Question 2.34: An inductive coil of resistance 10 Ω and inductance 20 mH ar...

R = 10  Ω,    L = 20 × 10^{–3} H,    C = 10 × 10^{–6} F

at resonance,                 X_{L} = X_{C}
from which we get resonance frequency

f_{0}=\frac{1}{2π\sqrt{LC}}=\frac{1}{6.28\sqrt{20  ×  10^{-3}  ×  10  ×  10^{-6}}}

= 356  Hz

At resonance, impedance of the circuit is equal to R. Therefore, the maximum current that will flow is equal to

I_{0}=\frac{V}{R}=\frac{50}{10}=5  A

To calculate the voltage drop across the circuit components, we calculate X_{L} and X_{C} at the resonance frequency first.

X_{L} = 2πf_{0}L = 2 × 3.14 × 356 × 20 ×10^{-3}  Ω

= 44.7  Ω

voltage drop across L,          V_{L} = I_{0} X_{L} = 5 × 44.7  = 223.5  V

voltage drop across R,         V_{R} = I_{0} R = 5 × 10 = 50  V

Note that the voltage drop across R is the same as the supply voltage, i.e., 50 V. Voltage drop across the capacitor should be the same as the voltage drop across inductive reactance X_{L}. Let us calculate V_{C}.

V_{C}=I_{0} X_{C}=\frac{I_{0}}{2πf_{0}C}

Substituting values

V_{C}=\frac{5}{6.28  ×  356  ×  10  ×  10^{-6}}

=\frac{5 ×  10^{3}}{6.28  ×  3.56  }=223.5  V

Thus,                                V_{L} = V_{C} = 223.5  V

Q-factor                             =\frac{ V_{L}}{V}=\frac{223.5}{50}=4.47