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))`

Answer

To solve the problem of determining the current in an L-R circuit after a long time when connected to a cell of e.m.f. E, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- The circuit consists of a resistor (R) and an inductor (L) connected to a battery with an electromotive force (e.m.f.) of E.
- At time t = 0, the circuit is switched on.

2. **Behavior of Inductor Over Time**:
- Initially, when the circuit is switched on, the inductor opposes the change in current due to its property of inductance.
- However, as time progresses, the inductor will eventually reach a steady state where it no longer opposes the current flow.

3. **Long Time Behavior**:
- After a long time, the inductor behaves like a short circuit. This means that the voltage across the inductor becomes zero (V_L = 0).
- In this steady state, the only component affecting the current is the resistor (R).

4. **Applying Ohm's Law**:
- In the steady state, the current (I) in the circuit can be calculated using Ohm's law:
\[ I = \frac{V}{R} \]
- Here, V is the voltage across the resistor, which is equal to the e.m.f. of the battery (E) since the inductor behaves as a short circuit.

5. **Calculating the Current**:
- Substituting the values, we get:
\[ I = \frac{E}{R} \]
- This is the current flowing through the circuit after a long time.

### Final Answer:
The current in the circuit after a long time will be \( I = \frac{E}{R} \).

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