The terminal voltage is `(E)/(2)` when a c urrent of 2A is flowing through `2Omega` resistance, then the internal resistance of cell is:-

Options

A. `1Omega`
B. `2Omega`
C. `3Omega`
D. `4Omega`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a circuit with a battery (with EMF \( E \)), a current of \( 2 \, \text{A} \) flowing through a \( 2 \, \Omega \) resistor, and the terminal voltage of the battery is given as \( \frac{E}{2} \). We need to find the internal resistance \( r \) of the battery.

### Step 2: Write Down the Known Values
- Current (\( I \)) = \( 2 \, \text{A} \)
- Resistance (\( R \)) = \( 2 \, \Omega \)
- Terminal Voltage (\( V \)) = \( \frac{E}{2} \)

### Step 3: Apply the Formula for Terminal Voltage
The terminal voltage \( V \) can be expressed using the formula:
\[ V = E - I \cdot r \]
Where:
- \( E \) = EMF of the battery
- \( r \) = internal resistance of the battery

### Step 4: Substitute Known Values into the Formula
Since we know that \( V = \frac{E}{2} \), we can substitute this into the equation:
\[ \frac{E}{2} = E - I \cdot r \]
Substituting \( I = 2 \, \text{A} \):
\[ \frac{E}{2} = E - 2r \]

### Step 5: Rearrange the Equation
Now, let's rearrange the equation to isolate \( r \):
\[ 2r = E - \frac{E}{2} \]
This simplifies to:
\[ 2r = \frac{E}{2} \]

### Step 6: Solve for Internal Resistance \( r \)
Now, divide both sides by 2:
\[ r = \frac{E}{4} \]

### Step 7: Find the Value of \( E \)
To find \( E \), we can use the voltage drop across the \( 2 \, \Omega \) resistor:
\[ V = I \cdot R = 2 \, \text{A} \cdot 2 \, \Omega = 4 \, \text{V} \]
Since \( V = \frac{E}{2} \):
\[ \frac{E}{2} = 4 \, \text{V} \]
Thus:
\[ E = 8 \, \text{V} \]

### Step 8: Substitute \( E \) Back to Find \( r \)
Now substitute \( E = 8 \, \text{V} \) back into the equation for \( r \):
\[ r = \frac{8 \, \text{V}}{4} = 2 \, \Omega \]

### Conclusion
The internal resistance of the cell is \( r = 2 \, \Omega \).

---