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

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(D) For $DC$ supply,the solenoid acts as a pure resistor $R$. Given $V = 100\, V$ and $I = 1\, A$,the resistance is $R = \frac{V}{I} = \frac{100}{1} =

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

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(D) For $DC$ supply,the solenoid acts as a pure resistor $R$. Given $V = 100\, V$ and $I = 1\, A$,the resistance is $R = \frac{V}{I} = \frac{100}{1} = 100\, \Omega$.For $AC$ supply,the impedance $Z$ is given by $Z = \frac{V_{rms}}{I_{rms}} = \frac{100}{0.5} = 200\, \Omega$.The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi f L$.Substituting the values: $200 = \sqrt{100^2 + (2 \cdot \pi \cdot 50 \cdot L)^2}$.Squaring both sides: $200^2 = 100^2 + (100\pi L)^2$.$40000 = 10000 + (100\pi L)^2$.$30000 = (100\pi L)^2$.Taking the square root: $100\pi L = \sqrt{30000} = 100\sqrt{3} \approx 173.2$.$L = \frac{173.2}{100\pi} \approx \frac{1.732}{3.14} \approx 0.55\, H$.Thus,the impedance is $200\, \Omega$ and the inductance is $0.55\, H$.

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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