An inductor having a resistance of 4 Ω and an inductance of 20 mH are connected across a 230 V, 50 Hz supply. What value of capacitance should be connected in parallel to the inductor to produce a resonance condition? What will be the value of the resonant current?

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Accepted Answer

[latex]\begin{aligned} & \mathrm{Z}_{\mathrm{L}}=\mathrm{R}+\mathrm{j} \omega \mathrm{L}, \omega=2 \pi \mathrm{f}=2 imes 3.14 imes 50=314 \ & \mathrm{Z}_{\mathrm{L}}=4+\mathrm{j} 314 imes 20 imes 10^{-3}=4+\mathrm{j} 6.28=7.44 \angle 57.5^{\circ} \Omega \ & \mathrm{I}_{\mathrm{L}}=\frac{\mathrm{V}}{\mathrm{Z}_{\mathrm{L}}}=\frac{230 \angle 0}{7.44 \angle 57.5^{\circ}}=30.9 \angle-57.5^{\circ} \mathrm{A} \end{aligned}[/latex]

For resonance, the current drawn by the capacitor in parallel must be equal to [latex]I_L \sin \phi_L[/latex] .

[latex]\begin{aligned} & \mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{L}} \sin \phi_{\mathrm{L}}=30.9 imes \sin 57.5^{\circ}=30.9 imes 0.843=26 \mathrm{~A} \ & \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}} ext { or, } \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{C}}}=\frac{230}{26}=8.84 \Omega \ & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{314 \mathrm{C}}=8.84 \Omega \end{aligned}[/latex]

or, [latex]\mathrm{C}=\frac{1}{314 imes 8.84} \mathrm{~F}=\frac{10^6}{314 imes 8.84} \mu \mathrm{F}[/latex]

= [latex]360.2 \mu \mathrm{F}[/latex]

Resonant current for parallel resonance is the minimum current which is the in-phase component, i.e., [latex]\mathrm{I}_{\mathrm{L}} \cos \phi_{\mathrm{L}}[/latex] .

[latex]\begin{aligned} \mathrm{I}_0 & =\mathrm{I}_{\mathrm{L}} \cos \phi_{\mathrm{L}}=30.9 imes \cos 57.5^{\circ}=30.9 imes 0.537 \ & =16.6 \mathrm{~A} \end{aligned}[/latex]

The phasor diagram representing the resonant condition has been shown in Fig. 3.72

Question 3.42: An inductor having a resistance of 4 Ω and an inductance of ......

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