A series combination of `L` and `R` is connected to a battery of emf `E` having negligible internal resistance. The final value of current depends upon

Options

A. Land R only
B. E And L only
C. E and R only
D. L, R and E only

Answer

To solve the problem, we need to determine the final value of the current in a series combination of an inductor (L) and a resistor (R) connected to a battery with an electromotive force (emf) E.

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- We have a series circuit consisting of an inductor (L) and a resistor (R) connected to a battery of emf E. The internal resistance of the battery is negligible.

2. **Apply Kirchhoff's Voltage Law**:
- According to Kirchhoff's law, the sum of the potential differences in a closed loop must equal zero. For our circuit, we can write:
\[ E - iR - L \frac{di}{dt} = 0 \]
where \( i \) is the current flowing through the circuit.

3. **Analyze the Current Over Time**:
- The current \( i \) in the circuit will not be constant initially; it will increase from zero to a maximum value as the inductor opposes the change in current. The equation governing the growth of current is given by:
\[ i(t) = \frac{E}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
- Here, \( e^{-\frac{R}{L}t} \) represents the exponential decay factor that approaches zero as time \( t \) approaches infinity.

4. **Determine the Final Current**:
- As time approaches infinity (\( t \to \infty \)), the exponential term \( e^{-\frac{R}{L}t} \) approaches zero. Therefore, the final steady-state current \( i \) becomes:
\[ i = \frac{E}{R} \]

5. **Conclusion**:
- The final value of the current depends only on the emf \( E \) and the resistance \( R \). The inductance \( L \) does not affect the final steady-state current; it only affects how quickly the current reaches that final value.

### Final Answer:
The final value of current depends upon \( E \) and \( R \) only.

### Options:
- E and L only
- E and R only (Correct)
- L and R only
- L, R, and E only