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

Explanation

To solve the problem, we need to find the internal resistance of the cell given the terminal voltage, current, and external resistance. Let's break it down step by step.

### Step 1: Understand the given information
We are given:
- Terminal voltage \( V = \frac{E}{2} \)
- Current \( I = 2 \, \text{A} \)
- External resistance \( R = 2 \, \Omega \)

### Step 2: Use Ohm's Law to express terminal voltage
According to Ohm's Law, the terminal voltage \( V \) can also be expressed as:
\[ V = I \times R \]
Substituting the values we have:
\[ V = 2 \, \text{A} \times 2 \, \Omega = 4 \, \text{V} \]

### Step 3: Set up the equation for terminal voltage
We know from the problem that:
\[ V = \frac{E}{2} \]
So we can equate the two expressions for \( V \):
\[ \frac{E}{2} = 4 \, \text{V} \]

### Step 4: Solve for \( E \)
To find \( E \), we multiply both sides of the equation by 2:
\[ E = 8 \, \text{V} \]

### Step 5: Relate terminal voltage to internal resistance
The terminal voltage can also be expressed in terms of the electromotive force (emf) \( E \) and the internal resistance \( r \):
\[ V = E - I \times r \]
Substituting the values we have:
\[ 4 = 8 - 2r \]

### Step 6: Solve for internal resistance \( r \)
Rearranging the equation to solve for \( r \):
\[ 2r = 8 - 4 \]
\[ 2r = 4 \]
\[ r = 2 \, \Omega \]

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

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