When a battery connected across a resistor of `16 ohm`, the voltage across the resistor is 12V. When the same battery is connected across a resistor of `10 ohm`, voltage across it is 11V. The internal resistance of the battery in ohm is

Options

A. `10/7`
B. `20/7`
C. `25/7`
D. `30/7`

Explanation

To find the internal resistance of the battery (r), we can use the information given about the two different resistors and the voltages across them. Let's break down the solution step by step.

### Step 1: Define the variables
Let:
- \( E \) = EMF (Electromotive Force) of the battery
- \( r \) = internal resistance of the battery
- \( R_1 = 16 \, \Omega \) (first resistor)
- \( R_2 = 10 \, \Omega \) (second resistor)
- \( V_1 = 12 \, V \) (voltage across the first resistor)
- \( V_2 = 11 \, V \) (voltage across the second resistor)

### Step 2: Calculate the current for the first resistor
Using Ohm's law, the current \( I_1 \) through the first resistor can be calculated as:
\[ I_1 = \frac{V_1}{R_1} = \frac{12 \, V}{16 \, \Omega} = 0.75 \, A \]

### Step 3: Write the KVL equation for the first resistor
According to Kirchhoff's Voltage Law (KVL), the total voltage in the loop is zero. For the first resistor:
\[ E - I_1 R_1 - I_1 r = 0 \]
Substituting \( I_1 \):
\[ E - 0.75 \times 16 - 0.75r = 0 \]
\[ E - 12 - 0.75r = 0 \quad \text{(Equation 1)} \]

### Step 4: Calculate the current for the second resistor
Using Ohm's law, the current \( I_2 \) through the second resistor can be calculated as:
\[ I_2 = \frac{V_2}{R_2} = \frac{11 \, V}{10 \, \Omega} = 1.1 \, A \]

### Step 5: Write the KVL equation for the second resistor
For the second resistor:
\[ E - I_2 R_2 - I_2 r = 0 \]
Substituting \( I_2 \):
\[ E - 1.1 \times 10 - 1.1r = 0 \]
\[ E - 11 - 1.1r = 0 \quad \text{(Equation 2)} \]

### Step 6: Solve the equations simultaneously
From Equation 1:
\[ E = 12 + 0.75r \quad \text{(1)} \]
From Equation 2:
\[ E = 11 + 1.1r \quad \text{(2)} \]

Setting the two expressions for \( E \) equal to each other:
\[ 12 + 0.75r = 11 + 1.1r \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 12 - 11 = 1.1r - 0.75r \]
\[ 1 = 0.35r \]

### Step 8: Solve for \( r \)
\[ r = \frac{1}{0.35} \approx 2.857 \, \Omega \]

### Step 9: Final answer
Thus, the internal resistance of the battery is approximately:
\[ r \approx 2.857 \, \Omega \]