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

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Question

(A) For $DC$,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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For $DC$,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 $AC$,the solenoid acts as an $LR$ circuit. Given $V = 100 \, V$ and $I = 0.5 \, A$,the impedance is $Z = \frac{V}{I} = \frac{100}{0.5} = 200 \, \Omega$.The impedance $Z$ is related to resistance $R$ and inductive reactance $X_L$ by $Z = \sqrt{R^2 + X_L^2}$.Substituting the values: $200 = \sqrt{100^2 + X_L^2} \Rightarrow 40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000 \Rightarrow X_L = \sqrt{30000} \approx 173.2 \, \Omega$.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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