Question 3.41: An inductive coil of resistance 5 Ω and inductive reactance ......
An inductive coil of resistance 5 Ω and inductive reactance 10 Ω is connected across a voltage of 230 V at 50 Hz. Calculate the value of the capacitor which when connected in parallel with the coil will bring down the magnitude of the circuit current to a minimum. Draw the phasor diagram.
Before a capacitor is connected, current flowing through the inductor, I_L is
\begin{aligned} I_L=\frac{V}{Z}=\frac{230}{5+\mathrm{j} 10}= & \frac{230}{11.18 \angle 64^{\circ}}=20.57 \angle-64^{\circ} \\ \cos \phi_{\mathrm{L}} & =\cos 64^{\circ}=0.438 \\ \sin \phi_{\mathrm{L}}= & \sin 64^{\circ}=0.895 \end{aligned}If a capacitor is now connected in parallel, it must draw a current I_C which will lead V by 90°. The magnitude of I_C must be equal to \mathrm{I}_{\mathrm{L}} \sin \phi_{\mathrm{L}} so that these two currents cancel each other. In such a case, the resultant current, I is the in-phase current, i.e., \mathrm{I}_{\mathrm{L}} \cos \phi_{\mathrm{L}} .
\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{L}} \sin \phi_{\mathrm{L}}=20.57 \times 0.895=18.4 \mathrm{~A}\\ \begin{aligned} & \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}} \text { or } \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{C}}}=\frac{230}{18.4}=12.5 \Omega \\ & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=12.5 \end{aligned}or, \mathrm{C}=\frac{1}{2 \pi \mathrm{f} \times 12.5}=\frac{1}{6.28 \times 50 \times 12.5} \mathrm{~F}=\frac{10^6}{314 \times 12.5} \mu \mathrm{F}
= 254.7 \mu \mathrm{F}
Magnitude of the in-phase current, i.e., the current which is in phase with the voltage, \mathrm{I}=\mathrm{I}_{\mathrm{L}} \cos \phi, is
\mathrm{I}=\mathrm{I}_{\mathrm{L}} \cos \phi_{\mathrm{L}}=20.57 \times 0.438=9 \mathrm{~A}This is the minimum current drawn by the circuit and is called the resonant current, I_0.
