(D) An inductor in a $d.c.$ circuit behaves as an open circuit at $t = 0$ (infinite resistance) and as a short circuit at steady state ($t \to \infty$

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If both the Assertion and Reason are incorrect.

(D) An inductor in a $d.c.$ circuit behaves as an open circuit at $t = 0$ (infinite resistance) and as a short circuit at steady state ($t \to \infty$,zero resistance).Since the resistance changes from infinity to zero,the assertion that it is 'always constant' is incorrect.In a steady state,an ideal inductor has zero resistance,so the statement that it is 'non-zero' is also incorrect.Therefore,both the Assertion and the Reason are incorrect.

Class 12 Physics Electromagnetic Induction R-L D.C. Circuit

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Assertion : The resistance offered by an inductor in a $d.c.$ circuit is always constant.
Reason : The resistance of an inductor in a steady state is non-zero.

AIIMS 2010 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Two resistances $20\,\Omega$ and $50\,\Omega$ and a pure inductance of $50\,H$ are connected to a $10\,V$ battery through a key as shown in the figure. The key is closed at $t = 0$. Find the final value of current in the $50\,\Omega$ resistor.

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An inductor of $2 \, H$ and a resistance of $10 \, \Omega$ are connected in series with a battery of $5 \, V$. The initial rate of change of current is ...... $A/s$.

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In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

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Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

Medium

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Assertion : The resistance offered by an inductor in a $d.c.$ circuit is always constant.Reason : The resistance of an inductor in a steady state is non-zero.

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Two resistances $20\,\Omega$ and $50\,\Omega$ and a pure inductance of $50\,H$ are connected to a $10\,V$ battery through a key as shown in the figure. The key is closed at $t = 0$. Find the final value of current in the $50\,\Omega$ resistor.

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

An inductor of $2 \, H$ and a resistance of $10 \, \Omega$ are connected in series with a battery of $5 \, V$. The initial rate of change of current is ...... $A/s$.

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:$(a)$ Just after the closing of the key.$(b)$ $A$ long time after the closing of the key.

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Assertion : The resistance offered by an inductor in a $d.c.$ circuit is always constant.
Reason : The resistance of an inductor in a steady state is non-zero.

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.