question_answer

A coil of 0.01 H inductance and \[1\Omega \] resistance is connected to 200 V 50 Hz AC supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current.

Answer:

            Given,   inductance,        L = 0.01 H Resistance,         \[R=1\Omega \] Voltage,             \[{{V}_{rms}}=200\,V\] and       Frequency,         f = 50 Hz Impedance of the circuit,             \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}=\sqrt{{{R}^{2}}+{{(2\pi fL)}^{2}}}\]             \[=\sqrt{{{1}^{2}}+{{(2\times 3.14\times 50\times 0.01)}^{2}}}\] or         \[Z=\sqrt{10.86}=3.3\,\Omega \] As,        \[\tan \phi =\frac{\omega L}{R}=\frac{2\pi fL}{R}\]             \[=\frac{2\times 3.14\times 50\times 0.01}{1}=3.14\] \[\Rightarrow \]   \[\phi ={{\tan }^{-1}}(3.14)=72.33{}^\circ \] \[\therefore \] Phase difference, \[\phi =\frac{72.33\times \pi }{180}\text{rad}\] Time lag between alternating voltage and current, \[\Delta \,t=\frac{\phi }{\omega }=\frac{72.33\pi }{180\times 2\pi \times 50}\]        \[[\because \,\omega =2\pi f]\]             \[\approx \frac{1}{250}s\]