When 100 V d.c. is applied across a solenoid | vedclass

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(A) For $d.c.$ supply,the inductor acts as a pure resistor:$R = \frac{V}{I} = \frac{100 \ V}{1 \ A} = 100 \ \Omega$.For $a.c.$ supply,the impedance $Z

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$200 \ \Omega, \frac{\sqrt{3}}{\pi} \ H$

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(A) For $d.c.$ supply,the inductor acts as a pure resistor:$R = \frac{V}{I} = \frac{100 \ V}{1 \ A} = 100 \ \Omega$.For $a.c.$ supply,the impedance $Z$ is given by:$Z = \frac{V}{I} = \frac{100 \ V}{0.5 \ A} = 200 \ \Omega$.The impedance of an $LR$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2 \pi f L$.Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.Squaring both sides: $40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.$X_L = \sqrt{30000} = 100\sqrt{3} \ \Omega$.Since $X_L = 2 \pi f L$,we have $100\sqrt{3} = 2 \pi (50) L$.$100\sqrt{3} = 100 \pi L \Rightarrow L = \frac{\sqrt{3}}{\pi} \ H$.

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

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