This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Calculate the quality factor (Q) of the tank circuit. The quality factor is defined as the ratio of the resonant frequency (f_r) to the bandwidth (BW). Given f_r = 12\,kHz = 12 × 10^3\,Hz and BW = 800\,Hz. Q = (f_r)/(BW) = (12 × 10^3\,Hz)/(800\,Hz) = 15 Step 2: Calculate the inductive reactance (X_L). For a series RLC circuit, the quality factor can also be expressed as the ratio of the inductive reactance to the winding resistance (R_w). Given R_w = 10\,. Q = (X_L)/(R_w) X_L = Q × R_w = 15 × 10\, = 150\, Step 3: Calculate the inductance (L). The inductive reactance is related to the inductance and resonant frequency by the formula: X_L = 2 f_r L Rearranging for L: L = (X_L)/(2 f_r) = (150\,)/(2 (12 × 10^3\,Hz)) L = (150)/(24000)\,H ≈ 0.0019894\,H L ≈ 1.99 mH Step 4: Calculate the capacitance (C). At resonance, the inductive reactance equals the capacitive reactance (X_C). X_C = X_L = 150\, The capacitive reactance is related to the capacitance and resonant frequency by the formula: X_C = (1)/(2 f_r C) Rearranging for C: C = (1)/(2 f_r X_C) = (1)/(2 (12 × 10^3\,Hz) (150\,)) C = (1)/(2 (1.8 × 10^6))\,F = (1)/(3.6 × 10^6)\,F C ≈ (1)/(11309733.55)\,F ≈ 8.8419 × 10^-8\,F C ≈ 0.0884 × 10^-6\,F C ≈ 0.0884 F 3 done, 2 left today. You're making progress.