Answer to Question #162150 in Electricity and Magnetism for Naomi

Question

An ac source of 500v amplitude and angular frequency 2000s^-1 is applied to an inductor of L =16mH. calculate (a)the inductive reactance (b) the rms current (c) the instantaneous power and it's maximum value (d) the maximum energy stored in the magnetic field of the inductor.

Answer 1

1. The inductive reactance is, by definition $\chi_L = \omega L =2\cdot 10^3\cdot 16 \cdot 10^{-3}= 32\Omega$
2. The RMS current is given by $\frac{U_{RMS}}{\chi_L}=\frac{U}{\sqrt{2} \cdot \chi_L} \approx 11 A$
3. The instantaneous power is given by $P = U\cdot I$, if we take $U(t) = U \cos (\omega t +\phi)$ then $I(t) = I \sin(\omega t+\phi)$ and thus $P = \frac{UI}{2} \sin(2\omega t + 2\phi)$. Thus $P_{max} = \frac{UI}{2} = \frac{U^2}{2\chi_L} =3906.25 W \approx 3.9 kW$
4. The energy stored is given by $E = \frac{LI^2}{2} = \frac{LI_{max}^2}{2} \sin^2(\omega t + \phi)$ and thus the maximum energy is given by $E = \frac{LI_{max}^2}{2} = \frac{U_{max}^2}{2\omega^2 L} \approx 1.95 J$