An inductor coil stores U energy when i` current is passed through it and dissipates energy at the rate of P. The time constant of the circuit, when this coil is connected across a battery of zero internal resistance is

Options

A. `(4U)/(P)`
B. `(U)/(P)`
C. `(2U)/(P)`
D. `(2P)/(U)`

Answer

To solve the problem, we need to find the time constant of a circuit that includes an inductor coil, which stores energy \( U \) when a current \( I \) is passed through it and dissipates energy at the rate \( P \).

### Step-by-Step Solution:

1. **Understanding Energy Stored in the Inductor:**
The energy \( U \) stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
where \( L \) is the inductance and \( I \) is the current flowing through the inductor.

2. **Power Dissipated in the Circuit:**
The power \( P \) dissipated in the circuit can be expressed as:
\[ P = I^2 R \]
where \( R \) is the resistance of the circuit.

3. **Rearranging the Energy Formula:**
From the energy formula, we can solve for \( L \):
\[ L = \frac{2U}{I^2} \]

4. **Rearranging the Power Formula:**
From the power formula, we can solve for \( R \):
\[ R = \frac{P}{I^2} \]

5. **Finding the Time Constant:**
The time constant \( \tau_L \) for an inductor is given by:
\[ \tau_L = \frac{L}{R} \]
Substituting the expressions for \( L \) and \( R \) into this formula:
\[ \tau_L = \frac{\frac{2U}{I^2}}{\frac{P}{I^2}} = \frac{2U}{P} \]

6. **Final Result:**
Thus, the time constant of the circuit is:
\[ \tau_L = \frac{2U}{P} \]

### Conclusion:
The time constant of the circuit when the inductor coil is connected across a battery of zero internal resistance is \( \frac{2U}{P} \).