An inductor of inductance L and another resistor of resistance R are connected in series with a battery of emf E and a switch. Switch is closed at t = 0. How much charge will pass through the battery in one time constant? Internal resistance of the battery is negligible.

Options

A. `(EL)/(eR^(2))`
B. `(EL)/(eR)`
C. `(2EL)/(eR^(2))`
D. `(EL)/(R^(2))`

Explanation

To solve the problem of how much charge will pass through the battery in one time constant in an LR circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit**:
We have an inductor (L) and a resistor (R) connected in series with a battery of emf (E) and a switch. When the switch is closed at time \( t = 0 \), current starts to flow through the circuit.

2. **Current in the Circuit**:
The current \( I(t) \) in an LR circuit can be expressed as:
\[ I(t) = \frac{E}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.

3. **Charge Flow**:
The charge \( Q \) that passes through the circuit in a time interval can be found by integrating the current over that time interval. The relationship between charge and current is given by:
\[ Q = \int_0^t I(t') dt' \]

4. **Substituting Current**:
Substitute the expression for \( I(t) \) into the integral:
\[ Q = \int_0^{\tau} \frac{E}{R} \left(1 - e^{-\frac{t'}{\tau}}\right) dt' \]

5. **Integrate**:
We can split the integral into two parts:
\[ Q = \frac{E}{R} \int_0^{\tau} dt' - \frac{E}{R} \int_0^{\tau} e^{-\frac{t'}{\tau}} dt' \]

The first integral evaluates to:
\[ \int_0^{\tau} dt' = \tau \]

The second integral can be evaluated using the formula for the integral of an exponential:
\[ \int e^{-ax} dx = -\frac{1}{a} e^{-ax} \]
Thus,
\[ \int_0^{\tau} e^{-\frac{t'}{\tau}} dt' = -\tau e^{-\frac{t'}{\tau}} \bigg|_0^{\tau} = -\tau \left(e^{-1} - 1\right) = \tau(1 - e^{-1}) \]

6. **Final Charge Calculation**:
Putting it all together:
\[ Q = \frac{E}{R} \left(\tau - \tau(1 - e^{-1})\right) \]
Simplifying gives:
\[ Q = \frac{E}{R} \tau e^{-1} \]
Since \( \tau = \frac{L}{R} \):
\[ Q = \frac{E L}{R^2} e^{-1} \]

### Conclusion:
The charge that passes through the battery in one time constant is:
\[ Q = \frac{E L}{R^2} (1 - e^{-1}) \]