1.

For the parallel resonant circuit given below, the cut-off frequencies are ____________(a) ∆f1 = 2.389 MHz; ∆f2 = 441.124 MHz(b) ∆f1 = 1.256 MHz; ∆f2 = 1.244 MHz(c) ∆f1 = 5.658 MHz; ∆f2 = 6.282 MHz(d) ∆f1 = 3.656 MHz; ∆f2 = 8.596 MHzThis question was posed to me by my school teacher while I was bunking the class.The query is from Problems of Parallel Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The CORRECT option is (b) ∆f1 = 1.256 MHz; ∆f2 = 1.244 MHz

Easiest explanation: 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

Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05

∴ ∆F = \(\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}\) = 12.5 kHz

Hence, BANDWIDTH = \(\frac{∆F}{2}\) = 6.25 kHz

∴ ∆f1 = f + \(\frac{∆F}{2}\) = 1.25 MHz + 6.25 kHz = 1.256 MHz

∴ ∆f1 = f – \(\frac{∆F}{2}\) = 1.25 MHz – 6.25 kHz = 1.244 MHz.


Discussion

No Comment Found

Related InterviewSolutions