

Question

The network in Fig. 12.32 has the following parameters:

[latex]V _{S}=120\underline{ /0^{\circ}} V , \quad G=0.01 S[/latex]
[latex]C=600 \mu F , \quad ext { and } \quad L=120 mH[/latex]

If the source operates at the resonant frequency of the network, compute all the branch currents.

Accepted Answer

The resonant frequency for the network is
[latex]\omega_{0}=\frac{1}{\sqrt{L C}}[/latex]
[latex]=\frac{1}{\sqrt{(120)\left(10^{-3} ight)(600)\left(10^{-6} ight)}}[/latex]
= 117.85 rad/s
At this frequency
[latex]Y _{C}=j \omega_{0} C=j 7.07 imes 10^{-2} S[/latex]
and
[latex]Y _{L}=-j\left(\frac{1}{\omega_{0} L} ight)=-j 7.07 imes 10^{-2} S[/latex]
The branch currents are then
[latex]I _{G}=G V _{S}=1.2 \underline{/ 0^{\circ}} A[/latex]
[latex]I _{C}= Y _{C} V _{S}=8.49 \underline{/ 90^{\circ}} A[/latex]
[latex]I _{L}= Y _{L} V _{S}=8.49 \underline{/-90^{\circ}} A[/latex]
and
[latex]I _{S}= I _{G}+ I _{C}+ I _{L}[/latex]
[latex]= I _{G}=1.2 \underline{/0^{\circ}} A[/latex]
As the analysis indicates, the source supplies only the losses in the resistive element. In addition, the source voltage and current are in phase and, therefore, the power factor is unity.

Question 12.13: The network in Fig. 12.32 has the following parameters: VS=1...

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