Question 2.36: 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
I_{L} = \frac{V}{Z} = \frac{230}{5 + j10 }=\frac{230}{11.18 ∠ 64° } = 20.57 ∠ -64°
\cos Φ_{L} = \cos 64° = 0.438
\sin Φ_{L} = \sin 64° = 0.895
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 I_{L} \sin Φ_{L} so that these two currents cancel each other. In such a case, the resultant current, I is the in-phase current, i.e., I_{L} \cos Φ_{L} .
I_{C} = I_{L} \sin Φ_{L} = 20.57 × 0.895 = 18.4 A
I_{C}= \frac{V}{X_{C}} or X_{C}=\frac{V}{I_{C}}=\frac{230}{18.4} =12.5 Ω
X_{C}=\frac{1}{ωC} = 12.5
or, C = \frac{1}{2πf × 12.5} = \frac{1}{6.28 × 50 × 12.5}F = \frac{10^{6}}{314 × 12.5}\mu F
=254.7 \mu F
Magnitude of the in-phase current, i.e., the current which is in phase with the voltage, I = I_{L } \cosΦ, is
I = I_{L } \cosΦ_{L} = 20.57 × 0.438 = 9 A
This is the minimum current drawn by the circuit and is called the resonant current, I_{0}.
