A virtual current of 4 , A and 50 , Hz flows | vedclass

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(B) Given: $I_{rms} = 4 \, A$,$f = 50 \, Hz$,$P = 240 \, W$,$V_{rms} = 100 \, V$. Power consumed in the coil is given by $P = I_{rms}^2 R$. $R = \frac

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$\frac{1}{5\pi} \, H$

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(B) Given: $I_{rms} = 4 \, A$,$f = 50 \, Hz$,$P = 240 \, W$,$V_{rms} = 100 \, V$. Power consumed in the coil is given by $P = I_{rms}^2 R$. $R = \frac{P}{I_{rms}^2} = \frac{240}{4^2} = \frac{240}{16} = 15 \, \Omega$. Impedance of the coil is $Z = \frac{V_{rms}}{I_{rms}} = \frac{100}{4} = 25 \, \Omega$. We know that $Z^2 = R^2 + X_L^2$,where $X_L = 2\pi f L$. $X_L = \sqrt{Z^2 - R^2} = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 \, \Omega$. Since $X_L = 2\pi f L$,we have $20 = 2 	imes \pi 	imes 50 	imes L$. $L = \frac{20}{100\pi} = \frac{1}{5\pi} \, H$.

$A$ virtual current of $4 \, A$ and $50 \, Hz$ flows in an $AC$ circuit containing a coil. The power consumed in the coil is $240 \, W$. If the virtual voltage across the coil is $100 \, V$,its inductance will be:

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