Inductive Reactance at Resonant Frequency in Series LCR Circuit

To find the inductive reactance of the series $LCR$ circuit at the resonant frequency, we proceed as follows:

1. Given Data:

• Resistance, $R = 20\ \Omega$
• Inductance, $L = 1.6\text{ H}$
• Capacitance, $C = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}$

2. Resonant Angular Frequency ($\omega_r$): In a series $LCR$ circuit, resonance occurs when the inductive reactance equals the capacitive reactance ($X_L = X_C$). The resonant angular frequency $\omega_r$ is given by: $\omega_r = \frac{1}{\sqrt{LC}}$

Substituting the given values into the formula: $\omega_r = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}}$ $\omega_r = \frac{1}{\sqrt{64 \times 10^{-6}}}$ $\omega_r = \frac{1}{8 \times 10^{-3}} = 125\text{ rad/s}$

3. Inductive Reactance ($X_L$) at Resonance: The inductive reactance $X_L$ at this resonant frequency is: $X_L = \omega_r L$

Substituting $\omega_r = 125\text{ rad/s}$ and $L = 1.6\text{ H}$: $X_L = 125 \times 1.6 = 200\ \Omega$

Alternatively, $X_L$ at resonance can be calculated directly as: $X_L = \sqrt{\frac{L}{C}} = \sqrt{\frac{1.6}{40 \times 10^{-6}}} = \sqrt{40000} = 200\ \Omega$

Final Answer: The inductive reactance at resonant frequency is 200 $\Omega$.