An L-R circuit consists of an inductance of 8mH and a resistance of `4 Omega` . The time constant of the circuit is :-

Options

A. 2ms
B. 12ms
C. 32ms
D. 500 s

Answer

To find the time constant of an L-R circuit, we can use the formula:

\[ \tau = \frac{L}{R} \]

where:
- \(\tau\) is the time constant,
- \(L\) is the inductance in henries (H),
- \(R\) is the resistance in ohms (Ω).

### Step-by-Step Solution:

1. **Identify the given values**:
- Inductance, \(L = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H}\)
- Resistance, \(R = 4 \, \Omega\)

2. **Substitute the values into the formula**:
\[ \tau = \frac{L}{R} = \frac{8 \times 10^{-3} \, \text{H}}{4 \, \Omega} \]

3. **Calculate the time constant**:
\[ \tau = \frac{8 \times 10^{-3}}{4} = 2 \times 10^{-3} \, \text{s} \]

4. **Convert to milliseconds**:
\[ \tau = 2 \, \text{ms} \]

### Final Answer:
The time constant of the circuit is \(2 \, \text{ms}\).

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