Question 2.37: An inductor having a resistance of 4 Ω and an inductance of ...

Z_{L} = R + j ωL,   ω = 2πf = 2 × 3.14 × 50 = 314

Z _{L} = 4 + j314 × 20 × 10^{-3} = 4 + j6.28 = 7.44 ∠57.5°  Ω

I _{L}=\frac{V}{Z_{L}} = \frac{230 ∠0}{7.44 ∠57.5°}=30.9 ∠– 57.5°  A

For resonance, the current drawn by the capacitor in parallel must be equal to I_{L} \sin Φ_{L}.

I_{C} =  I_{L} \sin Φ_{L} = 30.9 × \sin 57.5° = 30.9 × 0.843 = 26  A

I_{C}=\frac{V}{X_{C}}   or,   X_{C} = \frac{V}{I_{C}}=\frac{230}{26}=8.84  Ω

X_{C}=\frac{1}{ωC}=\frac{1}{314  C}=8.84  Ω

or,                                                   C=\frac{1}{314  ×  8.84}F=\frac{10^{6}}{314  ×  8.84}μF

= 360.2 μF

Resonant current for parallel resonance is the minimum current which is the in-phase component, i.e., I_{L} \cos Φ_{L} .

I_{0} = I_{L} \cos Φ_{L} = 30.9 × \cos 57.5° = 30.9 × 0.537

= 16.6 A

The phasor diagram representing the resonant condition has been shown in Fig. 2.79.