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

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Question

(C) For $DC$ source: The inductor acts as a short circuit (resistance $R$).$P = \frac{V^2}{R} \implies 20 = \frac{10^2}{R} \implies R = \frac{100}{20}

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) For $DC$ source: The inductor acts as a short circuit (resistance $R$).$P = \frac{V^2}{R} \implies 20 = \frac{10^2}{R} \implies R = \frac{100}{20} = 5 \, \Omega$.For $AC$ source: The impedance is $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi 
u L$.The power loss is $P = \frac{V^2 R}{Z^2} \implies 10 = \frac{10^2 	imes 5}{Z^2}$.$Z^2 = \frac{100 	imes 5}{10} = 50 \, \Omega^2$.Since $Z^2 = R^2 + (2\pi 
u L)^2$,we have $50 = 5^2 + (2\pi 
u 	imes 10 	imes 10^{-3})^2$.$50 - 25 = (2\pi 
u 	imes 10^{-2})^2$.$25 = (2\pi 
u 	imes 10^{-2})^2$.Taking the square root: $5 = 2\pi 
u 	imes 10^{-2}$.$
u = \frac{5}{2 	imes 3.1416 	imes 10^{-2}} \approx 79.58 \, 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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