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

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(A) For $DC$,the solenoid acts as a pure resistor $R$. Given $V = 100 \, V$ and $I = 1.0 \, A$,we have $R = \frac{V}{I} = \frac{100}{1} = 100 \, \Omeg

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

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(A) For $DC$,the solenoid acts as a pure resistor $R$. Given $V = 100 \, V$ and $I = 1.0 \, A$,we have $R = \frac{V}{I} = \frac{100}{1} = 100 \, \Omega$.For $AC$,the solenoid acts as an $LR$ circuit with impedance $Z$. Given $V = 100 \, V$ and $I = 0.5 \, A$,we have $Z = \frac{V}{I} = \frac{100}{0.5} = 200 \, \Omega$.The impedance is given by $Z = \sqrt{R^2 + X_L^2}$,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: $40000 = 10000 + (100\pi L)^2$.$30000 = (100\pi L)^2$.Taking the square root: $100\sqrt{3} = 100\pi L$.$L = \frac{\sqrt{3}}{\pi} \approx \frac{1.732}{3.1416} \approx 0.55 \, H$.

When $100 \, V$ $DC$ is supplied across a solenoid,a current of $1.0 \, A$ 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 $50 \, Hz$,then the impedance and inductance of the solenoid are:

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