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A coil has an inductance of $0.7\,henry$ and is joined in series with a resistance of $220\,\Omega $. When the alternating $emf$ of $220\,V$ at $50\,Hz$ is applied to it then the phase through which current lags behind the applied $emf$ and the wattless component of current in the circuit will be respectively
$30^o,\,\,1\,A$
$45^o,\,\,0.5\,A$
$60^o,\,\,1.5\,A$
none of these
Solution
$\mathrm{L}=0.7 \mathrm{\,H}, \mathrm{R}=220 \Omega, \mathrm{E}_{0}=220 \mathrm{\,V}, v=50 \mathrm{\,Hz}$
This is an $L – R$ circuit Phase difference,
$\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi v \mathrm{L}}{\mathrm{R}}$
$\left[\mathrm{X}_{\mathrm{L}}=2 \pi v \mathrm{L}=2 \times \frac{22}{7} \times 50 \times 0.7=220 \Omega\right]$
$=\frac{220}{220}=1$ or, $\phi=45^{\circ}$
Wattless component of current
$=I_{0} \sin \phi=\frac{I_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{E_{0}}{Z}$
$=\frac{1}{\sqrt{2}} \frac{220}{\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}} \cdot \frac{220}{\sqrt{220^{2}+220^{2}}}$
$=\frac{1}{2}=0.5 \mathrm{\,A}$
