A tank circuit has a resonant frequency of 12 kHz and a bandwidth of 800 Hz. If the winding resistance is 10 Ohms, calculate the inductance and capacitance.

The quality factor is defined as the ratio of the resonant frequency ($f_r$) to the bandwidth (BW).
Given $f_r = 12\,kHz = 12 \times 10^3\,Hz$ and $BW = 800\,Hz$.

$$Q = \frac{f_r}{BW} = \frac{12 \times 10^3\,Hz}{800\,Hz} = 15$$

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 $R_w = 10\,\Omega$.

$$Q = \frac{X_L}{R_w}$$ $$X_L = Q \times R_w = 15 \times 10\,\Omega = 150\,\Omega$$

The inductive reactance is related to the inductance and resonant frequency by the formula:

$$X_L = 2\pi f_r L$$

Rearranging for L:

$$L = \frac{X_L}{2\pi f_r} = \frac{150\,\Omega}{2\pi (12 \times 10^3\,Hz)}$$ $$L = \frac{150}{24000\pi}\,H \approx 0.0019894\,H$$ $$L \approx \boxed{\text{1.99 mH}}$$

At resonance, the inductive reactance equals the capacitive reactance ($X_C$).

$$X_C = X_L = 150\,\Omega$$

The capacitive reactance is related to the capacitance and resonant frequency by the formula:

$$X_C = \frac{1}{2\pi f_r C}$$

Rearranging for C:

$$C = \frac{1}{2\pi f_r X_C} = \frac{1}{2\pi (12 \times 10^3\,Hz) (150\,\Omega)}$$ $$C = \frac{1}{2\pi (1.8 \times 10^6)}\,F = \frac{1}{3.6\pi \times 10^6}\,F$$ $$C \approx \frac{1}{11309733.55}\,F \approx 8.8419 \times 10^{-8}\,F$$ $$C \approx 0.0884 \times 10^{-6}\,F$$ $$C \approx \boxed{\text{0.0884 \mu F}}$$