Question

In a series L-R growth circuit, if maximum current and maximum induced emf in an inductor of inductance 3mH are 2A and 6V respectively, then the time constant of the circuit is

• A. 1 ms.
• B. 1/3 ms.
• C. 1/6 ms
• D. 1/2 ms

Answer 1

The correct option is A 1 ms.

Maximum voltage induced across inductor is equal to $V=6Volt$.

$I_{max}=\frac{V}{R}$

$R=\frac{V}{I_{max}}=\frac{6}{2}=3\Omega$

Time constant $\tau =\frac{L}{R}=\frac{3m}{3}=1ms$