The self-inductance of a choke coil is 10, mH | vedclass

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(C) For $dc$ source: The inductor acts as a short circuit (resistance $R$ only). Power $P_{dc} = \frac{V^2}{R}$.Given $V = 10\, V$ and $P_{dc} = 20\,

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$80$

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(C) For $dc$ source: The inductor acts as a short circuit (resistance $R$ only). Power $P_{dc} = \frac{V^2}{R}$.Given $V = 10\, V$ and $P_{dc} = 20\, W$,so $R = \frac{10^2}{20} = \frac{100}{20} = 5\, \Omega$.For $ac$ source: The impedance $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi f L$.Power $P_{ac} = \frac{V^2 R}{Z^2} = \frac{V^2 R}{R^2 + X_L^2}$.Given $P_{ac} = 10\, W$,so $10 = \frac{10^2 	imes 5}{5^2 + X_L^2} \Rightarrow 10 = \frac{500}{25 + X_L^2}$.$25 + X_L^2 = 50 \Rightarrow X_L^2 = 25 \Rightarrow X_L = 5\, \Omega$.Since $X_L = 2\pi f L$,we have $5 = 2 	imes 3.14 	imes f 	imes (10 	imes 10^{-3})$.$f = \frac{5}{2 	imes 3.14 	imes 0.01} = \frac{5}{0.0628} \approx 79.6\, Hz \approx 80\, Hz$.

The self-inductance of a choke coil is $10\, mH$. When it is connected to a $10\, V$ $dc$ source,the power loss is $20\, W$. When it is connected to a $10\, V$ $ac$ source,the power loss is $10\, W$. The frequency of the $ac$ source is......$Hz$.

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