Question 13.4: A circuit has a coil of inductance 120 μH and resistance of ...

Given L = 120 μH, C = 253.3 pF and R = 15.7 Ω.

(a) The resonant frequency,           f_{O} =\frac{1}{2\pi \sqrt{LC} }= \frac{1}{2\pi \sqrt{120\times 10^{-6}\times 211\times 10^{-12} } }= 1  MHz

(b) The current at resonant frequency,

I_{o} =\frac{V}{R}= \frac{0.157}{15.7}= 0.01  A

Therefore, voltage across R,            V_{R} = I_{o} × R = 0.01 × 15.7 = 0.157  V

Voltage across L,                               V_{L} = I_{o} × X_{L} = I_{o} × (2 \pi f_{o} L)

= 0.01 × 2 \pi × 1 × 10^{6} × 120 × 10^{–6} = 7.54  V

Voltage across C,                                  V_{C} = I_{o} × X_{C} = I_{o} ×\frac{1}{2\pi f_{O}C}= \frac{0.01}{2\pi \times 1\times 10^{6} \times 211\times 10^{-12} }= 7.54  V

(c) The quality factor,                          Q =\frac{X_{L} }{R}= \frac{2\pi f_{O}L }{R} = \frac{2\pi \times 1\times 10^{6}\times 120\times 10^{-6} }{15.7} = 48.