When a coil is connected to AC supply of frequency $50 \mathrm{~Hz}$, a current of $4 \mathrm{~A}$ flows in…

When a coil is connected to AC supply of frequency $50 \mathrm{~Hz}$, a current of $4 \mathrm{~A}$ flows in it and it consumes $240 \mathrm{~W}$ power. If the potential difference across the coil is $100 \mathrm{~V}$, then the inductance value of the coil is
  1. $\mathrm{L}=(5 \pi) \mathrm{H}$
  2. $L=\frac{\pi}{5} \mathrm{H}$
  3. $L=\frac{1}{5 \pi} \mathrm{H}$
  4. $L=\frac{1}{25 \pi} \mathrm{H}$

Solution

$I^2 R=240 \mathrm{~W}, I=4 \mathrm{~A}$ $ \therefore \quad R=\frac{240}{16}=15 \Omega $ As, $V=I Z=I \sqrt{X_L^2+R^2} \Rightarrow \frac{V}{I}=\sqrt{X_L^2+R^2}$ $ \begin{array}{ll} \Rightarrow & \frac{V^2}{I^2}=X_L^2+R^2 \\ \text { or } & X_L^2=\frac{100 \times 100}{4 \times 4}-225=400 \\ \therefore & X_L=20=L \omega \\ \therefore & L=\frac{20}{2 \pi f}=\frac{20}{2 \pi \times 50}=\frac{1}{5 \pi} H \end{array} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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