A coil of `40 H` inductance is connected in series with a resistance of `8` ohm and the combination is joined to the terminals of a `2 V` battery. The time constant of the circuit

Options

A. 1/5 seconds
B. 40 seconds
C. 20 seconds
D. 5 seconds

Explanation

To find the time constant of the given LR circuit, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 40 H (Henry)
- Resistance (R) = 8 Ω (Ohm)

### Step 2: Understand the formula for the time constant
The time constant (τ) for an LR circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
where:
- \( \tau \) is the time constant,
- \( L \) is the inductance,
- \( R \) is the resistance.

### Step 3: Substitute the values into the formula
Now, we can substitute the given values into the formula:
\[ \tau = \frac{40 \, \text{H}}{8 \, \Omega} \]

### Step 4: Perform the calculation
Now, we perform the division:
\[ \tau = \frac{40}{8} = 5 \, \text{seconds} \]

### Step 5: Conclusion
The time constant of the circuit is \( 5 \, \text{seconds} \).

### Final Answer:
The time constant of the circuit is **5 seconds**.

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