For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________(a) VC = 16.306 V; VL = 16.306 V(b) VC = 11.268 V; VL = 11.268 V(c) VC = 16.306 V; VL = 16.306 V(d) VC = 14.441 V; VL = 14.441 VI have been asked this question at a job interview.This intriguing question comes from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory

For the series circuit given below, the value of the voltage across th

1.	For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________(a) VC = 16.306 V; VL = 16.306 V(b) VC = 11.268 V; VL = 11.268 V(c) VC = 16.306 V; VL = 16.306 V(d) VC = 14.441 V; VL = 14.441 VI have been asked this question at a job interview.This intriguing question comes from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (c) VC = 16.306 V; VL = 16.306 V Best explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive REACTANCE, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\) = \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ ZEQ = R = 47 Ω IT = \(\frac{V_{in}}{Z_{EQ}}= \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA ∴ Voltage ACROSS the capacitor, VC = XCIT = (2.168 kΩ)(7.521 mA) = 16.306 V ∴ Voltage across the INDUCTOR, VL = XLIT = (2.168 kΩ)(7.521 mA) = 16.306 V.

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