An R-L series circuit, where R = 10 Ω and L = 0.056 H, is connected to an AC supply of frequency 50 Hz. The magnitude of impedance of the circuit is:

Correct option is B

$$\text{The impedance of a series RL circuit is given by:} \\[4pt]Z = R + jX_L \\[4pt]\text{where } R \text{ is the resistance and } X_L \text{ is the inductive reactance.} \\[6pt]\text{The magnitude of the impedance is:} \\[4pt]|Z| = \sqrt{R^2 + X_L^2} \\[10pt]\textbf{Calculation:} \\[4pt]\text{Given:} \\[4pt]\bullet R = 10\,\Omega \\[4pt]\bullet L = 0.056\,H \\[4pt]\bullet f = 50\,Hz \\[8pt]\text{The angular frequency } \omega \text{ is:} \\[4pt]\omega = 2\pi f = 314\,\text{radians/second} \\[8pt]\text{The inductive reactance } X_L \text{ is:} \\[4pt]X_L = \omega \times L = 314 \times 0.056 = 17.584\,\Omega \\[8pt]\text{Now, the magnitude of the impedance:} \\[4pt]|Z| = \sqrt{(10)^2 + (17.584)^2} = \sqrt{100 + 309.26} = \sqrt{409.26} = 20.23\,\Omega$$​​