A student connects a long air-cored coil of m | vedclass

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Question

(A) When connected to a $D.C.$ source,the coil acts as a pure resistor $R$ because the inductive reactance $X_L = 2\pi fL$ is zero for $D.C.$ $(f=0)$.

Vedclass · Accepted Answer

$3$

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(A) When connected to a $D.C.$ source,the coil acts as a pure resistor $R$ because the inductive reactance $X_L = 2\pi fL$ is zero for $D.C.$ $(f=0)$.Using Ohm's law: $V = IR$$R = \frac{V}{I_{DC}} = \frac{100}{25} = 4\,\Omega$When connected to an $A.C.$ source,the coil acts as an $LR$ circuit with impedance $Z$.$Z = \frac{V}{I_{AC}} = \frac{100}{20} = 5\,\Omega$The impedance of an $LR$ circuit is given by $Z = \sqrt{X_L^2 + R^2}$.Substituting the values: $5 = \sqrt{X_L^2 + 4^2}$Squaring both sides: $25 = X_L^2 + 16$$X_L^2 = 25 - 16 = 9$$X_L = 3\,\Omega$Thus,the reactance of the coil is $3\,\Omega$.

$X=x$	$P(X=x)$
$0$	$0.4$
$1$	$0.3$
$2$	$0.1$
$3$	$0.1$
$4$	$0.1$

$A$ student connects a long air-cored coil of manganin wire to a $100\,V$ $D.C.$ supply and records a current of $25\,A$. When the same coil is connected across a $100\,V$,$50\,Hz$ $A.C.$ supply,the current reduces to $20\,A$. The reactance of the coil is....$\Omega $

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