A coil resistance R and inductance L is connected to a battery of emf E volt. The final current in the coil is

Options

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

Explanation

To find the final current in a coil with resistance \( R \) and inductance \( L \) connected to a battery of emf \( E \), we can follow these steps:

### Step 1: Understand the Circuit
The circuit consists of a coil with resistance \( R \) and inductance \( L \) connected to a battery with emf \( E \). When the switch is closed, the current starts to flow through the coil.

### Step 2: Apply Kirchhoff's Loop Rule
According to Kirchhoff's loop rule, the sum of the potential differences in a closed loop is zero. For our circuit, we can write:
\[ E = L \frac{di}{dt} + IR \]
where \( \frac{di}{dt} \) is the rate of change of current with respect to time, and \( IR \) is the voltage drop across the resistor.

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ L \frac{di}{dt} = E - IR \]

### Step 4: Separate Variables
We can separate the variables to facilitate integration:
\[ \frac{di}{E - IR} = \frac{dt}{L} \]

### Step 5: Integrate Both Sides
Integrating both sides, we need to consider the limits. Initially, at \( t = 0 \), the current \( i = 0 \), and at \( t \to \infty \), the current will reach its final value \( I_f \):
\[ \int_{0}^{I_f} \frac{di}{E - IR} = \int_{0}^{t} \frac{dt}{L} \]

### Step 6: Solve the Left Side Integral
The left side integral can be solved using the natural logarithm:
\[ -\frac{1}{R} \ln |E - IR| \bigg|_{0}^{I_f} = \frac{t}{L} \]
This gives:
\[ -\frac{1}{R} \left( \ln |E - I_f R| - \ln |E| \right) = \frac{t}{L} \]

### Step 7: Solve for Final Current
As \( t \to \infty \), the current \( I_f \) approaches a steady state where \( \frac{di}{dt} = 0 \). Thus, we can set \( E - I_f R = 0 \):
\[ I_f = \frac{E}{R} \]

### Final Answer
The final current in the coil is:
\[ I_f = \frac{E}{R} \]