In a series L-R growth circuit, if maximum current and maximum voltage across inductor of inductane 3mH are 2A and 6V respectively, the the time constant of the circuit is

Options

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

Answer

To solve the problem, we need to determine the time constant of a series L-R circuit given the inductance and the maximum current and voltage across the inductor.

**Step 1: Identify the given values.**
- Inductance (L) = 3 mH = 3 × 10^-3 H
- Maximum current (I₀) = 2 A
- Maximum voltage (V) = 6 V

**Step 2: Use Ohm's Law to find the resistance (R).**
Ohm's Law states that:
\[ V = I \cdot R \]
From this, we can rearrange to find R:
\[ R = \frac{V}{I} \]
Substituting the given values:
\[ R = \frac{6 \, \text{V}}{2 \, \text{A}} = 3 \, \Omega \]

**Step 3: Calculate the time constant (τ) of the circuit.**
The time constant (τ) for an L-R circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
Substituting the values we found:
\[ \tau = \frac{3 \times 10^{-3} \, \text{H}}{3 \, \Omega} = 1 \times 10^{-3} \, \text{s} = 1 \, \text{ms} \]

**Final Answer:**
The time constant of the circuit is **1 millisecond (ms)**.

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