For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` is

Options

A. `10/9Omega`
B. `9/10 Omega`
C. `11/9 Omega`
D. `5/9 Omega`

Explanation

To find the internal resistance of the cell, we can follow these steps:

### Step 1: Understand the given values
- The terminal potential difference when the circuit is open (no load) is \( E = 2.2 \, V \).
- The terminal potential difference when connected to a resistance \( R = 5 \, \Omega \) is \( V = 1.8 \, V \).

### Step 2: Calculate the current flowing through the circuit
When the resistance \( R \) is connected, we can calculate the current \( I \) using Ohm's law:
\[ I = \frac{V}{R} = \frac{1.8 \, V}{5 \, \Omega} = 0.36 \, A \]

### Step 3: Apply Kirchhoff's loop law
According to Kirchhoff's loop law, the sum of the potential differences in a closed loop is equal to zero. For our circuit, we can express this as:
\[ E - I \cdot R - I \cdot r = 0 \]
Where:
- \( E \) is the EMF of the cell (2.2 V),
- \( I \) is the current (0.36 A),
- \( R \) is the external resistance (5 Ω),
- \( r \) is the internal resistance of the cell.

Rearranging the equation gives us:
\[ E = I \cdot R + I \cdot r \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ 2.2 = (0.36 \cdot 5) + (0.36 \cdot r) \]

### Step 5: Simplify the equation
Calculating \( 0.36 \cdot 5 \):
\[ 0.36 \cdot 5 = 1.8 \]
So the equation becomes:
\[ 2.2 = 1.8 + 0.36 \cdot r \]

### Step 6: Solve for the internal resistance \( r \)
Subtract \( 1.8 \) from both sides:
\[ 2.2 - 1.8 = 0.36 \cdot r \]
\[ 0.4 = 0.36 \cdot r \]

Now, divide both sides by \( 0.36 \):
\[ r = \frac{0.4}{0.36} \approx 1.11 \, \Omega \]

### Step 7: Conclusion
The internal resistance of the cell is approximately \( 1.11 \, \Omega \).

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