When a resistor of `11 Omega` is connected in series with an electric cell, the current following in it is `0.5 A`. Instead, when a resistor of `5 Omega` is connected to the same electric cell in series, the current increases by `0.4 A` The internal resistance of the cell is

Options

A. `1.5 Omega`
B. `2 Omega`
C. `2.5 Omega`
D. `3.5 Omega`

Answer

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

### Step 1: Set up the equations for the two scenarios

1. **First scenario**: When a resistor of 11 Ω is connected, the current is 0.5 A.
- Let \( V \) be the voltage of the cell and \( r \) be the internal resistance of the cell.
- The total resistance in the circuit is \( R + r = 11 + r \).
- Using Ohm's law, we can write:
\[ I_1 = \frac{V}{R + r} \implies 0.5 = \frac{V}{11 + r} \quad \text{(Equation 1)} \]

2. **Second scenario**: When a resistor of 5 Ω is connected, the current increases by 0.4 A, making the new current 0.9 A.
- The total resistance in this case is \( R + r = 5 + r \).
- Again using Ohm's law:
\[ I_2 = \frac{V}{R + r} \implies 0.9 = \frac{V}{5 + r} \quad \text{(Equation 2)} \]

### Step 2: Solve the equations for \( V \)

From Equation 1:
\[ V = 0.5(11 + r) = 5.5 + 0.5r \quad \text{(1)} \]

From Equation 2:
\[ V = 0.9(5 + r) = 4.5 + 0.9r \quad \text{(2)} \]

### Step 3: Set the two expressions for \( V \) equal to each other

Since both expressions represent \( V \), we can set them equal:
\[ 5.5 + 0.5r = 4.5 + 0.9r \]

### Step 4: Rearrange the equation to solve for \( r \)

Rearranging gives:
\[ 5.5 - 4.5 = 0.9r - 0.5r \]
\[ 1 = 0.4r \]

### Step 5: Solve for \( r \)

Dividing both sides by 0.4:
\[ r = \frac{1}{0.4} = 2.5 \, \Omega \]

### Conclusion

The internal resistance of the cell is \( 2.5 \, \Omega \).

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