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  4. A coil of 40 henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is

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Q. A coil of $40$ henry inductance is connected in series with a resistance of $8$ ohm and the combination is joined to the terminals of a $2$ volt battery. The time constant of the circuit is

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Solution:

Time constant is $L / R$
Given, $L=40 H \,\& \,R =8 \,\Omega$
$\therefore T =40 / 8=5 \sec$