An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds

Options

A. `0.4 `
B. `0.2`
C. `0.125 `
D. `0.8`

Explanation

To solve the problem step by step, we will follow the logical sequence of calculations based on the information provided.

### Step 1: Calculate the Inductance (L)
The energy stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
Where:
- \( U \) is the energy stored (64 J)
- \( L \) is the inductance in Henrys
- \( I \) is the current (8 A)

Substituting the known values:
\[ 64 = \frac{1}{2} L (8^2) \]
\[ 64 = \frac{1}{2} L (64) \]
\[ 64 = 32L \]
Now, solving for \( L \):
\[ L = \frac{64}{32} = 2 \text{ H} \]

### Step 2: Calculate the Resistance (R)
The power dissipated in the circuit can be expressed as:
\[ P = I^2 R \]
Where:
- \( P \) is the power (640 W)
- \( I \) is the current (8 A)
- \( R \) is the resistance in Ohms

Substituting the known values:
\[ 640 = (8^2) R \]
\[ 640 = 64R \]
Now, solving for \( R \):
\[ R = \frac{640}{64} = 10 \text{ Ohms} \]

### Step 3: Calculate the Time Constant (τ)
The time constant \( \tau \) for an RL circuit is given by:
\[ \tau = \frac{L}{R} \]
Substituting the values we found:
\[ \tau = \frac{2 \text{ H}}{10 \text{ Ohms}} = 0.2 \text{ seconds} \]

### Final Answer
The time constant of the circuit is:
\[ \boxed{0.2 \text{ seconds}} \]