Answer 1

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:-

\[\tau = \frac{L}{R} = \frac{1}{2} = 0 . 5  s\]