An inductor coil stores `16` joules of energy and dissipates energy as heat at the rate of `320 W` when a current of 2 Amp is passed through it. When the coil is joined across a battery of emf `20V` and internal resistance `20Omega` the time constant for the circuit is `tau`. Find `100tau` ( in seconds).

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the Inductance (L)
We know that the energy (U) stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
Where:
- \( U = 16 \, \text{J} \) (energy stored)
- \( I = 2 \, \text{A} \) (current)

Substituting the values into the formula:
\[ 16 = \frac{1}{2} L (2^2) \]
\[ 16 = \frac{1}{2} L (4) \]
\[ 16 = 2L \]
Now, solving for \( L \):
\[ L = \frac{16}{2} = 8 \, \text{H} \]

### Step 2: Calculate the Total Resistance (R)
We know that the power (P) dissipated in the circuit can be expressed as:
\[ P = I^2 R \]
Where:
- \( P = 320 \, \text{W} \) (power dissipated)
- \( I = 2 \, \text{A} \) (current)

Substituting the values into the formula:
\[ 320 = (2^2) R \]
\[ 320 = 4R \]
Now, solving for \( R \):
\[ R = \frac{320}{4} = 80 \, \Omega \]

### Step 3: Calculate the Net Resistance
The net resistance in the circuit includes the internal resistance of the battery and the resistance calculated above:
- Internal resistance \( R_{internal} = 20 \, \Omega \)
- Resistance from power calculation \( R = 80 \, \Omega \)

Thus, the total resistance \( R_{net} \) is:
\[ R_{net} = R + R_{internal} = 80 + 20 = 100 \, \Omega \]

### Step 4: Calculate the Time Constant (τ)
The time constant \( \tau \) for an inductor is given by:
\[ \tau = \frac{L}{R_{net}} \]
Substituting the values we found:
\[ \tau = \frac{8}{100} = 0.08 \, \text{s} \]

### Step 5: Calculate \( 100\tau \)
Finally, we need to find \( 100\tau \):
\[ 100\tau = 100 \times 0.08 = 8 \, \text{s} \]

### Final Answer
Thus, the value of \( 100\tau \) is:
\[ \boxed{8} \]