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: Identify the given values
- Terminal potential difference when the circuit is open (E) = 2.2 V
- Terminal potential difference when the circuit is closed (V) = 1.8 V
- External resistance (R) = 5 Ω

### Step 2: Understand the relationship between EMF, terminal voltage, and internal resistance
When the circuit is closed, the terminal voltage (V) is given by the formula:
\[ V = E - I \cdot r \]
Where:
- \( E \) is the EMF of the cell,
- \( I \) is the current flowing through the circuit,
- \( r \) is the internal resistance of the cell.

### Step 3: Calculate the current (I) when the circuit is closed
Using Ohm's law, the current \( I \) can be calculated as:
\[ I = \frac{E}{R + r} \]
Substituting \( E = 2.2 \, V \) and \( R = 5 \, \Omega \):
\[ I = \frac{2.2}{5 + r} \]

### Step 4: Substitute the current into the terminal voltage equation
We know that when the circuit is closed, the terminal voltage is 1.8 V:
\[ 1.8 = 2.2 - I \cdot r \]
Substituting \( I \) from Step 3:
\[ 1.8 = 2.2 - \left(\frac{2.2}{5 + r}\right) \cdot r \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 1.8 = 2.2 - \frac{2.2r}{5 + r} \]
\[ 2.2 - 1.8 = \frac{2.2r}{5 + r} \]
\[ 0.4 = \frac{2.2r}{5 + r} \]

### Step 6: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 0.4(5 + r) = 2.2r \]
\[ 2 + 0.4r = 2.2r \]

### Step 7: Solve for r
Rearranging the equation:
\[ 2 = 2.2r - 0.4r \]
\[ 2 = 1.8r \]
\[ r = \frac{2}{1.8} = \frac{20}{18} = \frac{10}{9} \, \Omega \]

### Final Result
The internal resistance of the cell \( r \) is:
\[ r = \frac{10}{9} \, \Omega \]

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