When the resistance of `9 Omega` is connected at the ends of a battery, its potential difference decreases from 40 volt to 30 volt . The internal resistance of the battery is

Options

A. `6 Omega`
B. `3 Omega`
C. `9 Omega`
D. `15 Omega`

Answer

To find the internal resistance of the battery, we can use the following steps:

### Step 1: Identify the given values
- Initial potential difference (E) = 40 V (open circuit voltage)
- Final potential difference (V) = 30 V (voltage across the load)
- Load resistance (R) = 9 Ω

### Step 2: Use Ohm's Law to find the current
Ohm's law states that \( V = I \cdot R \). We can rearrange this to find the current (I) flowing through the circuit when the load is connected:
\[ I = \frac{V}{R} = \frac{30 \, \text{V}}{9 \, \Omega} = \frac{30}{9} = \frac{10}{3} \, \text{A} \]

### Step 3: Apply the formula for internal resistance
The formula relating the electromotive force (E), the terminal voltage (V), the current (I), and the internal resistance (r) of the battery is given by:
\[ E = V + I \cdot r \]
Substituting the known values into the equation:
\[ 40 \, \text{V} = 30 \, \text{V} + \left(\frac{10}{3} \, \text{A}\right) \cdot r \]

### Step 4: Solve for internal resistance (r)
Rearranging the equation to isolate r:
\[ 40 \, \text{V} - 30 \, \text{V} = \left(\frac{10}{3} \, \text{A}\right) \cdot r \]
\[ 10 \, \text{V} = \left(\frac{10}{3} \, \text{A}\right) \cdot r \]
Now, divide both sides by \(\frac{10}{3} \, \text{A}\):
\[ r = \frac{10 \, \text{V}}{\frac{10}{3} \, \text{A}} = 10 \, \text{V} \cdot \frac{3}{10 \, \text{A}} = 3 \, \Omega \]

### Final Answer
The internal resistance of the battery is \( r = 3 \, \Omega \).

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