When 80 V d.c. is applied across a solenoid, | vedclass

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(A) When $80 \ V$ $d.c.$ is applied,the solenoid acts as a pure resistor because the frequency of $d.c.$ is zero,so inductive reactance $X_L = 2 \pi f

Vedclass · Accepted Answer

$200 \ \Omega, 0.55 \ H$

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(A) When $80 \ V$ $d.c.$ is applied,the solenoid acts as a pure resistor because the frequency of $d.c.$ is zero,so inductive reactance $X_L = 2 \pi fL = 0$.Resistance $R = \frac{V}{I_{dc}} = \frac{80 \ V}{0.8 \ A} = 100 \ \Omega$.When $80 \ V$ $a.c.$ is applied,the impedance $Z$ is given by $Z = \frac{V}{I_{ac}} = \frac{80 \ V}{0.4 \ A} = 200 \ \Omega$.The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$.Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.$40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.$X_L = \sqrt{30000} \approx 173.2 \ \Omega$.Since $X_L = 2 \pi fL$,we have $L = \frac{X_L}{2 \pi f} = \frac{173.2}{2 	imes 3.14 	imes 50} = \frac{173.2}{314} \approx 0.55 \ H$.Thus,the impedance is $200 \ \Omega$ and the inductance is $0.55 \ H$.

When $80 \ V$ $d.c.$ is applied across a solenoid,a current of $0.8 \ A$ flows in it. When $80 \ V$ $a.c.$ is applied across the same solenoid,the current becomes $0.4 \ A$. If the frequency of $a.c.$ source is $50 \ Hz$,the impedance and inductance of the solenoid are nearly:

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