An L-R circuit has a cell of e.m.f. E , which is switched on at time t = 0. The current in the circuit after a long time will be

Options

A. zero
B. `(E)/(R)`
C. `(E)/(L)`
D. `(E)/(sqrt(L^(2)+R^(2)))`

Explanation

To find the current in an L-R circuit after a long time when a cell of e.m.f. \( E \) is switched on at time \( t = 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have an L-R circuit consisting of an inductor \( L \) and a resistor \( R \) connected in series with a battery of e.m.f. \( E \).

2. **Behavior of the Inductor**:
- At \( t = 0 \), when the circuit is first closed, the inductor behaves like an open circuit because it opposes the change in current. Therefore, the initial current \( I(0) = 0 \).

3. **Long Time Behavior**:
- As time progresses, the inductor will allow the current to increase. After a long time (as \( t \to \infty \)), the inductor will behave like a short circuit, meaning it will have no resistance to the current flowing through it.

4. **Current Calculation**:
- After a long time, the only resistance in the circuit is that of the resistor \( R \).
- Using Ohm's Law, the current \( I \) in the circuit can be calculated as:
\[ I = \frac{E}{R} \]

5. **Final Result**:
- Therefore, the current in the circuit after a long time will be:
\[ I = \frac{E}{R} \]