For the parallel resonant circuit given below, the current through the capacitor and inductor are _________(a) IC = 10.892 mA; IL = 12.732 mA(b) IC = 12.732 mA; IL = 10.892 mA(c) IC = 10.892 mA; IL = 10.892 mA(d) IC = 12.732 mA; IL = 12.732 mAThis question was posed to me during an interview.I want to ask this question from Problems of Parallel Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory

For the parallel resonant circuit given below, the current through the

1.	For the parallel resonant circuit given below, the current through the capacitor and inductor are _________(a) IC = 10.892 mA; IL = 12.732 mA(b) IC = 12.732 mA; IL = 10.892 mA(c) IC = 10.892 mA; IL = 10.892 mA(d) IC = 12.732 mA; IL = 12.732 mAThis question was posed to me during an interview.I want to ask this question from Problems of Parallel Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (d) IC = 12.732 mA; IL = 12.732 mA To explain: Resonant Frequency, f = \(\FRAC{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}\) = \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\) = \(\frac{1}{799×10^{-9}}\) = 1.25 MHz. Inductive Reactance, XL = 2πfL = (6.28) (1.25×10^6)(100×10^-6) = 785.394 Ω Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}\) = \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω. IC = \(\frac{V_A}{X_C} \) = \(\frac{10}{785.417} \) = 12.732 mA IL = \(\frac{V_A}{X_L} \) = \(\frac{10}{785.394} \) = 12.732 mA.

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