When 100, V DC is applied across a solenoid,a | vedclass

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(A) For $DC$ supply,the inductor acts as a pure resistor. Thus,the resistance $R$ is given by $R = \frac{V}{I_{DC}} = \frac{100}{1} = 100\, \Omega$.Fo

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$200\, \Omega$ and $0.55\, H$

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(A) For $DC$ supply,the inductor acts as a pure resistor. Thus,the resistance $R$ is given by $R = \frac{V}{I_{DC}} = \frac{100}{1} = 100\, \Omega$.For $AC$ supply,the impedance $Z$ is given by $Z = \frac{V}{I_{AC}} = \frac{100}{0.5} = 200\, \Omega$.The impedance of an $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Substituting the values: $200 = \sqrt{100^2 + X_L^2} \Rightarrow 40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.$X_L = \sqrt{30000} = 100\sqrt{3} \approx 173.2\, \Omega$.Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{173.2}{2 	imes 3.14 	imes 50} = \frac{173.2}{314} \approx 0.55\, H$.

When $100\, V$ $DC$ is applied across a solenoid,a current of $1\, A$ flows in it. When $100\, V$ $AC$ is applied across the same coil,the current drops to $0.5\, A$. If the frequency of the $AC$ source is $50\, Hz$,the impedance and inductance of the solenoid are:

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