A coil has an inductance of 10 , mH. When it | vedclass

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(C) For $dc$ source: $P = \frac{V^2}{R}$. Given $V = 10 \, V$ and $P = 20 \, W$,we have $R = \frac{V^2}{P} = \frac{10^2}{20} = \frac{100}{20} = 5 \, \

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$80$

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(C) For $dc$ source: $P = \frac{V^2}{R}$. Given $V = 10 \, V$ and $P = 20 \, W$,we have $R = \frac{V^2}{P} = \frac{10^2}{20} = \frac{100}{20} = 5 \, \Omega$.For $ac$ source: $P = I_{rms}^2 R = \left( \frac{V}{Z} ight)^2 R = \frac{V^2 R}{Z^2}$. Given $V = 10 \, V$ and $P = 10 \, W$,we have $Z^2 = \frac{V^2 R}{P} = \frac{10^2 	imes 5}{10} = 50 \, \Omega^2$.The impedance of an $RL$ circuit is $Z^2 = R^2 + X_L^2 = R^2 + (2 \pi 
u L)^2$.Substituting the values: $50 = 5^2 + (2 \pi 
u 	imes 10 	imes 10^{-3})^2$.$50 = 25 + (2 \pi 
u 	imes 10^{-2})^2$.$25 = (2 \pi 
u 	imes 10^{-2})^2$.Taking the square root: $5 = 2 \pi 
u 	imes 10^{-2}$.$
u = \frac{5}{2 \pi 	imes 10^{-2}} = \frac{500}{2 \pi} \approx \frac{500}{6.28} \approx 79.6 \, Hz \approx 80 \, Hz$.

$A$ coil has an inductance of $10 \, mH$. When it is connected to a $10 \, V \, dc$ source,the power dissipation is $20 \, W$. When it is connected to a $10 \, V \, ac$ source,the power dissipation is $10 \, W$. What is the frequency of the $ac$ source in $Hz$?

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