Question

An alternating voltage E=200 sin 300t is applied across a series combination of R=10 ohm and an inductor of 800mH. Calculate 1)impudence of the circuit 2)peak value of the current 3)power factor of the circuit.

Answer 1

$$\begin{gathered} \mathrm{EMF}: \\ \mathrm{E}=200\sin 300\mathrm{t} \\ \mathrm{Compare}\mathrm{with}\mathrm{E}={\mathrm{E}}_0\mathrm{sin\omega t} \\ {\mathrm{E}}_0=200\mathrm{V} \\ \mathrm{\omega }=300 \\ \left(1\right) \\ \mathrm{Inductive}\mathrm{reactance}: \\ {\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L} \\ =300\times 800\times {10}^{-3} \\ =240\mathrm{ohms} \\ \mathrm{Impedance}: \\ \mathrm{z}=\sqrt{{\mathrm{R}}^2+{{\mathrm{X}}_{\mathrm{L}}}^2} \\ =\sqrt{{10}^2+{240}^2} \\ =240.2\mathrm{ohms} \\ \left(2\right) \\ \mathrm{Peak}\mathrm{current}: \\ \mathrm{I}=\frac{{\mathrm{E}}_0}{\mathrm{Z}} \\ =\frac{200}{240.2} \\ =0.83\mathrm{A} \\ \left(3\right) \\ \mathrm{Power}\mathrm{factor}: \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\right) \\ ={\tan }^{-1}\left(\frac{240}{10}\right) \\ =87.6^0 \end{gathered}$$