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, \[\Delta \,t=\frac{\phi }{\omega }=\frac{72.33\pi }{180\times 2\pi \times 50}\] \[[\because \,\omega =2\pi f]\] \[\approx \frac{1}{250}s\]
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