A capacitor discharges through a resistance. The stored energy `U_0` in one capacitive time constant falls to

Options

A. `U_0//e^2`
B. `eU_0`
C. `U_0//e`
D. none of the above

Explanation

To solve the problem of how the stored energy in a capacitor discharging through a resistance falls to after one capacitive time constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Energy Stored in the Capacitor:**
The initial energy \( U_0 \) stored in a capacitor with capacitance \( C \) and charge \( Q_0 \) is given by the formula:
\[ U_0 = \frac{Q_0^2}{2C} \]

2. **Determine the Charge After One Time Constant:**
The charge \( Q \) on the capacitor at a time \( T \) is given by the equation:
\[ Q = Q_0 e^{-T/\tau} \]
where \( \tau = RC \) is the time constant of the circuit. For one time constant, \( T = \tau \):
\[ Q = Q_0 e^{-1} \]
Thus, at \( T = \tau \), the charge becomes:
\[ Q = \frac{Q_0}{e} \]

3. **Calculate the Energy Stored After One Time Constant:**
The energy \( U \) stored in the capacitor after one time constant can be calculated using the new charge \( Q \):
\[ U = \frac{Q^2}{2C} \]
Substituting \( Q = \frac{Q_0}{e} \):
\[ U = \frac{\left(\frac{Q_0}{e}\right)^2}{2C} = \frac{Q_0^2}{2e^2 C} \]

4. **Relate the New Energy to the Initial Energy:**
We can express the new energy \( U \) in terms of the initial energy \( U_0 \):
\[ U = \frac{1}{e^2} \cdot \frac{Q_0^2}{2C} = \frac{U_0}{e^2} \]

5. **Conclusion:**
Therefore, after one capacitive time constant, the stored energy in the capacitor falls to:
\[ U = \frac{U_0}{e^2} \]