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Question
An inductor-coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit.
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Solution
We know that time constant is the ratio of the self-inductance (L) of the coil to the resistance (R) of the circuit.
Given:-
Current in the circuit, i = 2 A
Emf of the battery, E = 4 V
Self-inductance of the coil, L = 1 H
Now,
Resistance of the coil:-
\[R = \frac{E}{i} = \frac{4}{2} = 2 \Omega\]
Time constant:-
\[\tau = \frac{L}{R} = \frac{1}{2} = 0 . 5 s\]
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