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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.
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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