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

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

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

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(A) $1$. For $DC$ supply,the inductor acts as a pure resistor. Therefore,the resistance $R$ is given by $R = \frac{V}{I_{DC}} = \frac{100}{1} = 100\,\Omega$.$2$. For $AC$ supply,the impedance $Z$ is given by $Z = \frac{V}{I_{AC}} = \frac{100}{0.5} = 200\,\Omega$.$3$. The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$. Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.$4$. Squaring both sides: $40000 = 10000 + X_L^2$,which gives $X_L^2 = 30000$,so $X_L = \sqrt{30000} \approx 173.2\,\Omega$.$5$. 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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